
imfosit 



1NGINEERING THERMODYNAMICS 



Published by the 

McGraw-Hill Book- Company 

NewYork 

Successors to theBookDepartments of the 

McGraw Publishing Company Hill Publishing Company 

. Publishers of Books for 
Electrical World The Engineering and Mining Journal 

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ENGINELix±L>cr 
THERMODYNAMICS 



BY 



CHARLES EDWARD LUCKE, PH.D., 

Professor of Mechanical Engineering in Columbia University, New York City 



McGRAW-HILL BOOK COMPANY 

239 WEST 39TH STREET, NEW YORK 

6 BOUVERIE STREET, LONDON, E. C. 

1912 



^> 



Copyright 1912, by the 
McGEAW-HILL BOOK COMPANY 



/JL-M/t 



THE SCIENTIFIC PRESS 

ROBERT DRUMMOND AND COMPANY 

BROOKLYN, N. Y. 



gd.A3209l8 



/ 



PREFACE 



Calculations about heat as a form of energy, and about work, another 
related form, both of them in connection witli changes in the condition of all 
sorts of substances that may give or take heat, and perform or receive work 
while changing condition, constitute the subject matter of this book. The 
treatment of the subject matter of this text is the result of personal experience 
in professional engineering practice and teaching students of engineering at 
Columbia Universit}'. 

Even a brief examination of the conditions surrounding changes in sub- 
stances as they gain or lose heat, do work or have work done on them, and of 
the corresponding relations between heat and work as forms of energy independ- 
ent of substances, will convince any one that the subject is one of great com- 
plexity. Accordingly the simplicity needed for practical use in the industries 
can be reached only by a consideration of a great mass of sub-topics and data. 

That the doing of work, and the changes in heat content of substances were 
related phenomena, and that these relations when formulated, would con- 
stitute a branch of science, was conceived about a half century ago, and the 
science was named Thermodynamics. The Engineer Rankine, who helped 
to create it, defined thermodynamics as " the reduction of the laws according 
to which such phenomena took place to a physical theory or corrected system 
of principles." Since Rankine's time thermodynamics has become a very 
highly developed science and has proved of great assistance in the formu- 
lation of modern physical chemistry, and to those branches of engineering 
that are concerned with heat. Unfortunately, as thermodynamics developed 
as a separate subject it did not render proportionate service to engineering, 
which itself developed even more rapidly in the same period under the guidance 
of men whose duty it was to create industrial apparatus and make it work 
properly, and who had little or no time to keep in touch with purely scientific 
advances or to interpret such advances for utilitarian ends. Thermodynamics 
proper is concerned with no numerical quantities nor with any particular 
substance nor for that matter with any actual substances whatever, but it is a 
physical theory of energy in relation to matter as a branch of natural philosophy. 
Engineering, however, is concerned with real substances, such as coal, steam, 
and gases and with numerical quantities, horse-powers of engines, temper- 
atures of steam, the heats of combustion of oils, so that alone, the principles 
of thermodynamic philosophy will not yield a solution of a practical problem, 



vi PREFACE 

be it one of design or one of analysis of test performance of actual heat machine 
or thermal apparatus. It is the province of engineering thermodynamics to 
guide numerical computation on thermal problems for real substances being 
treated in real apparatus. Its field, while including some of that of pure 
thermodynamics, extends far beyond the established provinces of that subject 
and extends to the interpretation of all pertinent principles and facts for purely 
useful purposes. Engineering thermodynamics, while using whatever prin- 
ciples of pure thermodynamics may help to solve its problems, must rely on 
a great mass of facts or relations that may not yet have risen to the dignity 
of thermodynamic laws. The workers in shops, factories, power plants or 
laboratories engaged in designing or operating to the best advantage machines 
and apparatus using heat with all sorts of substances, have developed great 
quantities of rules, methods and data that directly contribute to the ends sought. 
While for each class or type of apparatus there has grown up a separate set 
of data and methods in which much is common to several or all groups, not 
nearly so much assistance is rendered by one to another as should be by a proper 
use of engineering thermodynamics, which applies methods, principles and 
conclusion to all related problems. Classes of apparatus about which such 
groups of methods of analysis or synthesis, or collections of special data 
have developed and which it is province of engineering thermodynamics to 
unify so far as may be, include air compressors, and compressed air engines, 
reciprocating steam engines, steam turbines, steam boilers, coal-, oil- and gas- 
fired furnaces, gasifiers of coal and oil, gas producers, gas engines, complete 
steam or gas power plants, mechanical refrigeration and ice-making plants and 
chemical factory equipment, or more generally, machinery and apparatus for 
heating and cooling, evaporating and condensing, melting and freezing, moisten- 
ing and drying, gasification and combustiom. 

The nature of the subject and its division are better indicated by the 
classes of problems to be solved by its aid or the contributions expected of it 
than by the kinds of apparatus to which they apply. Probably its broadest 
contribution is the establishment of limits of possible performance of heat 
apparatus and machines. These limits will show what might be expected of 
a steam engine, gas engine or refrigerating machine when its mechanism is quite 
perfect and thus they become standards of reference with which actual per- 
formance can be compared, and a measure of the improvements yet possible. 
These same methods and practices are applicable to the analysis of the operat- 
ing performance of separate units and complete plants to discover the amount 
of energy being lost, how the total amount is divided between the different 
elements of the apparatus, which of the losses can be prevented and how, and 
finally which are absolutely unavoidable. This sort of analysis of the per- 
formance of thermal apparatus is the first step to be taken by the designer 
or manufacturer to improve the machine that he is creating for sale, and is 
essential to the purchaser and user of the machine, who cannot possibly keep 
it in the best operating condition without continually analyzing its performance 
and comparing results with thermodynamic possibilities, 



PREFACE vii 

The subject naturally divides into three general parts, the first dealing 
with the conditions surrounding the doing of work without any consideration 
of heat changes, the second heat gains and losses by substances without reference 
to work involved and the third, transformation of heat into work or work into 
heat in conjunction with changes in the condition of substances. The first 
part applies to the behavior of fluids in the cylinders of compressors and engines. 
The second part is concerned with the development of heat by combustion, 
its transmission from place to place, and the effect on the physical condition of 
solids, liquids, gases with their mixtures, solutions and reactions. The third 
part is fundamental to the efficient production of power by gases in internal 
combustion gas engines or compressed-air engines and by steam or other vapors 
in steam engines and turbines, and likewise as well to the production of 
mechanical refrigeration by ammonia, carbon dioxide and other vapors. 

Accordingly, the six chapters of the book treat these three parts in order. 
The first three chapters deal with work without any particular reference to 
heat, the second two with heat, without any particular reference to work, 
while the last is concerned with the relation between heat and work. After 
establishing in the first chapter the necessary units and basic principles for 
fixing quantities of work, the second chapter proceeds at once to the determina- 
tion of the work done in compressor cylinders and the third, the available work 
in engine cylinders, in terms of all the different variables that may determine 
the work for given dimensions of cylinder or for given quantities of fluid. There 
is established in these first three chapters a series of formulas directly applicable 
to a great variety of circumstances met with in ordinary practice. All are 
derived from a few simple principles and left in such form as to be readily 
available for numerical substitution. This permits of the solution of numerical 
problems on engine and compressor horse-power, fluid consumption or capacity 
with very little labor or time, although it has required the expansion of the 
subject over a considerable number of pages of book matter. A similar pro- 
cedure is followed in the succeeding chapters, formulas and data are developed 
and placed always with a view to the maximum clearness and utility. The 
essential unity of the entire subject has been preserved in that all the important 
related subjects are treated in the same consistent manner and at sufficient 
length to make them clear. When no general principles were available for 
a particular solution there has been no hesitation in reverting to specific data. 
The .subject could have been treated in a very much smaller space with less 
labor in . J)oo^_writing but necessitating far greater labor in numerical work 
on the part of the user. This same aim, that is, the saving of the user 7 s 
time and facilitating the arrival at numerical answers, is responsible for the 
insertion of a very considerable number of large tables, numerous original 
diagrams and charts, all calculated for the purpose and drawn to scale. 
These, however, take a great deal of room but are so extremely useful in 
everyday work as to justify any amount of space thus . taken up. For 
the sake of clearness all the steps in the derivation of any formula used 
are given, and numerical examples are added to illustrate their meaning 



viii PREFACE 

and application. This also requires a considerable amount of space but with- 
out it the limitations of the formulas would never be clear nor could a student 
learn the subject without material assistance. Similarly, space has been used 
in many parts of the book by writing formulas out in words instead of express- 
ing them in symbols. This saves a great deal of time and labor in hunting up 
the meaning of symbols by one who desires to use an unfamiliar formula involv- 
ing complex quantities, the meaning of which is often not clear when it is entirely 
symbolic. Thus, in the discussion of steam boiler capacity and efficiency, a 
dozen or more pages are taken up with formulas that could have been con- 
centrated in a single page were symbols used entirely, but only at the sacrifice 
of clearness and utility. Where in the derivation of a new formula or in the 
treatment of a new subject, reference to an old formula or statement is needed 
and important, repetition is resorted to, rather than mere reference, so that 
the new topic may be clear where presented, without constantly turning the 
pages of the book. It will be found, therefore, that while the size of the book 
is unusually large it will be less difficult to study than if it were short . 

As a text the book may be used for courses of practically any length, but 
it is not intended that in any course on the subject every page. of the book shall 
be used as assigned text. In the new graduate course in mechanical engineering 
at Columbia University, about three-fourths of the Subject matter of the book 
will be so used for a course of about one hundred and twenty periods of one 
hour each. All of the book matter not specifically assigned as text or reference 
in a course on engineering thermodynamics in any school may profitably be 
taken up in courses on other subjects, serving more or less as a basis for them. 
It is therefore adapted to courses on gas power, compressed air, steam turbines, 
steam power plants, steam engine design, mechanical refrigeration, heating 
and ventilation, chemical factory equipment, laboratory practice and research. 
Whenever a short course devoted to engineering thermodynamics alone is 
desired, the earlier sections of each chapter combined in some cases with the 
closing sections, may be assigned as text. In this manner a course of about 
thirty hours may be profitably pursued. This is a far better procedure than 
using a short text to fit a short course, as the student gets a better perspective, 
and may later return to omitted topics without difficulty. 

The preparation of the manuscript involves such a great amount of labor, 
that it would never have been undertaken without the assurance of assistance 
by Mr. E. D. Thurston, Jr., a fellow instructor at Columbia, in checking text 
and tables, calculating diagrams, writing problems and working examples. 
This help has been invaluable and is gratefully acknowledged. Recognition 
is also due for material aid rendered by Mr. T. M. Gunn in checking and in 
some cases deriving formulas, more especially those of the first three chapters. 
In spite, however, of all care to avoid errors it is too much to expect complete 
success in a new work of this character, but it is hoped that readers finding 
errors will point them out that future editions may be corrected. 

C. E. L. 

Columbia University, New York, September, 1912, 



CONTENTS 



Chapter I. Work and Power. General Principles 

PAGE 

1. Work defined 1 

2. Power defined 2 

3. Work in terms of pressure and volume 3 

4. Work of acceleration and resultant velocity 6 

5. Graphical representation of work 8 

6. Work by pressure volume change 10 

7. Work of expansion and compression 13 

8. Values of exponent s defining special cases of expansion or compression 20 

9. Work phases and cycles, positive, negative and net work 24 

10. Work determination by mean effective pressure 31 

11. Relation of pressure-volume diagrams to indicator cards 34 

12. To find the clearance 37 

13. Measurement of areas of PV diagrams and indicator cards 43 

14. Indicated horse-power 44 

15. Effective horse-power, brake horse-power, friction horse-power, mechanical efficiency, 

efficiency of transmission, thermal efficiency 47 

16. Specific displacement, quantity of fluid per hour, or per minute per I.H.P 49 

17. Velocity due to free expansion by PV method 52 

18. Weight of flow through nozzles by PV method 55 

19. Horse-power of nozzles and jets, by PV method 57 

Chapter II. Work of Compressors. Horse-power and Capacity of Air, Gas and 
Vapor Compressors, Blowing Engines and Dry Vacuum Pumps 

1. General description of structures and processes 73 

2. Standard reference diagrams or PV cycles for compressors and methods of analysis 

of compressor work and capacity 75. 

3. Single-stage compressor, no clearance, isothermal compression, Cycle I. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 81 

4. Single-stage compressor with clearance, isothermal compression, Cycle II. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 85 

5. Single-stage compressor, isothermal compression. Capacity, volumetric efficiency, 

work, mean effective pressure, horse-power and horse-power per cubic foot of 

substance, in terms of dimensions and cylinder clearance 87 

6. Single-stage compressor, no clearance, exponential compression, Cycle III. Work, 

capacity and work per cubic foot, in terms of pressures and volumes 91 

7. Single-stage compressor with clearance, exponential compression, Cycle IV. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 96 

8. Single-stage compressor, exponential compression. Relation between capacity, 

volumetric efficiency, work, mean effective pressure, horse-power and horse- 
power per cubic foot of substance and the dimensions of cylinder and clearance . . 98 

ix 



x CONTENTS 

PAGE 

9. Two-stage compressor, no clearance, perfect intercooling, exponential compression, 
best-receiver pressure, equality of stages. Cvcle V. Work and capacity in terms 
of pressures and volumes 103 

10. Two-stage compressor, with clearance, perfect intercooling exf c nential compression, 

best-receiver pressure, equality of stages, Cycle VI. Work and capacity in terms 

of pressures and volumes 109 

11. Two-stage compressor, any receiver pressure, exponential compression. Capacity, 

volumetric efficiency, work, mean effective pressure and horse-power, in terms of 
dimensions of cylinders and clearances 113 

12. Two-stage compressor, with best-receiver pressure, exponential compression. Capacity, 
volumetric efficiency, work, mean effective pressure and horse-power in terms 

of dimensions of cylinders and clearances 120 

13. Three-stage compressor, no clearance, perfect inter cooling exponential compres- 

bion, best two receiver pressures, equality of stages, Cycle VII. Work and 
capacity, in terms of pressures and volumes 125 

14. Three-stage compressor with clearance, perfect intercooling exponential compression, 

best-receiver pressures, equality of stages, Cycle VIII. Work and capacity in 
terms of pressures and volumes 131 

15. Three-stage compressor, any receiver pressure, exponential compression. Capacity, 

volumetric efficiency, work, mean effective pressure, and horse-power in terms 

of dimensions of cylinders and clearances 135 

16. Three-stage compressor with best-receiver pressures, exponential compression. Capac- 

ity, volumetric efficiency, work, mean effective pressure and horse-power in terms 

of dimensions of cylinders and clearances 143 

17. Comparative economy or efficiency of compressors 148 

18. Conditions of maximum work of compressors 151 

19. Compressor characteristics 153 

20. Work at partial capacity in compressors of variable capacity 160 

21. Graphic solution of compressor problems 168 

Chapter III. Work of Piston Engines. Horse-power and Consumption of 
Piston Engines Using Steam, Compressed Air, or any other Gas or Vapor 
under Pressure 

1. Action of fluid in single cylinders. General description of structure and processes . . 187 

2. Simple engines. Standard reference cycles or PV diagrams for the work of expansive 

fluids in a single cylinder 192 

3. Work of expansive fluid in single cylinder without clearance. Logarithmic expan- 

sion, Cycle I. Mean effective pressure, horse-power and consumption of simple 
engines 197 

4. Work of expansive fluid in single cylinder without clearance. Exponential expan- 

sion, Cycle II. Mean effective pressure, horse-power and consumption of simple 
engines 205 

5. Work of expansive fluid in single cylinder with clearance. Logarithmic expansion 

and compression; Cycle III. Mean effective pressure, horse-power, and con- 
sumption of simple engines 208 

6. Work of expansive fluid in single cylinder with clearance; exponential expansion 

and compression, Cycle IV. Mean effective pressure, horse-power and consumption 

of simple engines 219 

7. Action of fluid in multiple-expansion cylinders. General description of structure 

and processes 225 

8. Standard reference cycles or PV diagrams for the work of expansive fluids in two- 

cylinder compound engines 235 



CONTENTS xi 

PAGE 

9. Compound engine with infinite receiver. Logarithmic law. No clearance, Cycle 

V. General relations between pressures, dimensions, and work 256 

10. Compound engine with infinite receiver. Exponential law. No clearance, Cycle 

VI. General relations between pressures, dimensions and work 268 

11. Compound engine with finite receiver. Logarithmic law. No clearance, Cycle. 

VII. General relations between dimensions and work when H.P. exhaust and 
L.P. admission are not coincident 274 

12. Compound engine with finite receiver. Exponential law, no clearance, Cycle VIII. 

General relations between pressures, dimensions, and work, when high pressure 
Exhaust ivnd low-pressure admission are independent 287 

13. Compound engine without receiver. Logarithmic law. No clearance, Cycle IX. 

General relations between dimensions and work when high-pressure exhaust 
and low-pressure admission are coincident 292 

14. Compound engine without receiver. Exponential law. No clearance, Cycle X. 

General relations between dimensions and work when high-pressure exhaust 
and low-pressure admission are coincident 301 

15. Compound engine with infinite receiver. Logarithmic law, with clearance and. 

compression, Cycle XI. General relations between pressures, dimensions and 
work 306 

16. Compound engine with infinite receiver. Exponential law, with clearance and 

compression, Cycle XII. General relations between pressures, dimensions and 
work 319 

17. Compound engine with finite receiver. Logarithmic law, with clearance and com- 

pression, Cycle XIII. General relations between pressures, dimensions, and 
work when H.P. exhaust and L. P.. admission are independent 325 

18. Compound engine with finite receiver. Exponential law, with clearance and com- 

pression, Cycle XIV. General relations between pressures, dimensions, and 
work when H.P. exhaust and L.P. admission are independent 335 

19. Compound engine without receiver. Logarithmic law, with clearance and com- 

pression, Cycle XV. General relations between pressures, dimensions, and 
work when H.P. exhaust and L.P. admission are coincident 339 

20. Compound engine without receiver. Exponential law, with clearance and compres- 

sion, Cycle XVI. General relations between pressures, dimension, and work, 
.when H.P. exhaust and L.P. admission are coincident 346 

21. Triple-expansion engine with infinite receiver. Logarithmic law. No clearance, 

Cycle XVII. General relations between pressures, dimensions and work 349 

22. Multiple-expansion engine. General case. Any relation between cylinders and 

receiver. Determination of pressure-volume diagram and work, by graphic 
methods 357 

23. Mean effective pressure, engine horse-power, and work distribution and their vari- 

ation with valve movement and initial pressure. Diagram distortion and diagram 
factors. Mechanical efficiency 363 

24. Consumption of steam engines and its variation with valve movement and initial 

pressure. Best cut-off as affected by condensation and leakage 371 

25. Variation of steam consumption with engine load. The Willans line. Most eco- 

nomical load for more than one engine and best load division 381 

26. Graphical solution of problems on engine horse-power and cylinder sizes 387 



xii CONTENTS 



Chapter IV. Heat and Matter. Qualitative and Quantitative Relations 
between Heat Content of Substances and Physical-Chemical State 

PAGE 

1. Substances and heat effects important in engineering 398 

2. Classification of heating processes. Heat addition and abstraction with or without 

temperature change, qualitative relations 401 

3. Thermometry based on temperature change, heat effects. Thermometer and abso- 

lute temperature scales 407 

4. Calorimetry based on proportionality of heat effects to heat quantity. Units of 

heat and mechanical equivalent 415 

5. Temperature change relation to amount of heat for solids, liquids, gases and vapors 

not changing state. Specific heats 419 

6. Volume or density variation with temperature of solids, liquids, gases and vapors 

not changing state. Coefficients of expansion. Coefficients of pressure change 
for gases and vapors 435 

7. Pressure, volume, and temperature relations for gases. Perfect and real gases 438 

8. Gas density and specific volume and its relation to molecular weight and gas constant . 446 

9. Pressure and temperature relations for vapor of liquids or solids. Vaporization, 

sublimation and fusion curves. Boiling- and freezing-points for pure liquids 
and dilute solutions. Saturated and superheated vapors 451 

10. Change of state with amount of heat at constant temperature. Latent heats of 

fusion and vaporization. Total heats of vapors. Relation of specific volume 

of liquid and of vapor to the latent heat 467 

11. Gas and vapor mixtures. Partial and total gas and vapor pressures. Volume, 

weight and gas constant relations. Saturated mixtures. Humidity 481 

12. Absorption of gases by liquids and by solids. Relative volumes and weights with 

pressure and temperature. Heats of absorption and of dilution. Properties 

of aqua ammonia 493 

13. Combustion and related reactions. Relative weights and volumes of substances 

and elements before and after reaction 506 

14. Heats of reaction. Calorific power of combustible elements and of simple chemical 

compounds. B.T.U. per pound and per cubic foot 516 

15. Heat transmission processes. Factors of internal conduction, surface resistance, 

radiation and convection ' 528 

16. Heat transmission between separated fluids. Mean temperature differences, coeffi- 

cients of transmission 538 

17. Variation in coefficient of heat transmission due to kind of substance, character of 

separating wall and conditions of flow 555 



Chapter V. Heating by Combustion. Fuels, Furnaces, Gas Producers and 
Steam Boilers 

1. Origin of heat and transformation to useful form. Complexity of fuels as sources 

of heat. General classification, solid, liquid, gaseous, natural and artificial 644 

2. Natural solid fuels, wood peat, lignite, bituminous and anthracite coals. Chemical 

and physical properties. Classifications based on ultimate and proximate analysis 
and on behavior on heating 649 

3. Calorific power of coals and the combustible of coals. Calculation of calorific power 

from ultimate and proximate analyses. Calorific power of the volatile 662 

4. Mineral oil and natural gas fuel. Chemical and physical properties. Calorific 

power direct and as calculated for oils from ultimate analysis or from density, 
and for gas from sum of constituent gases 670 



CONTENTS xiii 

PAGE 

5. Charcoal, coke, coke oven and retort coal gas as products of heating wood and coal. 

Chemical, physical, and calorific properties per pound. Calorific power of gases 
per cubic foot in terms of constituent gases. Yield of gas and coke per pound 
coal 675 

6. Distillate oils, kerosene and gasoline, residue oils, and oil gas as products of heating 

mineral oils. Chemical, physical, and calorific properties. Calorific power of frac- 
tionated oils in terms of, (a) carbon and hydrogen; (b) density per pound, and 
estimated value per cubic foot vapor. Calorific power of oil gas per pound and per 
cubic foot in terms of constituent gases. Yield of distillates and oil gas 685 

7. Gasification of fixed carbon and coke by air-blast reactions, producing air gas, and 

blast-furnace gas. Comparative yield per pound coke and air. Sensible heat 
and heat of combustion of gas. Relation of constituents in gas. Efficiency 
of gasification 695 

8. Gasification of fixed carbon, coke, and coal previously heated, by steam-blast reac- 

tions, producing water gas. Composition and relation of constituents of water 
gas, yield per pound of steam and coal. Heat of combustion of gas and limitation 
of yield by negative heat of reaction .' 710 

9. Gasification of coals by steam and air blasts, resulting in producer gas. Composition 

and relation of constituents of producer gas, yield per pound of fixed carbon, air 
and steam. Modification of composition by addition of volatile of coal. Heat 
of combustion of gas, sensible heat, and efficiency of gasification. Horse-power 
of gas producers 719 

10. Combustion effects. Final temperature, volume and pressure for explosive and 

non-explosive combustion. Estimation of air weights and heat suppression 
due to CO in products from volumetric analysis 740 

11. Temperature of ignition and its variation with conditions. Limits of proportion 

air gas neutral, or detonating gas and neutral, for explosive combustion of mix- 
tures. Limits of adiabatic compression for self -ignition of mixtures 758 

12. Rate of combustion of solid fuels with draft. Propagation rates, uniform and 

detonating for explosive gaseous mixtures 765 

13. Steam-boiler evaporative capacity and horse-power. Horse-power units, equivalent 

rates of evaporation and of heat absorption. Factors of evaporation. Relation 
between absorption rates and rates of heat generation. Influence of heating and 
grate surface, calorific power of fuels, rates of combustion and furnace losses .... 773 

14. Steam-boiler efficiency, furnace and heating-surface efficiency. Heat balances and 

variation in heat distribution. Evaporation and losses per pound of fuel 796 

Chapter VI. Heat and Work. General Relations between Heat and Work. 
Thermal Efficiency of Steam, Gas, and Compressed-Air Engines. Flow 
of Expansive Fluids. Performance of Mechanical Refrigerating Systems 

1. General heat and work relations. Thermal cycles. Work and efficiency deter- 

mination by heat differences and ratios. Graphic method of temperature 
entropy heat diagram 874 

2. General energy equation between heat change, intrinsic energy change, and work 

done. Derived relations between physical constants for gases and for changes 

of state, solid to liquid, and liquid to vapor • 882 

3. Quantitative relations for primary thermal phases, algebraic, and graphic to PV, 

and T<!? coordinates. Constancy of PV, and T for gases and vapors, wet, dry 
and superheated 892 

4. Quantitative relations for secondary thermal phases. Adiabatics for gases and 

vapors. Constant quality, constant total heat, and logarithmic expansion fines 

for steam 904 



xiv CONTENTS 

PAGE 

5. Thermal cycles representative of heat-engine processes. Cyclic efficiency. A 

reference standard for engines and fuel-burning power systems. Classification 

of steam cycles 927 

6. The Rankine cycle. Work, mean effective pressure, jet velocity, water rate, heat 

consumption and efficiency of steam Cycle I. Adiabatic expansion, constant 
pressure, heat addition and abstraction, no compression 936 

7. Carnot cycle and derivatives. Work, mean effective pressure, water rate, heat 

consumption and efficiency of steam Cycle II. Adiabatic expansion and com- 
pression, constant pressure heat addition and abstraction . . , 957 

8. Gas cycles representative of ideal processes and standards of reference for gas 

engines 970 

9. Brown, Lenoir, Otto and Langen non-compression gas-cycles. Work, mean 

effective pressure, volume and pressure ranges, efficiency, heat and gas con- 
sumption 978 

10. Stirling and Ericsson gas cycles. Work, efficiency and derived quantities for 

isothermal compression with and without regenerators 993 

11. Otto, Atkinson, Bray ton, Diesel, and Carnot gas cycles. Work, efficiency and 

derived quantities for adiabatic compression gas cycles 1006 

12. Comparison of steam and gas cycles with the Rankine as standard for steam, and 

with the Otto and Diesel, as standard for gas. Relations of Otto and Diesel 

to Rankine cycle. Conditions for equal efficiency 1031 

13. Gas cycle performance as affected by variability of the specific heats of gases, 

applied to the Otto cycle 1035 

14. Actual performance of Otto and Diesel gas engines, and its relation to the cyclic 

Diagram factors for mean effective pressure and thermal efficiency. Effect 

of load on efficiency. Heat balance of gas engines alone and with gas producers 1042 

15. Actual performance of piston steam engines and steam turbines at their best load 

and its relation to the cyclic. Effect on efficiency of initial pressure, vacuum, 
superheat, jacketing, and reheating. Heat balances of steam power plants .... 1062 

16. Flow of hot water, steam and gases through orifices and nozzles. Velocity, weight 

per second, kinetic energy, and force of reaction of jets. Nozzle friction and 
reheating and coefficient of efflux. Relative proportions of series nozzles for 
turbines for proper division of work of expansion 1083 

17. Flow of expansive fluids under small pressure drops through orifices, valves, and 

Venturi tubes. Relation between loss of pressure and flow. Velocity heads 
and quantity of flow by Pitot tubes 1097 

18. Flow of gases and vapors in pipes, flues, ducts, and chimneys. Relation between 

quantity of flow and loss of pressure. Friction resistances. Draught and 
capacity of chimneys 1111 

19. Thermal efficiency of compressed-air engines alone and in combination with air 

compressors. Effect of preheating and reheating. Compressor suction heating, 

and volumetric efficiency. Wall action 1127 

20. Mechanical refrigeration, general description of processes and structures. Thermal 

cycles and refrigerating fluids. Limiting temperatures and pressures 1142 

21. Performance of mechanical refrigerating cycles and systems. Quantity of fluid 

circulated per minute per ton refrigeration, horse-power, and heat supplied 
per ton. Refrigeration per unit of work done and its relation to thermal effi- 
ciency of the system 1157 



LIST OF TABLES 



JO. PAGE 

1. Conversion table of units of distance 62 

2. Conversion table of units of surface 62 

3. Conversion table of units of volume , 62 

4. Conversion table of units of weights and force 63 

5. Conversion table of units of pressure 63 

6. Conversion table of units of work 64 

7. Conversion table of units of power 64 

8. Units of velocity 64 

9. Barometric heights, altitudes and pressures 65 

10. Values of s in the equation PV S - constant for various substances and conditions . . 67 

11. Horse-power per pound mean effective pressure 68 

12. Ratio of cut-offs in the two cylinders of the compound engine to give equal work 

for any receiver volume 284 

13. Piston positions for any crank angle 395 

14. Values for x for use in Heck's formula for missing water 396 

15. Some actual steam engine dimensions 396 

16. Fixed temperatures 411 

17. Fahrenheit temperatures by hydrogen and mercury thermometers 414 

18. Freezing-point of calcium chloride brine 425 

19. Specific heat of sodium chloride brine 427 

20. Specific heat and gas constants, 431 

21. The critical point 453 

22. Juhlin's data on the vapor pressure of ice 456 

23. Tamman's value on fusion pressure and temperature of water-ice 456 

24. Lowering of freezing-points 465 

25. Berthelot's data on heat for complete dilution of ammonia solutions 500 

26. Air required for combustion of various substances 515 

27. Radiation coefficients 535 

28. Coefficients of heat transfer 550 

29. Temperatures, Centigrade and Fahrenheit 571 

30. Heat and power conversion table 573 

31. Specific heat of solids 574 

32. Specific heats of liquids 576 

33. Baume-specific gravity scale t 577 

34. Specific heats of gases 578 

35. Coefficient of linear expansion of solids 580 

36. Coefficient of cubical expansion of solids 581 

37. Coefficient of volumetric expansion of gases and vapors at constant pressure 582 

xv 



xvi LIST OF TABLES 

NO, PAGE 

38. Coefficient of pressure rise of gases and vapors at constant volume 583 

39. Compressibility of gases by their isothermals 584 

40. Values of the gas constant R 584 

41. Density of gases 585 

42. International atomic weights 586 

43. Melting- or freezing-points 586 

44. Boiling-points ^ 588 

45. Latent heats of vaporization 590 

46. Latent heats of fusion 591 

47. Properties of saturated steam 592 

48. Properties of superheated steam 596 

49. Properties of saturated ammonia vapor 603 

50. Properties of saturated carbon dioxide vapor 618 

51. Relation between pressure, temperature and per cent NH 3 in solution 628 

52. Values of partial pressure of ammonia and water vapors for various temperatures 

and per cents of ammonia in solution 632 

53. Absorption of gases by liquids 634 

54. Absorption of air in water 635 

55. Heats of combustion of fuel elements and chemical compounds 636 

56. Internal thermal conductivity 639 

57. Relative thermal conductivity 642 

58. General classification of fuels , 648 

59. Comparison of cellulose and average wood compositions 650 

60. Classification of coals by composition . . 652 

61. Classification of coals by gas and coke qualities 654 

62. Composition of peats 655 

63. Composition of Austrian lignites 656 

64. Composition of English coking coals 658 

65. Wilkesbarre anthracite coal sizes and average ash content 659 

66. Density and calorific power of natural gas 673 

67. Products of wood distillation 676 

68. Products of peat distillation 678 

69. Products of bituminous coal distillation 680 

70. Gas yield of English cannel coals 682 

71. Comparison of coke oven and retort coal gas 682 

72. Relation between oxygen in coal and hydrocarbon in gas 684 

73. Density and calorific power of coke oven gas 684 

74. Average distillation products of crude mineral oils 686 

75. American mineral oil products .' 687 

76. U.S. gasolene and kerosene bearing crude oils 688 

77. Calorific power of gasolenes and kerosenes 691 

78. Properties of oil-gas 693 

79. Yield of retort oil gas 694 

80. Density and calorific power of oil gas 694 

81. Boudouard's equilibrium relations for CO and C0 2 with temperature 697 

82. Change of 2 in air to CO and C0 2 at 1472° F 699 

83. Composition of hypothetical air gas, general 704 

84. Composition of hypothetical air gas, no C0 2 and no CO 705 

85. Density and calorific power of blast furnace gas 708 

86. Water gas characteristics with bed temperature 710 

87. Composition of hypothetical water gas, general 714 

88. Composition of hypothetical water gas ? no CO? and no CO 715 



LIST OF TABLES xvii 

NO. PAGE 

89. Density and calorific power of water gas *..... 718 

90. Composition of hypothetical producer gas from fixed carbon 725 

91. Density and calorific power of producer gas 737 

92. Characteristics of explosive mixtures of oil gas and air 747 

93. Calculated ignition temperatures for producer gas 761 

94. Compressions commonly used in gas engines 762 

95. Ignition temperatures 763 

96. Variation of ignition temperature of charcoal with distillation temperature 763 

97. Per cent detonating mixture at explosive limits of proportion 764 

98. Velocity of detonating or explosive waves 766 

99. Rate of propagation, feet per second, for different proportions in gaseous mixtures . 768 

100. Rates of combustion for coal 769 

101. Constants of proportion for rate of coal combustion for use in Eq. (848) 771 

102. Boiler efficiency summaries 799 

103. Three examples of heat balance for boilers 800 

104. Composition and calorific power of characteristic coals 818 

105. Combustible and volatile of coals, lignites and peats 826 

106. Paraffines from Pennsylvania petroleums 835 

107. Calorific power of mineral oils by calorimeter and calculation by density formula 

of Sherman and Kropff 836 

108. Properties of mineral oils 838 

109. Composition of natural gases 841 

110. Composition of coke oven and retort coal gas 842 

111. Composition of U. S. coke 846 

112. Fractionation tests of kerosenes and petroleums 847 

113. Fractionation tests of gasolenes 851 

114. Composition of blast-furnace gas and air gas 853 

115. Rate of formation of CO from C0 2 and carbon 855 

116. Composition of water gas 857 

117. Composition of producer gas 858 

118. Gas producer tests 864 

119. Composition of oil producer gas 866 

120. Composition of powdered coal producer gas 866 

121. Calorific powers of best air-gas mixtures 867 

122. Composition of boiler-flue gases 868 

123. Limits of proportions of explosive air-gas mixtures 869 

124. Rate of combustion of coal with draft 870 

125. Rate of combustion of coal 871 

126. Values of s for adiabatic expansion of steam 912 

127. Values of s for adiabatic expansion of steam determined from initial and final volumes 

only 913 

128 1042 

129. Diagram factors for Otto cycle gas engines 1046 

130. Mechanical efficiencies of gas engines 1050 

131. Allowable compression for gas engines 1050 

132. Mean effective pressure factors for Otto cycle engines 1053 

133. Guldner's values of Otto engine real volumetric efficiency with estimated mean 

suction resistances 1055 

134. Comparative heat balances of gas producer and engine plants 1057 

135. Heat balances of gas producer plants 1060 

136. Heat balances of gas and oil engines 1060 

137. Steam plant heat balances 1063 



xviii LIST OF TABLES 

NO. PAGE 

138. Efficiency factors for reciprocating steam engines and turbines 1064 

139. Steam turbine efficiency and efficiency factors with varying vacuum and with 

steam approximately at constant initial pressure 1071 

140. Efficiency factors for low-pressure steam in piston engines 1074 

141. Coefficient of discharge for various air pressure and diameters of orifice (Durley). 1101 

142. Values of C for air flow (Weisbach) ] 101 

143. Flow change resistance factors Fr (Reitschel) 1121 



TABLE OF SYMBOLS 



A =area in square feet. 

= constant, in formula for most economical load of a steam engine, Chapter III. 

= constant, in pipe flow formula, Chapter VI. 

= excess air per pound of coal, Chapter V. 

= pounds of ammonia dissolved per pound of weak liquor, Chapter IV. 
a = area in square inches. 

= coefficient of linear expansion, Chapter IV. 

= constant in equation for the ratio of cylinder sizes for equal work distribution in com- 
pound engine, Chapter III. 

= constant in equation for change in intrinsic energy, Chapter VI. 

= constant in equation for specific heat, Chapter IV. 

= cubic feet of air per cubic foot of gas in explosive mixtures, Chapter V. 

= effective area of piston, square inches, Chapter I. 



B = constant in equation for the most economical load of the steam engine, Chapter III. 
= constant in equation for flow in pipes, Chapter VI. 
Be\=Baume. 
B.H.P. = brake horse-power, Chapters III and VI. 

= boiler horse-power, Chapter V. 
B.T.U. = British thermal unit. 

6 = constant in equation for change in intrinsic energy, Chapter VI. 
= constant in equation for specific heat, Chapter IV. 
(bk.pr.) =back pressure in pounds per square inch. 



C = Centigrade. 
= circumference or perimeter of ducts in equations for flow, Chapter VI. 
= constant. 

= heat suppression factor, Chapter V. 
= ratio of pressure after compression to that before compression in gas engine cycles, 

Chapter VI. 
= specific heat, Chapter IV. 
Cc = per cent of ammonia in weak liquor, Chapter VI. 
C p = specific heat at constant pressure. 
Cs = per cent of ammonia in rich liquor, Chapter VI. 
C s = specific heat of water, Chapter VI. 
C v = specific heat at constant volume. 
CI = clearance expressed in cubic feet, 
c = clearance expressed as a fraction of the displacement 
= constant, 
cu.ft. = cubic foot, 
cu.in. = cubic inch. 



D = constant in equations for pipe flow, Chapter VI. 

= density, Chapter IV. 

= diameter of pipe in feet, Chapter VI. 

= displacement in cubic feet. 
D s = specific displacement, Chapter I. 

xix 



xx TABLE OF SYMBOLS 

d = constant in equation for change in intrinsic energy, Chapter VI. 
= diameter of a cylinder in inches, Chapter I. 
= diameter of pipe in inches, Chapter VI. 
= differential, 
(del.pr.) = delivery pressure in pounds per square inch, Chapter II. 



E = constant in equation for pipe flow, Chapter VI. 
= external latent heat, Chapter IV. 
= thermal efficiency, Chapter VI. 
Eb = thermal efficiency referred to brake horse-power, Chapter III. 
Eb— boiler efficiency, Chapter V. k 
Ef = furnace efficiency, Chapter V. 

Ej = thermal efficiency referred to indicated horse-power, Chapter III. 
E m = mechanical efficiency, Chapter III. 
E s = heating surface efficiency, Chapter V. 
Ev = volumetric efficiency (apparent), Chapter VI. 
Ey r = volumetric efficiency (true), Chapter VI. 

e = as a subscript to log to designate base e. 
= constant in equation for change in intrinsic energy, Chapter VI. 

ei= ratio of true volumetric efficiency to hypothetical, Chapter II. 

ei = ratio of true volumetric efficiency to apparent, Chapter III. 

e 3 = ratio of true indicated horse-power to hypothetical, Chapter II. 



F = constant in equation for pipe flow, Chapter VI. 
= diagram factor for gas engine indicator cards, Chapter VI. 
= Fahrenheit. 
= force in pounds. 
Ff = friction factor, FfX velocity head = loss due to friction, Chapter VI. 
Fr = resistance factor, Fr X velocity head = loss due to resistances, Chapter VI. 
Fs= special resistances to flow in equations for chimney draft, Chapter VI. 
/= constant in equation for changes in intrinsic energy, Chapter VI. 
= function, 
ft. = foot, 
ft.-lb. = foot-pound. 



G = constant in equation for pipe flow, Chapter VI. 
= weight of gas per hour in equation for chimney flow, Chapter VI. 
G m = maximum weight of gases in equation for chimney flow, Chapter VI. 
G. S. = grate surface. 

g = acceleration due to gravity, 32.2 (approx.) feet per second, per second. 



H = as a subscript to denote high pressure cylinder. 
= heat per pound of dry saturated vapor above 32° F. 
= heat per cubic foot gas. 
=heat transmitted, Chapter IV. 
= height of column of hot gases in feet, Chapter VI. 
= pressure or head in feet of fluid, Chapter VI. 
Ha = difference in pressure on two sides of an orifice in feet of air, Chapter VI. 
Hg = equivalent head of hot gases, Chapter VI. 
Hm = pressure in feet of mercury, Chapter VI. 
Hw = pressure in feet of water, Chapter VI. 
H.P. =high pressure. 

= horse-power, Chapter I. 



TABLE OF SYMBOLS xxi 

H.S. = heating surface. 
(H.P.cap.)=high pressure cylinder capacity, Chapter III. 
h = heat of superheat. 
Iim = difference in pressure on two sides of an orifice in inches of mercury, Chapter VI. 
hw — difference in pressure on two sides of an orifice in inches of water, Chapter VI. 



7 = as a subscript to denote intermediate cylinder, Chapter III. 
I.H.P. = indicated horse-power. 

in. =inch. 
(in.pr.) = initial pressure in pounds per square inch. 



J = Joule's equivalent = 778 (approx.) foot-pounds per B.T.U. 



K — coefficient of thermal conductivity, Chapter IV. 
= constant. 
= proportionality coefficient in equation for draft, Chapter VI. 

K e = engine constant =~rr^ in expression for horse-power. Chapter III. 
33000 



L=as a subscript to denote low-pressure cylinder. 
= distance in feet. 
= latent heat, Chapters IV and VI. 
=length of stroke in feet, Chapter I. 
L = per cent of heat in fuel lost in furnace, Chapter V. 
L.P. =low pressure. 
(L.P. Cap. )= low-pressure capacity, Chapter II. 
1 = constant, Chapter III. 
= length, Chapter IV. 
lb. = pound. 

log = logarithm to the base 10. 
log e = logarithm to the base e. 



M = mass. 
(M.E.P.) =mean effective pressure, pounds per square foot. 
m = constant, Chapter III. 

=mean hydraulic radius = . 

perimeter 

= molecular weight, Chapter IV. 

= ratio of initial pressure to that end of expansion in Otto and Langen gas cycle, 
Chapter VI. 
(m.b.p.) =mean back pressure in pounds per square inch. 
(m.e.p.) =mean effective pressure in pounds per square inch, 
(mi. p.) =mean forward pressure in pounds per square inch. 



N = constant, Chapter III. 

= revolutions per minute. 
n = cycles per minute. 

= constant, Chapter III. 

= cubic foot of neutral per cubic foot of gaseous mixture, Chapter V. 

= number of degrees exposed on thermometer stem, Chapter IV. 

= ratio of volume after expansion to volume before in Atkinson gas cycle, Chapter IV. 

= specific volume of dry saturated steam, Chapter VI, 



xxii TABLE OF SYMBOLS 

0= volume of receiver of compound engine in cubic feet, Chapter III. 
P = draft in pounds per square foot, Chapter VI. 
= load in kilowatts, Chapter III. 
= pressure in pounds per square foot. 
Pf = static pressure in pounds per square foot lost in wall friction. Chapter VI. 
Pr = static pressure in pounds per square foot lost in changes of cross-section, etc. 

Chapter VI. 

Pv = velocity head in pounds per square foot. 

p= pressure in pounds per square inch. 

7? e = mean exhaust pressure, Chapter VI. 

p s =mean suction pressure, Chapter VI. 

pv= partial pressure of water vapor in air, Chapter VI. 



Q = quantity of heat or energy in B.T.U. gained or lost by a body passing from one state to 
another. 
Qi'=heat added from fire in Stirling and Ericsson cycles, Chapter VI. 
Qi"=heat added from regenerator in Stirling and Ericsson cycles, Chapter VI. 
Q 2 '=heat abstracted by water jacket in Stirling and Ericsson cycles, Chapter VI. 
$2"= heat abstracted by regenerator in Stirling and Ericsson cycles, Chapter VI. 
q = quantity of heat per pound of liquid above 32° F. 



R = ratio of heating surface to grate surface, Chapter V. 

= gas constant. 
Rc= ratio of cylinder sizes in two-stage air compressor or compound engine, Chapters II 

and III. 
Rh= ratio of expansion in high-pressure cylinder, Chapter III. 
Rl= ratio of expansion in low-pressure cylinder, Chapter III. 
Rp = ratio of initial to back pressure, Chapters III and VI. 
R p = ratio of delivery to supply pressure, Chapter II. 
Rv = ratio of larger volume to smaller volume. 

r=rate of flame propagation in explosive mixtures, Chapter V. 
rp= pressure differences (maximum— minimum) in gas cycles, Chapter VI. 
ry= volume differences (maximum — minimum) in gas cycles, Chapter VI. 
(rec.pr.) = receiver pressure in pounds per square inch, Chapter III. 
(rel.pr.) = release pressure in pounds per square inch, Chapter III. 



S = per cent of ammonia in solution, Chapter IV. 
= piston speed, Chapter I. 

= pounds of steam per pound of air in producer blast, Chapter V. 
= specific heat, Chapter IV. 

= specific heat of superheated steam, Chapter VI. 
(Sup.Vol.) = volume of steam supplied to the cylinder per stroke, Chapter III. 
s = general exponent of V in expansion or compression of gases. 
sp.gr. = specific gravity, 
sp.ht. = specific heat, 
sq.ft. = square foot. 
sq.in.= square inch, 
(sup.pr.) = supply pressure, in pounds per square inch. 



T = temperature, degrees absolute. 
Tc — temperature of air, Chapter VI. 
Th = temperature of gases in chimney, Chapter VI. 
f = temperature in degrees scale, 



TABLE OF SYMBOLS xxiii 

£7 = rate of heat transfer in B.T.U. per square foot per hour per degree difference in tem- 
perature, Chapter IV. 
U= intrinsic energy, Chapter VI. 
u = velocity in feet per second. 
% = velocity in feet per minute, Chapter VI. 



V = volume in cubic feet. 
Va — cubic feet per pound air, Chapter VI. 
Vg = cubic feet per pound, gas, Chapter VI. 
Vl = volume of liquid in cubic feet per pound. 
Vs= volume of solid in cubic feet per pound. 
Vv = volume of vapor in cubic feet per pound. 

v = volume, Chapter IV. 



W = work in foot-pounds. 
VV.R. = water rate. 

w = pounds of water per pound of ammonia in solution, Chapter IV. 
= weight in pounds. 
wr= pounds of rich liquor per pound of ammonia, Chapter VI. 



X = compression in the steam engine as a fraction of the stroke, Chapter III. 

i ■ heat added T7T 

= 1H , Chapter VI. 

temperature at beginning of addition X specific heat at constant volume 

x = constant in the expression for missing water, Chapter III. 

= fraction of liquid made from solid or vapor made from liquid, Chapter VI. 

= per cent of carbon burned to CO>, Chapter V. 

= per cent of nozzle reheat, Chapter VI. 

= per cent of steam remaining in high-pressure cylinder of compound engine at any point 

of the exhaust stroke, Chapter III. 
= quantity of heat added in generator of absorption system in addition to the amount of 

heat of absorption of 1 lb. of ammonia, Chapter VI. 
= ratio of low-pressure admission volume to high-pressure admission volume, Chapter III. 



Y = total steam used per hour by an engine, Chapter III. 

1 , heat added 

= 1 + — — , Chapter VI. 

temperature at beginning of addition X specific heat at constant pressure 

y = per cent of vane reheat Chapter VI. 

= ratio of the volume of receiver to that of the high-pressure cylinder of the compound 

engine, Chapter III. 



Z= fraction of the stroke of the steam engine completed at cut-off, Chapter III. 

., . heat added from regenerator _, 

= 1+7 — . , Chapter VI. 

temperature at beginning of addition X specific heat at constant volume 

Z'= hypothetical best value of Z. 

., , heat added from regenerator _, TTT 

= 1 +- , Chapter VI. 

temperature at beginning of addition X specific heat at constant pressure 

2=ratio of R.P.M. to cycles per minute. 



=an angle, Chapter I. 

coefficient of cubical expansion, Chapter III. 

constant in the equation for latent heat, Chapter VI. 



xxiv TABLE OF SYMBOLS 

= constant in equation for variable specific heat at constant volume, Chapter VI. 
a' = constant in equation for variable specific heat at constant pressure, Chapter VI. 



constant in equation for latent heat, Chapter VI. 

fraction of fuel heat available for raising temperature, Chapter V. 



Y = constant in equation for latent heat, Chapter VI. 

= ratio of cross-section to perimeter, Chapter IV. 

. . . sp. ht. at const, press. 

= special value for s for adiabatic expansion or compression = : 

sp. ht. at const, vol. 

y'= ratio of specific heat at constant pressure to specific heat at constant volume when each 

is a variable, Chapter VI. 



A = increment. 

8 = density in pounds per cubic foot. 
8c = density in cold gases in equations for chimney draft, Chapter VI. 
8# = density of hot gases in equations for chimney draft, Chapter VI. 



C, = coefficient of friction, Chapter VI. 



\x = material coefficient in heat transfer expression, Chapter IV. 



p = internal thermal resistance, Chapter IV. 



2 = summation. 

c = surface thermal resistance, Chapter IV. 



«c = time in seconds. 

<£ = entropy, Chapter VI. 
$ = entropy, Chapter VI. 



Note. A small letter when used as a subscript to a capital in general refers to a point 
on a diagram, e.g., P a designates pressure at the point A. Two small letters used as sub- 
scripts together, refer in general to a quantity between two points, e.g., Wab designates 
work done from point A to point B. 



ENGINEERING THERMODYNAMICS 



CHAPTER I 
WORK AND POWER. GENERAL PRINCIPLES. 

1. Work Defined. Work, in the popular sense of performance of any labor, 
is not a sufficiently precise term for use in computations, but the analytical 
mechanics has given a technical meaning to the word which is definite and which 
is adopted in all thermodynamic analysis. The mechanical definition of work is 
mathematical inasmuch as work is always a product of forces opposing motion 
and distance swept through, the force entering with the product being limited 
to that acting in the direction of the motion. The unit of distance in the 
English system is the foot, and of force the pound, so that the unit of work is 
the foot-pound. In the metric system the distance unit is the meter and the 
force unit the kilogramme, making the work unit the kilogrammeter. Thus, 
the lifting of one pound weight one foot requires the expenditure of one foot- 
pound of work, and the falling of one pound through one foot will perform one 
foot-pound of work. It is not only by lifting and falling weights that work is 
expended or done; for if any piece of mechanism be moved through a distance 
of one foot, whether in a straight or curved path, and its movement be resisted 
by a force of one pound, there will be performed one foot-pound of work against 
the resistance. It is frequently necessary to transform work from one sys- 
tem of units to the other, in which case the factors given at the end of this 
Chapter are useful. 

Work is used in the negative as well as in the positive sense, as the force 
considered resists or produces the motion, and there may be both positive and 
negative work done at the same time; similar distinctions may be drawn with 
reference to the place or location of the point of application of the force. Con- 
sider, for example, the piston rod of a direct-acting pump in which a certain 
force acting on the steam end causes motion against some less or equal force 
acting at the water end. Then the work at the steam end of the pump may be 
considered to be positive and at the water end negative, so far as the move- 
ment of the rod is concerned; when, however, this same movement causes a 
movement of the water, work done at the water end (although negative with 
reference to the rod motion, since it opposes that motion) is positive with refer- 



2 . ENGINEEEING THERMODYNAMICS 

ence to the water, since it causes this motion. It may also be said that the 
steam does work on the steam end of the rod and the water end of the rod does 
work on the water, so that one end receives and the other delivers work, the rod 
acting as a transmitter or that the work performed at the steam end is the input 
and that at the water end the output work. 

(See the end of Chapter I for Tables I, II, III, IV, and VI, Units of 
Distance, of Surface, of Volume, of Weight and Force, and of Work.) 

Example. An elevator weighing 2000 lbs. is raised 80 ft. How much work is done in 
foot-pounds? 

Foot-pounds = force Xdistance 

= 2000 X80 = 160,000 ft.-lbs. 

Ans. 160,000 ft.-lbs. 

Prob. 1. A pump lifts 150 gallons of water to a height of 250 ft. How much work does 
it do? m 

Prob. 2. By means of a jack a piece of machinery weighing 10 tons is raised i in. What 
is the work done? 

Prob. 3. A rifle bullet weighing 2 oz. travels vertically upward li miles. What work was 
done in foot-pounds? 

Prob. 4. A cubic foot of water falls 50 ft. in reaching a water-wheel. How much work can 
it do? 

Prob. 5. A piston of an elevator is 12 ins. in diameter and has acting on it a pressure of 
80 lbs. per square inch. What work is done per foot of travel? 

Prob. 6. It has been found that a horse can exert 75 lbs. pull when going 7 miles per 
hour. How much work can be done per minute? 

Prob. 7. How much work is done by an engine which raises a 10-ton casting 50 ft.? 

Prob. 8. The pressure of the air on front of a train is 50 lbs. per square foot when the 
speed is 50 miles per hour. If the train presents an area of 50 sq.ft., what work is done in 
overcoming wind resistance? 

Prob. 9. The pressure in a 10-inch gun during the time of firing is 2000 lbs. per square 
inch. How much work is done in ejecting the projectile if the gun is 33 ft. 4 ins. long? 

2. Power Defined. Power is defined as the rate of working or the work 
done in a given time interval, thus introducing a third unit of mechanics, time, 
so that power will always be expressed as a quotient, the numerator being a prod- 
uct of force and distance, and the denominator time. This is in opposition 
to the popular use of the word, which is very hazy, but is most often applied to 
the capability of performing much work] or the exertion of great force, thus, 
popularly, a powerful man is one who is strong, but in the technical sense a man 
would be powerful only when he could do much work continuously and rapidly. 
An engine has large power when it can perform against resistance many foot- 
pounds per minute. The unit of power in the English system is the horse-power, 
or the performance of 550 foot-pounds per second or 33,000 foot-pounds per 
minute, or 1,980,000 foot-pounds per hour. In the metric system the horse- 
power is termed cheval-vapeur, and is the performance of 75 killogrammeters 
= 542 \ foot-pounds per second, or 4500 kilogrammeters = 32,549 foot-pounds 



WORK AND POWER. 3 

per minute, or 270,000 kilogrammeters = 1,952,932 foot-pounds per hour. 
Table VII at the end of Chapter I gives conversion factors for power units. 

Example. The piston of a steam engine travels 600 ft. per minute and the mean force 
of steam acting upon it is 65,000 lbs. What is the horse-power? 

TT foot-pounds per minute 
Horse-power = ^^ 

r ..distance 

force X — -. 

t ime 

~ 337OOO 

^65,000X600_ 1182Hp 

33,000 

Prob. 1. The draw-bar pull of a locomotive is 3000 lbs. when the train is traveling 50 
miles per hour. What horse-power is being developed? 

Prob. 2. A mine cage weighing 2 tons is lifted up a 2000-ft. shaft in 40 seconds. What 
horse-power will be required if the weight of the cable is neglected? 

Prob. 3. By direct pull on a cable it is found possible to lift 4 tons 20 ft. per second. With 
a differential pulley 40 tons may be lifted 3 ft. per second. What is the difference in power 
required? 

Prob. 4. A horse exerts a pull of 100 lbs. on a load. How fast must the load be moved to 
develop one horse-power? 

Prob'. 5. The resistance offered to a ship at a speed of 12 knots was 39,700 lbs. What 
horse-power must be available to maintain this speed? (One knot is a speed of one nautical 
mile per hour.) 

Prob. 6. It is estimated that 100,000 cu. ft. of water go over a fall 60 ft. high every 
second. What horse-power is going to waste? 

Prob. 7. The force acting on a piston of a pump is 80,000 lbs. If the piston speed is 150 
ft. per minute, what is the horse-power? 

Prob. 8. To draw a set of plows 2 5 miles per hour requires a draw-bar pull of 10,000 
lbs. What must be the horse-power of a tractor to accomplish this? 

Prob. 9. The horse-power to draw a car up a grade is the sum of the power necessary to 
pull it on a level and that necessary to lift it vertically the same number of feet as it rises on 
the grade. What will be the horse 7 power required to draw a car 20 miles per hour up a 12 per 
cent grade if the car weighs 2500 lbs. and the draw-bar pull on the level is 250 lbs.? 

3. Work in Terms of Pressure and Volume. Another of the definitions 
of mechanics fixes pressure as force per unit area so that pressure is always a 
quotient, the numerator being force and the denominator area, or length to 
the second power. If, therefore, the pressure of a fluid be known, and accord- 
ing to hydromechanics it acts equally and normally over all surface in contact 
with it, then the force acting in a given direction against any surface will be 
the product of the pressure and the projected area of the surface, the projection 
being on a plane at right angles to the direction considered. In the case of pis- 
tons and plungers the line of direction is the axis of the cylinder, and the pro- 
jected area is the area of the piston less the area of any rod passing completely 
through the fluid that may be so placed. When this plane area moves in a' 



4 ENGINEERING THERMODYNAMICS 

direction perpendicular to itself, the product of its area and the distance will be 
the volume swept through, and if a piston be involved the volume is technically 
the displacement of the piston. Accordingly, work may be expressed in three 
ways, as follows: 

Work = force X distance ; 

Work = pressure X area X distance ; 

Work = pressure X volume. 

The product should always be in foot-pounds, but will be, only when appro- 
priate units are chosen for the factors. These necessary factors are given as 
follows: 

Work in foot-pounds = force in lbs. X distance in ft. 

= pressure in lbs. per sq.ft. X area in sq.ft. X distance in ft 
= pressure in lbs. per sq.in. X area in sq.in. X distance in ft 
= pressure in lbs. per sq.ft. X volume in cu.ft. 
= pressure in lbs. per sq.in. X 144 X volume in cu.ft. 

As pressures are in practice expressed in terms not only as above, but also 
in heights of columns of common fluids and in atmospheres, both in English and 
metric systems, it is convenient for calculation to set down factors of equivalence 
as in Table V, at the end of the Chapter. 

In thermodynamic computations the pressure volume product as an expres- 
sion for work is most useful, as the substances used are always vapors and gases, 
which, as will be explained later in more detail, have the valuable property of 
changing volume indefinitely with or without change of pressure according 
to the mode of treatment. Every such increase of volume gives, as a conse- 
quence, some work, since the pressure never reaches zero, so that to derive work 
from vapors and gases they are treated in such a way as will allow them to change 
volume considerably with as much pressure acting as possible. 

It should be noted that true pressures are always absolute, that is, measured 
above a perfect vacuum or counted from zero, while most pressure gages and 
other devices for measuring pressure, such as indicators, give results measured 
above or below atmospheric pressure, or as commonly stated, above or below 
atmosphere. In all problems involving work of gases and vapors, the absolute 
values of the pressures must be used; hence, if a gage or indicator measure- 
ment is being considered, the pressure of the atmosphere found by means of the 
barometer must be added to the pressure above atmosphere in order to obtain the 
absolute or true pressures. When the pressures are below atmosphere the 
combination with the barometric reading will depend on the record. If a record 
be taken by an indicator it will be in pounds per square inch below atmosphere 
and must be subtracted from the barometric equivalent in the same units to 
give the absolute pressure in pounds per square inch. When, however, a 
vacuum gage reads in inches of mercury below atmosphere, as such gages 
do, the difference between its reading and the barometric gives the absolute 



WORK AND POWER. 5 

pressure in inches of mercury directly, which can be converted to the desired 
units by the proper factors. 

While it is true that the barometer is continually fluctuating at every place, 
it frequently happens that standards for various altitudes enter into calculations, 
and to facilitate such work, values are given for the standard barometer at various 
altitudes with equivalent pressures in pounds per square inch in Table IX. 

Frequently in practice, pressures are given without a definite statement 
of what units are used. Such a custom frequently leads to ambiguity, but it 
is often possible to interpret them correctly from a knowledge of the nature of the 
problem in hand. For instance, steam pressures stated by a man in ordinary prac- 
tice as being 100 lbs. may mean 100 lbs. per square inch gage (above atmosphere),, 
but may be 100 lbs. per square inch absolute. Steam pressures are then most 
commonly stated per square inch and should be designated as either gage or abso- 
lute. Pressures of compressed air are commonly expressed in the same units as 
steam, either gage or absolute, though sometimes in atmospheres. Steam pressures 
below atmosphere may be stated as a vacuum of so many inches of mercury, 
meaning that the pressure is less than atmosphere by that amount, or may 
be given as a pressure of so many inches of mercury absolute, or as so many 
pounds per square inch absolute. The pressures of gases stored in tanks under 
high pressure are frequently recorded in atmospheres, due to the convenience 
of computation of quantities on this basis. Pressures of air obtained by blowers 
or fans are usually given by the manufacturers of such apparatus in ounces 
per square inch above (or below) atmosphere. Such pressures and also differ- 
ences of pressure of air due to chimney draft, or forced draft, and the pressure 
of illuminating gas in city mains, are commonly stated in inches of water, each 
inch of water being equivalent to 5.196 lbs. per square foot. The pressure of 
water in city mains or other pressure pipes may be stated either in pounds per 
square inch or in feet of water head. 

Example. A piston on which the mean pressure is 60 lbs. per square inch sweeps through 
a volume of 300 cu.ft. What is the work done? 

W=PxV, where V= cu.ft. and P=lbs. per sq.ft. 
.*. 17=60x144x300 =2,592,000 ft. -lbs. 

Prob. 1. The mean pressure acting per square inch when a mass of air changes in 
volume from 10 cu.ft. to 50 cu.ft. is 40 lbs. per square inch. How much work is done? 

Prob. 2. An engine is required to develop 30 H.P. If the volume swept through per 
minute is 150 cu.ft., what must the mean pressure be? 

Prob. 3. The mean effective pressure in compressing air from one to five atmospheres is 
28.7 lbs. per square inch. How many horse-power are required to compress 1000 cu.ft. of 
free air per minute? 

Prob. 4. At an altitude of 4100 ft. a pressure gage showed the pressure on one side of 
a piston to be 50 lbs. per square inch while the pressure on the opposite side is 3 lbs. per 
square inch absolute. What pressure was tending to move the piston? 

Prob. 5. At an altitude of 1 mile the mean pressure in a gas engine cylinder during the 
suction stroke was found to be 12 lbs. per square inch absolute. What work was done 
by the engine to draw in a charge if the cylinder was 5 ins. in diameter and the stroke 6 ins.? 



6 ENGINEERING THERMODYNAMICS 

Prob. 6. After explosion the piston of the above engine was forced out so that the gas 
volume was five times that at the beginning of the stroke. What must the M.E.P. have 
been to get 20,000 ft.-lbs. of work? 

Prob. 7. On entering a heating oven cold air expands to twice its volume. What 
work is done per cubic foot of air? 

Prob. 8. A projectile is forced from a gun by a constant air pressure of 1000 lbs. per 
square inch. Before it begins to move there is \ a cu.ft. of air in the barrel, and at the instant 
it leaves the barrel the volume is 10 cu.ft. What work was done on the projectile? 

Prob. 9. Water is forced from a tank against a head of 75 ft. by rilling the tank with 
compressed air. How much work is done in emptying a tank containing 1000 cu.ft.? 

4. Work of Acceleration and Resultant Velocity. When a force acting 
on a mass is opposed by an equal resistance there may be no motion at all, or 
there may be motion of constant velocity. Any differences, however, between 
the two opposing forces will cause a change of velocity so long as the difference 
lasts, and this difference between the two forces may be itself considered as the 
only active force. Observations on unresisted falling bodies show that they 
increase in velocity 32.16 ft. per second for each second they are free to fall, 
and this quantity is universally denoted by g. If then, a body have any 
velocity, u\, and be acted on by a force equal to its own weight in the direction 
of its motion for a time, t seconds, it will have a velocity U2 after that time. 

u 2 = ui+gt (1) 

It may be that the force acting is not equal to the weight of the body, in which 
case the acceleration will be different and so also the final velocity, due to the 
action of the force, but the force producing any acceleration will be to the 
weight of the body as the actual acceleration is to the gravitational acceleration. 
So that 

Actual acceler ation force _ actual acceleration 

Weight of body or gravitational force gravitational acceleration (g)' 
and 

Actual accelerating force = t——. — —. = — ^—. f-r X actual acceleration. 

gravitational acceleration (g) 

or 

„ w . v change of velocity 

Force = mass X acceleration = mass X —r. v. , — , 

time of change 

F=Mx ^^l (2 ) 

T 

The work performed in accelerating a body is the product of the resistance 
met into the distance covered, L, while the resistance, or the above-defined force, 
acts, or while the velocity is being increased. This distance is the product of 
the time of action and the mean velocity, or the distance in feet, 

u 2 -\-ui , Q v 

L= — s — t (3) 



WORK AND POWER. 7 

The work-is the product of Eqs. (2) and (3), or, work of acceleration is 

^ _ M(U 2 -Ui) ^ y (U2 + Ui)T 

W ~ r X 2 

where w is the weight in pounds. Exactly the same result will be obtained by 
the calculus when the acceleration is variable, so that Eq. (4) is of universal 
application. 

The work performed in accelerating a body depends on nothing but its mass 
and the initial and final velocities, and is in every case equal to the product of 
half the mass and the difference between the squares of the initial and final 
velocities, or the product of the weight divided by 64-4 an d the difference between 
the squares of the initial and final velocities. 

It frequently happens that the velocity due to the reception of work is desired, 
and this is the case with nozzle flow in injectors and turbines, where the steam 
performs work upon itself and so acquires a velocity. In all such cases the 
velocity due to the reception of the work energy is 



M2 



2 , 64.32PT _ 



where W is work in foot-pounds and w, as before, is weight in pounds. Or if 
the initial velocity be zero, as it frequently is, 



u 2 



Jm-Jf*x& ( 6) 

\ w \ w 



For conversion of velocity units, Table VIII, at the end of the Chapter, 
is useful. 

Example. A force of 100 lbs. acts for 5 seconds on a body weighing 10 lbs.; if the 
original velocity of the body was 5 ft. per second, what will be the final velocity, the 
distance traveled and the work done? 

rp_ M(Uz— U j) m 

r ~ > 

T 

im _ 10 Q 2 -5) . 

U 2 = 1615 ft. per second; 
S-(*±*s),-4O60ft. 

W = M(tte '~ M '' ) =405,000 ft.-lbs. 

Prob. 1. A stone weighing \ lb. is dropped from a height of 1 mile. With what veloc- 
ity and in what length of time will it strike if the air resistance is zero? 

Prob. 2. A car moving 20 miles per hour and weighing 25 tons is brought to rest in 
500 ft. What is the negative acceleration, the time required to stop, and the work done? 



8 ENGINEERING THERMODYNAMICS 

Prob. 3. Steam escapes through an opening with a velocity of half a mile per second. 
How many foot-pounds of energy were imparted to each pound of it to accomplish this? 

Prob. 4. A weight of 100 lbs. is projected upward with a constant force of 200 lbs. 
How much further will it have gone at the end of 10 seconds than if it had been merely 
falling under the influence of gravity for the same period of time? 

Prob. 5. A projectile weighing 100 lbs. is dropped from an aeroplane at the height of 
J mile. How soon will it strike, neglecting air resis ance? 

Prob. 6. A water-wheel is kept in motion by a jet of water impinging on flat vanes. 
The velocity of the vanes is one-half that of the jet. The jet discharges 1000 lbs. of 
water per minute with a velocity of 200 ft. per second. Assuming no losses, what is 
amount of the work done? 

Prob. 7. With the wind blowing 30 miles per hour, how much work could a 12-ft. 
windmill perform if 25 per cent of the available work were utilized. 

Note. The weight of a cubic foot of air may be taken as .075 lb. 

Prob. 8. An engine has a piston speed of 600 ft. per minute and runs at 150 R.P.M. 
If the reciprocating parts weigh 500 lbs., how much work is done in accelerating the 
piston during each stroke? 

Prob. 9. A flywheel with rim 10 ft. in diameter to center of section and weighing 5 
tons, revolves at a rate of 150 R.P.M. ; 100,000 ft.-lbs. of work are expended on it. How 
much will the speed change? 

5. Graphical Representation of Work. As work is always a product of 
force and distance or pressure and volume, it may be graphically expressed by 



5 

4 
3 


B 




















C 












































































2 


























1 






















D 





A 12 3 4 5 

Distances in Feet 

Fig. 1. — Constant Force, Work Diagram, Force-Distance Coordinates. 



an area on a diagram having as coordinates the factors of the product. It is 
customary in such representations to use the horizontal distances for volumes 
and the vertical for pressures, which, if laid off to appropriate scale and 
in proper units, will give foot-pounds of work directly by the area enclosed. 
Thus in Fig. 1, if a force of 5 lbs. (A B) act through a distance of 5 ft. (BC) 
there will be performed 25 foot-pounds of work as indicated by the area of the 



WORK AND POWER 



9 



rectangle A BCD, which encloses 25 unit rectangles, each representing one foot- 
pound of work. 

If a steam cylinder piston suffers a displacement of 5 cu.ft. under the steam 
pressure (absolute) of 5 lbs. per square foot then the operation which results 
in the performance of 25 foot-pounds of work is represented by the diagram 
Fig. 2, ABCD. 





B 




















C 




b 


























4 


























3 


























)i 


























1 






















D 





A 1 2 3 4 5 

Volumes in Cubic Feet 

Fig. 2. — Constant Pressure Work Diagram, Pressure- Volume Coordinates. 

Prob. 1. Following the method given for Fig. 1, draw a diagram for the example of 
Section 3 

Prob. 2. By means of a diagram, show that the work done by a pressure of 1000 lbs. 
per square foot traversing a distance of 10 ft. is 10,000 ft.-lbs. 

Prob. 3. Draw a diagram for the case of a volume change from 1 to 10 cu.ft. while the 
pressure acting is 20 lbs. per square inch. 

Prob. 4. Draw a pressure volume diagram for the case of forcing a piston out of a 
cylinder by a water pressure of 15,000 lbs. per square foot, the volume of the cylinder at 
the start is J cu.ft. and at the end 6 cu.ft. Make a diagram to scale and report work 
per square inch of diagram. 

Prob. 5. A pump draws in water at a constant suction pressure of 14 lbs. and dis- 
charges it at a constant delivery pressure of 150 lbs. per sq.in. Considering the pump 
barrel to be empty at beginning of suction and end of delivery and to contain 3 cu.ft. 
when full, draw the diagram for this case and find the foot-pounds of work done. 

Prob. 6. In raising a weight a man pulls on a rope with a constant force of 80 lbs. 
If the weight is lifted 40 ft., find from a diagram the work done. 

Prob. 7. In working a windlass a force of 100 lbs. is applied at the end of a 6-ft. 
lever, the drum of th windlass being 1 ft. in diameter. Draw a work diagram for 
work applied and for work done in lifting if there be no loss in the windlass. 

Prob. 8. The steam and water pistons of a pump are on the same rod and the area 
of the former is twice that of the latter, the stroke being 3 ft. Show by a diagram 
that the work done in the two cylinders is the same if losses be neglected. 

Prob. 9. An engine exerts a draw-bar pull of 8000 lbs. at speed of 25 miles an hour. 
A change in grade occurs and speed increases to 40 miles per hour and the pull decreases 
to 5000 lbs. Show by a diagram the change in horse-power. 



10 



ENGINEERING THERMODYNAMICS 



6. Work by Pressure Volume Change. Suppose that instead of being 
constant the pressure were irregular and, being measured at intervals of 1 cu.ft. 
displacement, found to be as follows: 



Pressure. 


Displacement 


Lbs. per Sq.Ft. 


Volume. Cu.Ft. 


100 





125 


1 


150 


2 


100 


3 


75 


4 


50 


5 



175 



150 



125 



100 



75 



50 





































C 


















A' 


E 


i^" 






















k^ 


-i 


B' 






\ 


D 






























E 






























F 
































H 




















G 







2 

^ 25 



12 3 4 5 6 

Volumes in Cubic Feet 
Fig. 3. — Work Diagram, Pressure-Volume Coordinates. Discontinuous Pressure-Volume 

Relations. 

This condition might be plotted as in Fig. 3, A, B, C, D, E, F, G, H. The 
work done will be the area under the line joining the observation points. In 
the absence of exact data on the nature of the pressure variations between the 
two observation points A and B, sl variety of assumptions might be made as to 
the precise evaluation of this area, as follows: 

(a) The pressure may have remained constant at its original value for the 
first cubic foot of displacement, as shown dotted A-B' and then suddenly have 
risen to B. In this case the work done for this step would be 100 foot-pounds. 

(b) Immediately after the measurement at A the pressure may have risen 
to A' and remained constant during displacement A' to B, in which case the 
work done would be 125 foot-pounds. 

(c) The pressure may have risen regularly along the solid line A B, in which 



case the work area is a trapezoid and has the value 
pounds. 



100+125 



XI = 112.5 foot- 



WORK AND POWER 



11 



It thus appears that for the exact evaluation of work done by pressure 
volume change, continuous data are necessary on the value of pressure with 
respect to the volume. If such continuous data, obtained by measurement or 
otherwise, be plotted, there will result a continuous line technically termed the 
pressure-volume curve for the process. Such a curve for a pressure volume 
change starting at 1 cu.ft. and 45 lbs. per square foot, and ending at 7 cu.ft., 
and 30 lbs. per square foot, is represented by Fig. 4, A, B, C, D, E. 

The work done during this displacement under continuously varying pressure 
is likewise the area between the curve and the horizontal axis when pressures are 
laid off vertically, and will be in foot-pounds if the scale of pressures is pounds 
per square foot and volumes, cubic feet. Such an irregular area can be divided 
into small vertical rectangular strips, each so narrow that the pressure is sensibly 
constant, however much it may differ in different strips. The area of the 
rectangle is PA 7, each having the width AV and the height P, and the work 



3 60 

OQ 

S 50 

co 40 
§30 



10 



£ ° 









































^"B 




















































A 
























































C 


































































E 
























D 



! 3*4 5 

Volumes in Cubic Feet 



Fig. 4. — Work Diagram, Pressure- Volume Coordinates. Continuous Pressure- Volume 

Relations. 



area will be exactly evaluated if the strips are narrow enough to fulfill the 
conditions of sensibly constant pressure in any one. This condition is true only 
for infinitely narrow strips having the width dV and height P, so that each has 
the area PdV and the whole area or work done is 



W 



-/■ 



PdV. 



(7) 



This is the general algebraic expression for work done by any sort of continuous 
pressure volume change. It thus appears that whenever there are available 
sufficient data to plot a continuous curve representing a pressure volume change, 
the work can be found by evaluating the area lying under the curve and bounded 
by the curve coordinates and the axis of volumes. The work done may be 
found by actual measurement of the area or by algebraic solution of Eq. (7), 
which can be integrated only when there is a known algebraic relation between 
the pressure and the corresponding volume of the expansive fluid, gas or vapor. 

Prob. 1. Draw the diagrams for the following cases : (a) The pressure in a cylinder 
12 ins. in diameter was found to vary at different parts of an 18-in. stroke as follows: 



12 



ENGINEEEING THEEMODYNAMICS 



Pressure in Pounds 


Per Cent of 


per Sq.In. 


Stroke. 


100 





100 


10 


100 


30 


100 


50 


83.3 


60 


71.5 


70 


62.5 


80 


55.5 


90 


50.0 


100 



{b) On a gas engine diagram the following pressures were found for parts of stroke. 



In 


Out 


V 


V 


V 


V 


V 


P 


0.25 cu.ft. 
0.20 " 
0.14 " 
0.10 " 


14.7 
19.5 
29.7 
45.2 


0.1 

0.102 

0.104 

0.106 

0.108 

0.11 

0.12 


45.2 
79.7 
123.2 
157.7 
181.7 
188.2 
166.2 


0.13 
0.15 
0.17 
0.19 
0.21 
0.23 


146.2 
116.7 
95.7 
80.7 
68.7 
58.7 



Prob. 2. Steam at a pressure of 100 lbs.. per square inch absolute is admitted to a 
cylinder containing .1 cu.ft. of steam at the same pressure, until the cylinder contains 
1 cu.ft., when the supply valve closes and the volume increases so that the product of 
pressure and volume is constant until a pressure of 30 lbs. is reached. The exhaust 
valve is opened, the pressure drops to 10 lbs. and steam is forced out until the volume 
becomes 1 cu.ft., when the exhaust valve closes and the remaining steam decreases in 
volume so that product of pressure and volume is constant until the original point 
is reached. Draw the pressure volume diagram for this case. 

Prob. 3. During an air compressor stroke the pressures and volumes were as 
follows : 



Volume in 


Tressure in Lbs. 


Cu.Ft. 


Sq.In. 


2.0 


14.0 


1.8 


15.5 


1.6 


17.5 


1.4 


20.0 


1.2 


23.3 


1.0 


28.0 


0.8 


28.0 


0.4 


28.0 


0.0 


28.0 



Draw the diagram to a suitable scale to give work area in foot-pounds directly. 
Prob. 4. Draw the diagrams for last two problems of Section 3. 



WORK AND POWER 13 

7. Work of Expansion and Compression. Any given quantity of gas or 
vapor confined and not subject to extraordinary thermal changes such as 
explosion, will suffer regular pressure changes for each unit of volume change, 
or conversely, suffer a regular volume change for each unit of pressure change, 
so that pressure change is dependent on volume change and vice versa. When 
the volume of a mass of gas or vapor, V\, is allowed to increase to V2 by the 
movement of a piston in a cylinder, the pressure will regularly increase or 
decrease from Pi to P2, and experience has shown that no matter what the gas 
or vapor or the thermal conditions, if steady, the volumes and pressures will 
have the relation for the same mass, 

P lVl s + P 2 V 2 s = K, (8) 

or the product of the pressure and s power of the volume of a given mass 
will always be the same. The exponent s may have any value, but usually 
lies between 1 and 1.5 for conditions met in practice. 
The precise value of s for any given case depends on 

(a) The substance. 

(b) The thermal conditions surrounding expansion or compression, s being 
different if the substance receives heat from, or loses heat to, external sur- 
roundings, or neither receives nor loses. 

(c) The condition of vapors as to moisture or superheat when vapors are 
under treatment. 

Some commonly used values of s are given in Table X at the end of this 
chapter for various substances subjected to different thermal conditions dur- 
ing expansion or compression. 

Not only does Eq. (8) express the general law of expansion, but it likewise 
expresses the law of compression for decreasing volumes in the cylinder with 
corresponding rise in pressure. Expansion in a cylinder fitted with a piston 
is called balanced expansion because the pressure over the piston area is 
balanced by resistance to piston movement and the mass of gas or vapor is 
substantially at rest, the work of expansion being imparted to the piston and 
resisting mechanism attached to it. On the other hand when the gas or vapor 
under pressure passes through a nozzle orifice to a region of lower pressure the 
falling pressure is accompanied by increasing volumes as before, but the work 
of expansion is imparted not to a piston, because there is none, but to the fluid 
itself, accelerating it until a velocity has been acquired as a resultant of the 
work energy received. Such expansion is termed free expansion and the law of 
Eq. (8) applies as well to free as to balanced expansion. This equation, then, 
is of very great value, as it is a convenient basis for computations of the work 
done in expansion or compression in cylinders and nozzles of all sorts involv- 
ing every gas or vapor substance. Some expansion curves for different values 
of s are plotted to scale in Fig. 5, and the corresponding compression curves in 
Fig. 6= in which 

Curve A has the exponent s = 
Curved " " s = .5 



14 



ENGINEERING THERMODYNAMICS 



Curve C has the exponent 8 = 1.0 



Curve D 
Curve E 
Curve F 
Curve G 
Curve H 



8 = 1.1 

8=1.2 

s=1.3 
s = 1.4 

8 = 1.5 




Fig. 5. — Comparison of Expansion Lines having Different Values j)f s. 
The volume after expansion is given by 



v 2 =v 1 



(9) 



so that the final volume depends on the original volume, on the ratio of the two 
pressures and on the value of the exponent. Similarly, the pressure after 
expansion 

p »- p *%'' ■ (10) 



WOEK AND POWER 



15 



depends on the original pressure, on the ratio of the two volumes and on 
the exponent. 

The general equation for the work of expansion or compression can now be 
integrated by means of the Eq. (8), which fixes the relation between pressures 
and volumes. From Eq. (8), 

p=* 




9 10 11 12 13 14 15 16 17 18 19 

Volumes in Cubic Feet 
Fig. 6. — Comparison of Compression Curves having Different Values of 

which, substituted in Eq.(7), gives 

J V s ' 
but as K is a constant, 



W = K 



fdV 

J V s ' 



(11) 



The integral of Eq. (11) will have two forms: 

(1) When s is equal to one, in which case P±Vi =P2V2=K 1 ; 

(2) When s is not equal to one. 



16 



ENGINEERING THERMODYNAMICS 



Taking first the case when s is equal to one, 

' v *dV 



W = K 1 



fr*dV 



Whence 



W = KU0geY 1 



= PlV 1 \0g e 
= P 2 V 2 log e 



Yl 
Vl 

Yl 
Vl 



= K 1 log e j± 
= Pi7ilo & g 

= P2V 2 l0g e ^ 



(a) 
(b) 
(c) 
(d) 

w 
■if) 



When s = 1 . 



(12) 



Eqs. (12) are all equal and set down in different forms for convenience in 
computation; in them 

V 2 = largest volume = initial vol. for compression = final vol. for expansion. 

P2 = smallest pressure = initial pres. for compression = final pres. for expansion. 

Vi = smallest volume = final vol. for compression = initial vol. for expansion. 

Pi = largest pressure = final pres. for compression = initial pres. for expansion. 

These Eqs. (12) all indicate that the work of expansion and compression of 
this class is dependent only on the ratio of pressures or volumes at the beginning 
and end of the process, and the PV product at either beginning or end, this 
product being of constant value. 

When the exponent s is not equal to one, the equation takes the form, 



r v * dv • r v * 

W = K\ ^-=K\ V~ s dV 

JVx V s JVx 

-£i[r*-'-v*-] 



As s is greater than one, the denominator and exponents will be negative, so, 
changing the form to secure positive values, 



W 



s-lVF,*- 1 



1 



This can be put in a still more convenient form. Multiplying and dividing by 
1 1 



or 



TV 



-^[(r'-] 



_/v : 

s-l Vi 



f] 



WORK AND POWER 



17 



Substituting the value of 2£ = P2lY = PilV, 



w 1 P2V2* 
W 8-1 TV -1 



wr-^m-mi 



Whence 



r-^[(?r-.'] »i 
-si-©-] *> 



■ . When s^l 



(13) 



Eqs. (13) gives the work for this class of expansion and compression in terms 
of pressure ratios and volume ratios and in them 

V2 — largest volume = initial vol. for compression = final vol. for expansion; 
P2 = smallest pressure = initial pres. for compression = final pres. for expansion ; 
Vi = smallest volume = final vol. for compression = initial vol. for expansion; 
Pi = largest pressure = final pres. for compression = initial pres. for expansion. 

The work of expansion or compression of this class is dependent according 
to Eqs. (13), upon the ratio of pressures or volumes at beginning and end of the 
process, the exponent, and on the pressure volume product appropriately taken. 
It should be remembered that for the result to be in foot-pounds appropriate 
units should be used and all pressures taken absolute. Examination of Eqs. 
(12) and (13), for the work done by expansion or compression of both classes, 
shows that it is dependent on the initial and final values of pressures and volumes 
and on the exponent s, which defines the law of variation of pressure with 
volume between the initial and final states. 

Example 1. Method of calculating Diagram, Figs. 5 and 6. Consider the curve 
for which s = 1.4 as typical of the group. 

Assumed Data. Vi =1.0 cu.ft. Pi =20,000 lbs. per square foot. 

8 = 1.4. 

PiVi s =K =20,000 Xl 14 =20,000. 



Then 



For any other value of P, V was found from the relation, 



"-© 7 



18 

Let P* =6000, 
then 

or 



ENGINEERING THERMODYNAMICS 



.715 



715 



log 3.33 = .5224 

.715 X. 5224 = .373 =logT*. 
/. F*=2.36. 

A series of points, as shown below, were found, through which the curve was drawn. 



p 


20,000 
P • 


. 20,000 
log p . 


1 . 20,000 
— log— p— ^ 


V 




18000 


1.111 


0.0453 


0.032 


1.08 




14000 


1.430 


0.1553 


0.111 


1.30 




10000 


2.000 


0.3010 


0.214 


1.64 




6000 


3.330 


0.5224 


0.373 


2.36 




2000 


10.000 


1.0000 


0.714 


5.18 




1000 


20.000 


1.3010 


0.930 


8.51 





Curves for other values of s were similarly drawn. Starting at a common volume 
of 20 cu. ft. the compression curves of Fig. 6 were determined by the same methods. 

Example 2. A pound of air at 32° F. and under atmospheric pressure is compressed 
to a pressure of five times the original. What will be the final volume and the work 
done if s = l and if s = 1.4? The volume of 1 lb. of air at 32° F. and one atmosphere 
is 12.4 cu.ft. approx. 



Fors = l, 



7 ?= Pi 
Vi P 2 



Pl -5 

pT 5j 



12.4 



Vi 



V 2 = 12 A cu.ft.; 



5, whence Vi =2.48 cu.ft. 



Pi 



Tf=P 2 F 2 log e ^ = 2116Xl2.41og e 5; 
P2 



=2116x12.4x1.61=42,300 foot-pounds. 



For s = 1.4, 



F 2 



r(ar - *g-«> n -«>- n - 



5 may be raised to the .71 power by means of logarithms as follows: (5)' 71 is equal to 
the number whose logarithm is .71 log 5. 



WORK AND POWER 19 

Log 5 =.699, .71 X.699 =.4963, and number of which this is the logarithm is 3.13, 



hence, 



s — 1 



Fi = F 2 ^3.13 or Fi=3.96; 

s-l 



2 

2116X12.4 



/pa-- 2116X12.4 T fi 



X. 583 =38,200 ft.-lbs. 



.4 
The value of W can also be found by any other form of equation (13) such as, 



"-sMir] 



The value of Vi being found as before, the work expression becomes after numerical 
substitution 



_ 10,580X3.96 
W = — \ 



[-(Dl 



As the quantity to be raised to the .4 power is less than one, students may find it 
easier to use the reciprocal as follows: 



/3.9CA 
U2.4/ 



— 

V3.96/ 



A = r4r - .632 



12.4\ - 4 (3.13)-* 1.58 



Hence 



w = 10,580 X3.96 (1 _ 632) =S8200 ft _ lbg 



Prob. 1. Find Vi and W for Example 2 if s = 1.2 and 1.3. 

Prob. 2. If a pound of air were compressed from a pressure of 1 lb. per square inch 
absolute to 15 lbs. per square inch absolute find Vi and W when s = 1 and 1.4. V 2 =180 
cu.ft. What would be the H.P. to compress 1 lb. of air per minute? 

Prob. 3. Air expands so that 8 = 1. If Pi = 10,000 lbs. per square foot, Vi = 10 cu.ft. 
and V 2 = 100 cu.ft. and the expansion takes place in 20 seconds, wha' is the H.P. devel- 
oped? 

Prob. 4. 100 cu.ft. of air at atmospheric pressure is compressed in a cylinder to a pres- 
sure of 8 atmospheres and then expelled against this constant pressure. Find graphically 
and by calculation the foot-pounds of work done for the case where s = 1 and for the case 
where s = 1.4. 

Prob. 5. At an altitude of 4000 ft., a r is compressed to a pressure of 60 lbs. per 
sq.in. gage. Find he H.P. required to compress 1000 cu.ft. of free air per minute. 

Prob. 6. From the algebraic equation show how much work is done for a volume 
change of 1 to 4, provided pressure is originally 1000 lbs. per square foot when 

(a) PV°=K lt 
(6) PV=K 2 , 
(c) PF 2 = Z 3 . 



20 ENGINEEEING THEEMODYNAMICS 

Prob. 7. A vacuum pump compresses air from 1 lb. per square inch absolute to 15 
lbs. per square inch absolute and discharges it. An air compressor compresses air from 
atmosphere to 15 atmospheres and discharges it. Compare the work done for equal 
initial volumes, s = 1.4. 

Prob. 8. For steam expanding according to the saturation law, compare the work 
done by 1 lb. expanding from 150 lbs. per square inch absolute to 15 lbs. per square inch 
absolute with, the work of the same quantity expanding from 15 lbs. to 1 lb. per square 
inch absolute. 

Note. 1 lb. of steam occupies 3 cu.ft. at 150 lbs. per square inch absolute. 

Prob. 9. Two air compressors of the same size compress air adiabatically from atmos- 
phere to 100 lbs. gage and discharge it. One is at sea level, the other at 10,000 ft. ele- 
vation. Compare the work in the two cases. 

8. Values of Exponent s Defining Special Cases of Expansion or Compres- 
sion. There are three general methods of finding s for the definition of particular 
cases of expansion or compression to allow of the solution of numerical problems. 
The first is experimental, the second and third thermodynamic. If by measure- 
ment the pressures and volumes of a series of points on an expansion or com- 
pression curve, obtained by test with appropriate instruments, for example, 
the indicator, be set down in a table and they be compared in pairs, values of 
s can be found as follows: Calling the points A, B, C, etc., then, 

PaVa S = PbVb S , 

and 

log Pa+S log Fa = log Pb + S log Vb, 

or 

«(l0g Vb — log Fa) = log Pa — log Pb, 

hence 

log P a - log P b 
l0gF 6 -l0gF a W 



or 

. los © (b) f 



(14) 



'«< ft) 



According to Eq. (14a), if the difference between the logarithms of the pressures 
at B and A be divided by the differences between the logarithms of the volumes 
at A and B respectively, the quotient will be s. According to Eq. (146), the 
logarithm of the ratio of pressures, B to A , divided by the logarithm of the ratio 
of volumes, A to B respectively will also give s. It is interesting to note that 
if the logarithms of the pressures be plotted vertically and logarithms of volumes 
horizontally as in Fig. 7, then the line AC equal to the intercept on the horizontal 
axis represents the difference between the logarithms of volumes or, 

cZ=log Va-'og Vb, 



WORK AND POWER 



21 



and similarly 






CB = logPb- logP, 


Hence 






CB ■ 

s = = tan a, 

CA 



or the slope of the line indicates the value of s. This is a particularly valuable 
method, as it indicates at a glance the constancy or variability of s, and there 
are many cases of practice where s does vary. Should s be constant the line 
will be straight; should it be variable the line will be curved, but can generally 





B' 






\ 










3SVa~ 


Lo-Vb 








kT~ 




L 


l 


B 






.8 
.0 
.4 
.2 




; 


> 










S 


=tana 


BC 
~CA 




C' 


f 


( 




C 


CX { 




A 










Q? 

$ 










\ 










1 

£2 

Q. 




















if 

i-3 



















.0 .2 A .6 .8 1. 

Log. V 

Fig. 7. — Graphic Method of Finding s, from Logarithms of Pressures and Volumes. 



be divided into parts, each of which is substantially straight and each will 
have a different s. It is sometimes most convenient to take only the beginning 
and end of the curve and to use the value of s corresponding to these points, 
neglecting intermediate values. 

A second method for finding s for a given compression or expansion line 
by means of areas is indicated in a note in Section 17 of this Chapter that 
is omitted here because it depends on formulas not yet derived. It is by this 
sort of study of experimental data that most of the valuable values of s have 
been obtained. There is, however, another method of finding a value for 
s by purely thermodynamic analysis based on certain fundamental hypo- 
theses, and the value is as useful as the hypotheses are fair or true to the 
facts of a particular case. 

One of the most common hypotheses of this sort is that the gas or vapor 
undergoing expansion or compression shall neither receive any heat from, 
nor give up any to bodies external to itself during the process, and such a process 
is given the name adiabatic. Whether adiabatic processes are possible in actual 
cylinders or nozzles does not affect the analysis with which pure thermody- 
namics is concerned. By certain mathematical transformations, to be carried 
out later, and based on a fundamental thermodynamic proposition, the adia- 



22 ENGINEEEING THEKMODYNAMICS 

batic hypothesis will lead to a value of s, the use of which gives results valuable 
as a basis of reference, and which when compared with an actual case will per- 
mit of a determination of how far the real case has departed from the adiabatic 
condition, and how much heat has been received or lost at any part of the 
process. The particular value of s which exists in an adiabatic change is repre- 
sented by the symbol y. 

Another common hypothesis on which another value of s can be derived, 
is that gases in expansion or compression shall remain at a constant temperature, 
thus giving rise to the name isothermal. This is generally confined to gases 
and superheated vapors, as it is difficult to conceive of a case of isothermal or con- 
stant temperature expansion or compression of wet vapors, as will be seen later. 

In the study of vapors, which, it must be understood, may be dry or wet, 
that is, containing liquid, a common hypothesis is that during the expansion or 
compression they shall remain just barely dry or that they shall receive or lose 
just enough heat to keep any vapor from condensing, or but no more than 
sufficient to keep any moisture that tends to form always evaporated. Expan- 
sion or compression according to this hypothesis is said to follow the saturation 
law, and the substance to remain saturated. It will appear from this thermal 
analysis later that the value of s for the isothermal hypothesis is the same for all 
gases and equal to one, but for the adiabatic hypothesis s=y will have a 
different value for different substances, though several may have the same 
value, while for vapors y will be found to be a variable for any one, its value 
depending not only on the substance, but on the temperatures, pressures and 
wetness. 

When gases or vapors are suffered to expand in cylinders and nozzles or 
caused to compress, it is often difficult and sometimes impossible or perhaps 
undesirable to avoid interference with the adiabatic conditions for vapors 
and gases, with the isothermal for gases or with the saturation law for vapors, 
yet the work to be done and the horse-power developed cannot be predicted 
without a known value of s, which for such cases must be found by experi- 
ence. A frequent cause of interference with these predictions, which should 
be noted, is leakage in cylinders, which, of course, causes the mass under 
treatment to vary. 

According to these methods those values of s have been found which are 
given in Table X, at the end of the Chapter. Mixtures of common gases such 
as constitute natural, producer, - blast furnace or illuminating gas, alone or 
with air or products of combustion, such as used in internal combustion engines, 
have values of s that can be calculated from the elementary gases or measured 
under actual conditions. 

All vapors, except those considerably overheated, have variable exponents 
for adiabatic expansion and compression. This fact makes the exact solution 
of problems of work for wet vapors, expanding or compressing, which form the 
bulk of the practical cases, impossible by such methods as have been described. 
This class of cases can be treated with precision only by strictly thermal 
methods, to be described later. 



WORK AND POWER 



23 



Prob. 1. By plotting the values for the logarithms of the following pressures and vol- 
umes see if the value for s is constant, and if not find the mean value in each case. 



V 
10 

11 

12 



11 
12 



(a) Gas Engine Compression 

p v v y 

45.2 13 32.2 . 18 

39.7 14 29.7 20 

35.7 16 24.7 25 



(6) Gas Engine Expansion 

p V p 

13 146.2 

188.2 15 116.7 

166.2 17 65.7 



V 
19 
21 
23 



V 
21.0 
19.5 
14.7 



V 

80.7 
68.7 
58.7 



V 
2.242 
2.994 
4.556 



(c) Steam Expansion 
p V p 



203.3 

145.8 

89.9 



7.338 52.5 
12.44 28.8 

22.68 14.7 



Prob. 2. By plotting the values for the logarithms of the volumes and pressures on 
the expansion and compression curves of the following cards, find value for s. * 



20- 



Atrhosphere 




A-tmosphere 



24 



ENGINEERING THERMODYNAMICS 



140 
120 
100 



40 




Atmosphere 




Atmosphere 



Prob. 3. From the steam tables at the end of Chapter IV. select the pressures and 
volumes for dry-saturated steam and find the value of s between 

(a) 150 lbs. per square inch and 1 lb. per square inch. 
(6) 15 
Prob. 4. Find for superheated steam at 150 lbs. per square inch and with 100° 
of superheat expanding to 100 lbs. per square inch without losing any superheat, the 
corresponding value of s, using tabular data. 

Prob. 5. From the ammonia table data for dry-saturated vapor find the value of s 
between 

(a) 150 lbs. per square inch and 1 lb. per square inch. 

(b) 15 

9. Work Phases and Cycles, Positive and Negative and Net Work. Accord- 
ing to the preceding it is easy to calculate or predict numerically the work of 
expansion or compression whenever the conditions are sufficiently definite to 
permit of the selection of the appropriate s. It very seldom happens, however, 
that the most important processes are single processes or that the work of 
expansion or compression is of interest by itself. For example, before expansion 
can begin in a steam cylinder steam must be first admitted, and in air com- 
pressors air must be drawn in before it can be compressed. Similarly, after 
expansion in a steam cylinder there must be an expulsion of used vapor before 
another admission and expansion can take place, while in the air compressor 
after compression the compressed air must be expelled before more can enter 
for treatment. The whole series of operations is a matter of more concern 
than any one alone, and must be treated as a whole. The effect can be most 
easily found by the summation of the separate effects, and this method of 
summation will be found of universal application. 

The whole series of processes taking place and involving pressure volume 
changes is called a cycle, any one of them a phase. It is apparent that there 
can be only a limited number of phases so definite as to permit of the mathe- 



WOEK AND POWER 25 

matical treatment necessary for prediction of work, but it is equally clear that 
there may be a far greater number of combinations of phases constituting 
cycles. Before proceeding to analyze the action of steam or gas in a cylinder 
it is necessary first to determine on structural, thermal or any other logical 
grounds, what series of separate processes will be involved, in what order, and 
the pressure volume characteristics of each. Then and then only, can the 
cycle as a whole be treated. These phases or separate and characteristic proc- 
esses affecting the work done or involving pressure volume changes are divisible 
into two classes so far as the causes producing them are concerned, the first 
thermal and the second mechanical. It requires no particular knowledge of 
thermodynamics to realize that if air be confined in a cylinder with a free piston 
and is heated, that the volume will increase while pressure remains constant, 
since the piston will move out with the slightest excess of pressure inside over 
what is outside. This is a pressure constant, volume increasing, phase, and 
is thermal since it is a heat effect. If an ample supply of steam be available 
from a boiler held at a constant pressure by the manipulation of dampers and 
fires by the fireman and the steam be admitted to a cylinder with a piston, 
the piston will move out, the pressure remaining constant and volume increas- 
ing. This is also a pressure constant, volume increasing phase, exactly as before, 
but is mechanical because it is due to a transportation of steam from the boiler 
to the cylinder, although in another sense it may be considered as thermal if 
the boiler, pipe and cylinder be considered as one part during the admission. 
A similar constant pressure phase will result when a compressor piston is 
forcibly drawn out, slightly reducing the pressure and permitting the outside 
atmosphere to push air in, to follow the piston, and again after compression of 
air to a slight excess, the opening of valves to storage tanks or pipe lines having 
a constant pressure will allow the air to flow out or be pushed out of the cylinder 
at constant pressure. These two constant pressure phases are strictly mechan- 
ical, as both represent transmission of the mass. If a cylinder contain water 
and heat be applied without permitting any piston movement, there will be 
a rise of pressure at constant volume, a similar constant volume pressure rise 
phase will result from the heating of a contained mass of gas or vapor under the 
same circumstances, both of these being strictly thermal. 

However much the causes of the various characteristic phases may differ, 
the work effects of similar ones is the same and at present only work effects 
are under consideration. For example, all constant volume phases do no 
work as work cannot be done without change of volume. 

The consideration of the strictly thermal phases is one of the principal 
problems of thermodynamics, for by this means the relation between the work 
done to the heat necessary to produce the phase changes is established, and a 
basis laid for determining the ratio of work to heat, or efficiency. For the 
present it is sufficient to note that the work effects of any phase will depend 
only on the pressure volume changes which characterize it. 

Consider a cycle Fig. 8, consisting of (AB), admission of 2 cu.ft of steam at 
a constant pressure of 200,000 lbs. per square foot, to a cylinder originally 



26 



ENGINEEEING THERMODYNAMICS 



containing nothing, followed by (BC), expansion with s = l, to a pressure of 
20,000 lbs. per square foot; (CF), constant volume change of pressure, and 
(FG), constant pressure exhaust at 10,000 lbs. per square foot. These opera- 
tions are plotted to scale. Starting at zero volume, because the cylinder 



200000 




4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

Volumes in Cubic Feet 

Fig. 8.— Analysis of Work Diagram for Admission Expansion and Exhaust of Engine without 

Clearance. 

originally contains nothing, and at a pressure of 200,000 lbs. per square foot, 
the line AB, ending at volume 2 cu.ft,, represents admission and the cross- 
hatched area under AB represents the 400,000 ft.-lbs. of work done during 
admission. At B the admission ceases by closure of a valve and the 2 cu.ft. 
of steam at the original pressure expands with lowering pressure according 
to the law 



So that when 



PaV a = P b V b = 200,000X2 = 400,000 ft.-lbs., 

7=4 cu.ft, p= ^5^= 100,000 lbs> per sqft . 



7 = 5 cu.ft.. p = *9MW = 80,000 lbs. per sq.ft.; 
7 = 10 cu.ft, P = 4Q ^ QQQ = 40,000 lbs. per sq.ft. 



This continues until 7 = 20 at point C, at which time P= 4Q ^ Q0 = 20,000 

lbs. per square foot, and the work done during expansion is the cross-hatched 
area JBCD under the expansion curve BC, the value of which can be found by 
measuring the diagram or by using the formula Eq. (12), 

Wbc = PbV b \0ge~> 
Vb 



WORK AND POWER 27 

which on substitution gives 

Wbc = 400,000 loge 10 = 400,000X2.3; 
= 920,000 ft.-lbs. 

This completes the stroke and the work for the stroke can be found by addition 
of the numerical values, 

Wa b = 400,000 ft.-lbs.; 

Tf 6c = 920,000 ft.-lbs. ; 

Wab+Wbc=l,320,Q0O ft.-lbs. 

It is often more convenient to find an algebraic expression for the whole, 
which for this case will be, 

W ab = P a V a = P b V b ; 

W bc = P b V b \og e ~; 

Vb 

Wac = W ab +W be = P b V b (l+\og e ~j , 

= 400,000(l+log e 10) =400,000X3.3 = 1,320,000 ft.-lbs. 

On the return of the piston it encounters a resistance due to a constant pressure 
of 10,000 lbs. per square inch, opposing its motion; it must, therefore, do work 
on the steam in expelling it. Before the return stroke begins, however, the 
pressure drops by the opening of the exhaust valve from the terminal pressure 
of the expansion curve to the exhaust or back pressure along the constant volume 
line, CF, of course, doing no work, after which the return stroke begins, the 
pressure volume line being FG and the work of the stroke being represented by 
the cross-hatched area DFGH, 

Wta = PfVf = 10,000 X 20 = 200,000 ft.-lbs. 

This is negative work, as it is done in opposition to the movement of the piston. 
The cycle is completed by admission of steam at constant zero volume, raising 
the pressure along GA. The net work is the difference between the positive 
and negative work, or algebraically 

W=P b V b (l+logey)-PfVf, 

= 1,320,000 - 200,000 = l,120,00(f ft.-lbs. 



28 



ENGINEERING THERMODYNAMICS 



Consider now a cycle of an air compressor, Fig. 9. Admission or suction 
is represented by AB, compression by BC, delivery by CD and constant volume 
drop in pressure after delivery by DA. The work of admission is represented 
by the area ABFE or algebraically by 

Wab = PbVb, 

the work of compression by the area FBCG, or algebraically since s = 1.4 by 

1 



W bc = 



PhV, 



by b 



-1 



(sr-4 



n r A 


B 




r ^ : Woi-lt'of -X'dVniiion - - ±d 
































































































------ _ _. ... „ . . .. -J-. .. _ . 










^ 6000 3 - s s- 


- --- =t=..i============» 














m 4500 3 :: 
















co 3000 " 3 - If 


:..:.: ::..:: !~: _. .. : :::::±: 


































E ::._ :::::± :_:__: iii : : :±_ .:...: 


Mi!!!! 1 b. L. 



1 2 



3 t 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

Volumes in Cubic Feet 



Fig. 9. — Analysis of Work Diagram for Admission Compression and Delivery of Compressor 

without Clearance. 

the work of delivery by the area CDEG, or algebraically 

Wca = PcVc 

The positive work is that assisting the motion of the piston during suction; 
the area ABFE or algebraically PbVb. The negative work, that in opposi- 
tion to the motion, is the sum of the compression and delivery work, the area 
FBCDE, or algebraically, 



w, + w.=p c y c+ ^[(;y -i]. 



The net work is the difference and is negative, as such a cycle is mainly resistant, 
and to execute it the piston must be driven with expenditure of work on the 
gas. The value of the net work is, 

W=W bc +W cd -Wa h 



-*".^[(£f-> 



fflf-*] 



P b V b , 



WORK AND POWER 



29 



an expression which will be simplified in the chapter on compressors. This 
net work is represented by the area ABCD, which is the area enclosed by the 
cycle itself independent of the axes of coordinates. 

It might seem from the two examples given as if net work could be 
obtained without the tedious problem of summation, and this is in a sense true 
if the cycle is plotted to scale or an algebraic expression be available, but 
these processes are practically equivalent to summation of phase results. It 
might also seem that the work area would always be that enclosed by the 
cycle, and this is true with a very important limitation, which enters when 
the cycle has loops. If, for example, as in Fig. 10, steam admitted A to B, 
expanded along BC to a pressure C, then on opening the exhaust the pressure 
instead of falling to the back pressure or exhaust line as in Fig. 8, would here 




1 2 3 4 5 6 7 8 9 10 11 12 18 14 15 16 17 18 19 20 

Volumes in Cubic Feet 
Fig. 10. — Analysis of Work Diagram for Engine with Over-expansion Negative Work Loop. 

rise along CD, as the back pressure is higher than the terminal expansion pres- 
sure, after which exhaust will take place at constant back pressure along DE. 
The forward stroke work is that under AB and BC or ABCEG, the return 
stroke work is the area DEGH and the net work is 

Area ABCEG -Area. DEGH. 

As the area HGECX is common to both terms of the difference, the net work 
may be set down as equal to 

Area ABXH-CDX. 



It may be set down then in general for looped cycles that the net work area 
is the difference between that of the two loops. If, however, the method laid 
down for the treatment of any cycle be adhered to there need not be any dis- 
tinction drawn between ordinary and looped cycles, that is, in finding the work 



30 



ENGINEERING THERMODYNAMICS 



of a cycle divide it into characteristic phases and group them into positive and 
negative, find the work for each and take the algebraic sum. 

Special cases of cycles and their characteristics for steam compressors and 
gas engine cylinders, as well as nozzle expansion, will be taken up later in more 
detail and will constitute the subject matter of the next two chapters. 

Example 1. Method of calculating Diagram, Fig. 8. 

Pa =200,000 lbs. per square foot. 
P b =P a 
Vd = Vr. " \ P c =20,000 " 

P/ = 10,000 " " 

Pe=Pf. 



Va 


=0 cu.ft. 


v b 


=2 


u 


Va 


= V C 


a 


V e 


= 2 


u 


S 


= 1 


4 



To obtain point C. 



PcV c =P h V h 



or 



v _P b V b _ 200,000 X2_ 9n 
Vc ~~PT~ 20,000 ■ ' 



/. F c =20 and P c = 20,000. 

Intermediate points B to C are obtained by assuming various pressures and 
finding the corresponding volumes as for V c . 

Example 2. Method of calculating Diagram, Fig. 9. 



Assumed data ■ 


'7 =Ocu.ft. ' 

7 6 =20 " 

v d =o ll 

s = lA 


• 




r P a =2116 lbs. per square foot. 

P*=Pa 

P c = 14,812 " 
,Pa=Pc 


) obtain point C, 








iW-^PcFc 1 - 4 o 


r V c = 


-Vr± 


W-MW 


^=7 
P b '' 


log 7 =.845 


and 


.715 


1 
X.845=log^V' 4 =.6105, 



or 



Therefore, 



F c =4.02. 



F c =4.02, and P c = 14,812. 



Intermediate values B to C may be found by assuming pressures and finding volumes cor- 
responding as for V c . 

Prob. 1. Steam at 150 lbs. per square inch absolute pressure is admitted into a cylin- 
der in which the volume is originally zero until the volume is 2 cu.ft., when the valve is 
closed and expansion begins and continues until the volume is 8 cu.ft., then exhaust 
valve opens and the pressure falls to 10 lbs. absolute and steam is entirely swept out. 
Draw the diagram and find the net work done. 



WORK AND POWER 31 

Prob. 2. A piston moving forward in a cylinder draws in 10 cuit. of C0 2 at a pressure 
of .9 of an atmosphere at sea level and then compresses it adiabatically until the pressure 
rises to 9 atmospheres and discharges it at constant pressure. Draw the diagram and 
find net work done. 

Prob. 3. A cylinder 18 ins. in diameter and 24 ins. piston stroke receives steam at 
100 lbs. per square inch absolute pressure for f of the stroke. It then expands to the 
end of the stroke and is exhausted at atmospheric pressure. Draw the diagram and 
find the H.P. if the engine makes 100 strokes per minute. 

Prob. 4. Two compressors without clearance each with a cylinder displacement of 
2 cu.ft. draw in air at 14 lbs. per square inch absolute and compress it to 80 lbs. per 
square inch absolute before delivery. Find the difference in H.P. per 1000 cu.ft. of 
free air per minute if one is compressing isothermally and the other adiabatically. 
Draw diagram for each case. 

Prob. 5. A quantity of air 5 cu.ft. in volume and at atmospheric pressure is compressed 
in a cylinder by the movement of a piston until the pressure is 50 lbs. per square inch 
gage. If the air be heated the pressure will rise, as in an explosion. In this case the 
piston remains stationary, while the air is heated until the pressure reaches 200 lbs. per 
square inch gage. It then expands adiabatically to the original volume when the 
pressure is reduced to atmosphere with no change in volume. Draw the diagram, and 
find the work done. 

Prob. 6. The Brayton cycle is oik 4 in which gas is compressed adiabatically and then, 
by the addition of heat, the gas is made to expand without change of pressure. Adi- 
abatic expansion then follows to original pressure and the cycle ends by decrease in volume 
to original amount without change of pressure. Draw such a cycle starting with 5 cu.ft. 
of air at atmospheric pressure, compressing to 4 atmospheres, expanding at constant 
pressure to 5 cu.ft., expanding adiabatically to original pressure and finally ending at 
original point. Find also, work done. 

Prob. 7. In the Ericsson cycle air is expanded at constant temperature, cooled at 
constant pressure, compressed at constant temperature and receives heat at constant 
volume. Draw a diagram for the case where 5 cu.ft. at atmospheric pressure are com- 
pressed to 1 cu.ft., heated until volume is 8 cu.ft., expanded to atmosphere and then 
cooled to original volume. Find the work. 

Prob. 8. In the Stirling cycle constant volume heating and cooling replace that at 
constant pressure in the Ericsson. Draw diagram starting with 5 cu.ft. and atmsopheric 
pressure compressing to 1 cu.ft. and then after allowing the pressure to double, expand 
to original volume and cool to atmosphere. Find the work. 

Prob. 9. The Joule cycle consists of adiabatic compression and expansion and con- 
stant pressure heating and cooling. Assuming data as in last problem draw the 
diagram and find the work. 

Prob. 10. The Carnot cycle consists of isothermal expansion, adiabatic expansion, 
isothermal compression and adiabatic compression. Draw the diagram for this cycle 
and find the work. 

10. Work Determination by Mean Effective Pressure. While the methods 
already described are useful for finding the work done in foot-pounds for a defined 
cycle with known pressure and volume limits, they are not, as a rule, convenient 
for the calculation of the work done in a cylinder of given dimensions. As 
work done can always be represented by an area, this area divided by its length 
will give its mean height. If the area be in foot-pounds with coordinates 



32 ENGINEERING THERMODYNAMICS 

pounds per square foot, and cubic feet, then the division of area in foot-pounds 
by length in cubic feet will give the mean height or the mean pressure in 
pounds per square foot. Again, dividing the work of the cycle into forward- 
stroke work and back-stroke work, or the respective foot-pound areas divided 
by the length of the diagram in cubic feet, will give the mean forward pressure 
and the mean back pressure. The difference between mean forward pressure 
and mean back pressure will give the mean effective pressure, or that average 
pressure which if maintained for one stroke would do the same work as the 
cycle no matter how many strokes the cycle itself may have required for its 
execution, which is very convenient considering the fact that most gas engines 
require four strokes to complete one cycle. The mean effective pressure may 
also be found directly from the enclosed cycle area, taking proper account of 
loops, as representative of net work by dividing this net work area by the length 
of the diagram in appropriate units. This method is especially convenient 
when the diagram is drawn to odd scales so that areas do not give foot-pounds 
directly, for no matter what the scale the mean height of the diagram, when 
multiplied by the pressure scale factor, represents the mean effective pressure. 
This mean height can always be found in inches for any scale of diagram by 
finding the area of the diagram in square inches and by dividing by the length 
in inches, and this mean height in inches multiplied by the scale of pressures 
in whatever units may be used will give the mean effective pressure in the same 
units. 

Mean pressures, forward, back or effective, are found and used in two general 
ways; first, algebraically, and second graphically and generally in this case 
from test records. By the first method, formulas, based on some assumed 
laws for the phases, can be found, and the mean effective pressure and its value 
predicted. This permits of the prediction of work that may be done by a given 
quantity of gas or vapor, or the work per cycle in a cylinder, or finally the horse- 
power of a machine, of which the cylinder is a part, operating at a given speed 
and all without any diagram measurement whatever. By the second method, 
a diagram of pressures in the cylinder at each point of the stroke can be obtained 
by the indicator, yielding information on the scale of pressures. The net work 
area measured in square inches, when divided by the length in inches, gives 
the mean height in inches, which, multiplied by the pressure scale per inch 
of height, gives the mean effective pressure in the same units, which are 
usually pounds per square inch in practice. 

As an example of the algebraic method of prediction, consider the cycle 
represented by Fig. 8. The forward work is represented by 

Forward work = P b Vb( l+log e -~ 

the length of the diagram representing the volume swept through in the per- 
formance of this work is V c , hence 

Mean forward pressure = — ~ ( 1 + log c ~ 

V c \ V e 



WORK AND POWER 33 

But PbVb = P c V c by the law of this particular expansion curve, hence 
Mean forward pressure = P C ( l+log e ~ j. 

As the back pressure is constant its mean value is this constant value, hence 

Constant (mean) back pressure = Pf. 
By subtraction 

Mean effective pressure = P C ( l+log e -^ ) — P f 

= 3.3P C -P f ; 

= 3.3X20,000-10,000; 

= 66,000-10,000 = 56,000 lbs. per sq.ft. 

The work done in foot-pounds is the mean effective pressure in pounds per 
square foot, multiplied by the displacement in cubic feet. 

17 = 56,000X20 = 1,120,000 ft.-lbs. as before. 
D 




Fig. 1-1. — Gas-Engine Indicator Card. For Determination of Mean Effective Pressure 

without Volume Scale. 

As an example of the determination of mean effective pressure from a test 
or indicator diagram of unknown scale except for pressures, and without axes 
of coordinates, consider Fig. 11, which represents a gas engine cycle in four 
strokes, the precise significance of the lines being immaterial now. The 
pressure scale is 180 lbs. per square inch, per inch of height. 

By measurement of the areas in square inches it is found that 

Large loop area CDEXC =2.6 sq.in. 

Small loop area ABXA =0.5 sq.in. 

Net cycle area =2.1 sq.in. 

Length of diagram =3.5 in. 

Mean height of net work cycle = 0.6 in. 

Mean effective pressure = 120 X .06 = 72 lbs. per square inch. 



34 ENGINEERING THERMODYNAMICS 

It is quite immaterial whether this diagram were obtained from a large or 
a small cylinder; no matter what the size, the same diagram might be secured 
and truly represent the pressure volume changes therein. If this particular 
cylinder happened to have a diameter of 10 ins. and a stroke of 12 ins. the 
work per stroke can be found. The area of the cylinder will be 78.54 sq.ins., 
hence the average force on the piston is 72 lbs. per square inch X 78.54 sq.ins. = 
5654.88 lbs., and the stroke being 1 ft. the work per stroke is 5654.88 ft.-lbs. 
Both of these methods are used in practical work and that one is adopted in 
any particular case which will yield results by the least labor. 

Prob. 1. An indicator card from an air compressor is found to have an area of 3.11 
sq. ins., while the length is 2\ ins. and scale of spring is given as 80 lbs. per square inch 
per inch height. What is m.e.p. and what would be the horse-power if the compressor 
ran with a piston speed of 250 ft. per minute and had a piston 9 ins. in diameter? 

Prob. 2. For the same machine another card was taken with a 60-lb. spring and had 
an area of 4.12 sq.ins. How does this compare with first card, the two having the same 
length? 

Prob. 3. A steam engine having a cylinder 18 ins. in diameter and a stroke of 24 ins., 
takes in \ cu.ft. of steam at 100 lbs. absolute, allows it to expand and exhausts at atmos- 
pheric pressure. An indicator card taken from the same engine showed a length of 3 
ins., an area of .91 sq.in. when an 80-lb. spring is used. How does the actual m.e.p. 
compare with the computed? 

Prob. 4. Find m.e.p. by the algebraic method of prediction for, 

(a) Brayton cycle; 

(b) Carnot cycle; 

(c) Stirling cycle; 

(d) Ericsson cycle; 

(e) Joule cycle. 

(See problems following Section 9). 

11. Relation of Pressure-Volume Diagrams to Indicator Cards. The 
Indicator. When a work cycle or diagram of pressure volume changes is drawn 
to scale with pressures and volumes as coordinates, it is termed a pressure 
volume or PV diagram, and may be obtained by plotting point by point from 
the algebraic expression for the law of each phase or by modifying the indicator 
card. The indicator card is that diagram of pressures and stroke obtained 
by applying the indicator to a cylinder in operation. This instrument consists 
essentially of a small cylinder in which a finely finished piston moves freely 
without appreciable friction, with a spring to oppose its motion, a pencil mechan- 
ism to record the extent of the motion, and a drum carrying paper which is moved 
in proportion to the engine piston movement. The indicator cylinder is open 
at the bottom and fitted with a ground union joint for attachment to the main 
cylinder through a special cock, which when open permits all the varying 
pressures in the main cylinder to act on the indicator piston, and when closed 
to the main cylinder opens the indicator cylinder to the atmosphere. The 



WORK AND POWER 35 

upper side of the indicator piston being always open to the atmosphere, its 
movement will be the result of the difference between the pressure in the main 
cylinder and atmospheric pressure. A helical spring, carefully calibrated and, 
therefore, of known scale, is fixed between the indicator piston and open cap or 
head of its cylinder, so that whenever the pressure in the main cylinder exceeds 
atmosphere the indicator piston moves toward the open head of the indicator 
cylinder, compressing the spring. Pressures in the main cylinder if less than 
atmosphere will cause the indicator piston to move the other way, extending 
the spring. This compression and extension of the spring is found in the 
calibration of the spring to correspond to a definite number of pounds per 
square inch above or below atmosphere per inch of spring distortion, so that 
the extent of the piston movement measures the pressure above or below atmos- 
phere. A piston rod projects outward through the cylinder cap and moves a 
series of levers and links carrying a pencil point, the object of the linkage being 
to multiply the piston movement, but in direct proportion, giving a large 
movement to the pencil for a small piston movement. A cylinder drum carry- 
ing a sheet of paper is pivoted to the cylinder frame so that the pencil move- 
ment will draw on the paper a straight line parallel to the axis of the drum, if 
drum is stationary, or perpendicular to it if drum rotates and pencil is sta- 
tionary. The height of such lines then above or below a zero or datum line, 
which is the atmospheric line drawn with the cock closed, measures the pressure 
of the fluid under study. The springs have scale numbers which give the 
pressure in pounds per square inch per inch of pencil movement. This paper- 
carrying drum is not fixed, but arranged to rotate about its axis, being pulled 
out by a cord attached to the piston or some connecting part through a pro- 
portional reducing motion so as to draw out the cord an amount slightly less 
than the circumference of the drum no matter what the piston movement. 
After having been thus drawn out a coiled spring inside the drum draws it back 
on the return stroke. By this mechanism it is clear that, due to the combined 
movement of the pencil up and down, in proportion to the pressure, and that 
of the drum and paper across the pencil in proportion to the piston movement, 
a diagram will be drawn whose ordinates represent pressures above and below 
atmosphere and abscissae, piston stroke completed at the same time, or dis- 
placement volume swept through. It must be clearly understood that such 
indicator diagrams or cards do not give the true or absolute pressures nor the 
true volumes of steam or gas in the cylinder, but only the pressures above or 
below atmosphere and the changes of volume of the fluid corresponding to the 
piston movement. Of course, if there is no gas or steam in the cylinder at the 
beginning of the stroke, the true volume of the fluid will be always equal to the 
displacement, but no such cylinder can be made. 

While the indicator card is sufficient for the determination of mean effective 
pressure and work per stroke, its lack of axes of coordinates of pressure and 
volume prevents any study of the laws of its curves. That such study is 
important must be clear, for without it no data or constants such as the exponent 
6* can be obtained for prediction of results in other similar cases, nor can the 



36 



ENGINEERING THERMODYNAMICS 






presence of leaks be detected, or the gain or loss of heat during the various 
processes studied. In short, the most valuable analysis of the operations is 
impossible. 

To convert the indicator card, which is only a diagram of stroke or displace- 
ment on which are shown pressures above and below atmosphere into a pres- 
sure volume diagram, there must first be found (a) the relation of true or abso- 
lute pressures to gage pressures, which involves the pressure equivalent of 
the barometer, and (6) the relation of displacement volumes to true volumes 
of vapor or gas present, which involves the clearance or inactive volume of the 
cylinder. The conversion of gage to absolute pressures by the barometer 
reading has already been explained, Section 3, while the conversion of displace- 
ment volumes to true fluid volumes is made by adding to the displacement 
volume the constant value in the same units of the clearance, which is usually 
the result of irregularity of form at the cylinder ends dictated by structural 
necessities of valves, and of linear clearance or free distance between the pis- 
ton at the end of its stroke and the heads of the cylinder to avoid any possi- 
bility of touching due to wear or looseness of the bearings. 




Fig. 12. — Ammonia Compressor Indicator Card with Coordinates of Pressures and Volumes 
Added to Convert it into a Pressure- Volume Diagram. 

Let ABCD, Fig. 12, represent an indicator card from an ammonia compressor 
on which EF is the atmospheric line. The cylinder bore is 14 ins., stroke 
22 ins., and the scale of the indicator spring 100, barometer 28 ins., and measured 
clearance 32 cu.in. According to Table IX, 28 ins. of mercury corresponds to 
13.753 lbs. per square inch, and as 100 lbs. per square inch, according to the spring 
scale, corresponds to 1 in. of height on the diagram, 1 lb. per square inch cor- 
responds to 0.01 in. of height, or 13.75 lbs. per square inch atmospheric pres- 
sure to .137 in. of height. The zero of pressures then on the diagram must 
lie .137 in. below the line EF. Lay off then a line MH, this distance below 
EF. This will be the position of the axis of volume coordinates. 

Actual measurement of the space in the cylinder with the piston at the end 
of its stroke gave the clearance volume of 32 cu. ins. As the bore is 14 
ins. the piston area is 153.94 sq. ins. which in connection with the stroke 



WORK AND POWER 37 

of 22 ins. gives a displacement volume of 22X153.94-3386.68 cu. ins. 

32 
Compared with this the clearance volume is fi - = .94 per cent of the 

displacement. It should be noted here that clearance is generally expressed 
in per cent of displacement volume. Just touching the diagram at the ends 
drop two lines at right angles to the atmospheric line intersecting the axis of 
volumes previously found at G and H. The intercept GH then represents the 
displacement, or 3386.68 cu.in. or 1.96 cu.ft. Lay off to the left of G, .0094, or 
in round numbers 1/100 of GH, fixing the point M, MG representing the clearance 
to scale, and a vertical through M the axis of pressures. The axes of coordinates 
are now placed to scale with the diagram but no scale marked thereon. The 
pressure scale can be laid off by starting at M and marking off inch points 
each representing 100 lbs. per square inch. Pounds per square foot can also 
be marked by a separate scale 144 times as large. As the length of the diagram 
is 2.94 ins. and displacement 1.96 cu.ft., 1 in. of horizontal distance corresponds 
to .667 cu.ft. or 1 cu.ft to 1.50 ins. of distance. Lay off then from M dis- 
tances of 1.50 ins. and mark the first 1 cu.ft. and the second 2 cu.ft., dividing 
the intervals into fractions. A similar scale of volumes in cubic inches might 
also be obtained. 

By this process any indicator card may be converted into a pressure volume 
diagram for study and analysis, but there will always be required the two factors 
of true atmospheric pressure to find one axis of coordinates and the clearance 
volume to find the other. 

Prob. 1. If in cards Nos. 1 and 2, Section 8, the clearances are 5 per cent and 3 per 
cent respectively of the displacement, convert the cards to PV diagrams on the same 
base to scales of 4 ins. to 1 cu.ft. and 1 in. to 1000 lbs. per square foot, for cylinders 
9 j ins. and 14 1 ins. respectively in diameter and stroke 12 ins. 

Prob. 2. Do the same for cards Nos. 3 and 4, if clearances are 7 per cent and 4 per 
cent respectively, for cylinders 10 ins. and 17 ins. respectively in diameter and stroke 
12 ins. 

Prob. 3. Do the same for cards Nos. 7 and 8 if clearances are 12 per cent and 8 per 
cent respectively and cylinders 12 ins. and 19 ins. in diameter and stroke 24 ins. 

12. To Find the Clearance. There are two general methods for the find- 
ing of clearance, the first a direct volumetric measurement of the space itself 
by filling with measured liquid and the second a determination by algebraic 
or graphic means from the location of two points on the expansion or com- 
pression curves of the indicator card based on an assumed law for the curves. 

The first method of direct measurement is the only one that offers even a 
promise of accuracy, but even this is difficult to carry out because of the 
tendency of the measuring liquid to leak past piston or valves, which makes 
the result too large if the liquid be measured before the filling of the clearance 
space and too small if the liquid be measured after filling and drawing off. 
There is also a tendency in the latter case for some of the liquid to remain 
inside the space, besides the possibility in all cases of the failure to completely 
fill the space due to air pockets at high places. 



.100 




























A" 


A' 












V 












80 


a'a 


A 












\ 








































B" 


% 
















\b 








00 


B" 


B\E 
















\ 


\ 






40 




1 



















































10 



Displacement 
Fig. 13. 





































A" A 










\ 






















A" 


MA 








\ 






















































B" 


B'\ 














SB 
















B" 


B' \ 


B 


























10 






\ 



























































10 12 



Displacement 
Fig. 14. 



A" 








a\a 












































































B" 








B' 










-.B 






































A" 








A^^ 


^A 


























t 






B 








B 










1 


B 


























[_ 







30 



I 

S 20 

.2 



10 



1 2 

Displacement 

Fig. 15. 

Diagrams Illustrating Location of Clearance Line from Expansion or Compression Lines 

of Known Laws. 



WORK AND POWER 39 

By the second general method any two points, A and B, on an expansion or 
compression curve, Figs. 13, 14, 15, may be selected and horizontals drawn to 
the vertical line indicating the beginning of the stroke. The points A' and B' 
are distant from the unlocated axis an amount A'A" = B'B", representing the 
clearance. 

Let the clearance volume = CI. ; 
" the displacement up to A = D a ; 
" the displacement up to B = D b ; 
" the whole displacement =D; 
" s be the exponent in PV* = constant, which defines the law of the curve. 

Then in general, 

P a V a S = PbVb S . 

But 

V a = D a +Cl, 

and 

Vb = Db+Cl, 
hence 

PJ (Da+Cl) =P b hDb+Cl), 
or 

CI \Pa s -Pv)= P) D b - P) D a , 

whence the clearance in whatever units the displacement may be measured will be 

ni _ PbsD b -PgsD a 

iV -Pv 
or 






P b s 

and CI, in per cent of the whole displacement will be 

i 

\Pj D 



Db /Pa\s Do 



Clearance as a fraction of displacement =c 



When s= 1 this takes the form 



( 


pj 


A, 
D 


\pJd 



Clearance in fraction of displacement = c = ^— — . . . (15) 



40 ENGINEERING THERMODYNAMICS 

To use such an expression it is only necessary to measure off the atmospheric 
pressure below the atmospheric line, draw verticals at ends of the diagram 
and use the length of the horizontals and verticals to the points in the formula, 
each horizontal representing one D and each vertical a P. 

Graphic methods for the location of the axis of pressures, and hence the 
clearance, depend on the properties of the curves as derived from analytics. 
For example, when s= 1, 

PaVa = PbVb, 

which is the equation of the equilateral hyperbola, a fact that gives a common 
name to the law, i.e., hyperbolic expansion or hyperbolic compression. Two 
common characteristics of this curve may be used either separately or together, 
the proof of which need not be given here, first that the diagonal of the rectangle 
having two opposite corners on the curve when drawn through the other two 
corners will pass through the origin of coordinates, and second, that the other 
diagonal drawn through two points of the curve and extended to intersect the 
axes of coordinates will have equal intercepts between each point and the 
nearest axis cut. 

According to the first principle, lay off, Fig. 16, the vacuum line or axis 
of volume XY and selecting any two points A and B, construct the rectangle 
ACBD. Draw the diagonal CDE and erect at E the axis of pressures EZ, 
then will EZ and EY be the axes of coordinates. According to the second 
principle, proceed as before to locate the axes of volumes XY and select two 
points, A and B, Fig. 17. Draw a straight line through these points, which 
represents the other diagonal of the rectangle ACBD, producing it to inter- 
sect XY at M and lay off AN = lfM. Then will the vertical NE be the axis 
of pressures. It should be noted that these two graphic methods apply only 
when s = l; other methods must be used when s is not equal to 1. 

A method of finding the axis of zero volume is based upon the slope of the 
exponential curve, 

PV s =c. 

Differentiation with respect to V gives 



p,y.-i+7«g=0, 



or 

Ps MP 



whence 



+ dV °' 



F= *> P 



'(-Si (16) 

\ dV) 



WORK AND POWER 



41 



In other words, the true volume at any given point on the known curve 
may be found by dividing the product of P and s by the tangent or the slope 
of the line at the given point, with the sign changed. This method gives 
results dependent for their accuracy upon the determination of the tangent to 
the curve, which is sometimes difficult. 





Z 






























70 




\ 




























60 




\ 




























50 
to 

E 

3 40 




\ 






























\ 


A 




C 






















^30 


\ 


/ 








> 


/ 
























20 




4 


/ 


N 


B 












Atmospheric 


Lino 
















/ 




























10 

X o 


/ 


^ — 


























Y 








1 




3 




"> 








) 


11 


1 


a 



Displacement 
Fig. 16. 




5 7 

Displacement 

Fig. 17. 

Graphic Methods of Locating the Clearance Line for Logarithmic or Hyperbolic Expansion 

and Compression Curves. 



The following graphical solution is dependent upon the principle just given, 
and while not mathematically exact, gives results so near correct that the 
error is not easily measured. The curve ACB, Fig. 18, is first known experi- 
mentally or otherwise and therefore the value of s, and the axis FV from 
which pressures are measured is located. Assume that the axis of zero volume, 



42 



ENGINEERING THERMODYNAMICS 



KP, is not known but must be found. Selecting any two convenient points, 
A and B, on the curve, complete the rectangle AHBG with sides parallel and 
perpendicular to FV. The diagonal HG cuts the curve at C and the horizontal 
axis at E. From C drop the perpendicular CD. If now the distance DE 
be multiplied by the exponent s, and laid off DK, and the vertical KP erected, 
this may be taken as the zero volume axis. 

It cannot be too strongly stated that methods for the finding of clearance 
or the location of the axes of pressures from the indicator card, much as they 
have been used in practice, are inaccurate and practically useless unless it is 
positively known beforehand just what value s has, since the assumed value 









\ 




















PI 






\ 


























\ 


Jk 






H 
























/ 
/ 
/ 
/ 
/ 
























/ 
/ 

k / 


/ 






















/ 


Vc 
i \ 






















/ 


£ 


i 
i 

.j. 

i 




B 
















v 


/ 

/p( 


d? } 


i 
j 




















/P( 








1 / 
j / 


<:P( 


d?' 


i 

i 

>! 

ID 
















F ] 


K 


*'/ 


i>r\ 





Fig. 18. — Graphic Method of Locating the Clearance Line for Exponential Expansion and 

Compression Curves. 

of s enters into the work, and s for the actual diagram, as already explained, 
is affected by the substance, leakage, by moisture or wetness of vapor and by 
all heat interchange or exchange between the gas of vapor and its container. 



Prob. 1. If in card No. 6, Section 8, compression follows the law PV S =K, 
where s = 1.4 and the barometer reads 29.9 ins. of mercury, locate the axes algebraically 
and graphically. 

Prob. 2. If in card No. 3, Section 8, expansion follows the law PV S =K, 
where s = 1.37 and the barometer reads 29.8 ins. of mercury, locate the axes algebraically 
and graphically. 

Prob. 3. If in card No. 5, Section 8, expansion follows the law PV S =K, 
where s = l and the barometer reads 27.5 ins. of mercury, locate axes algebraically and 
graphically. 



WORK AND POWER 



43 



13. Measurement of Areas of PV Diagrams and Indicator Cards. Areas 
of pressure volume diagrams or indicator cards must be evaluated for the 
determination of work or mean effective pressure, except when calculation by 
formula and hypothesis is possible. There are two general methods applicable 
to both the indicator card and PV diagram, that of average heights, and the 
planimeter measure, besides a third approximate but very useful method, 
especially applicable to plotted curves on cross-section paper. 

The third method assumes that the diagram may be divided into strips of 
equal width as in Fig. 19, which is very easily done if the diagram is plotted on 
cross-section paper. At the end of each strip, a line is drawn perpendicular to 
the axis of the strip, such that the area intercepted inside the figure is apparently 
equal to that outside the figure. If this line is correctly located, the area of the 
rectangular strip will equal the area of the strip bounded by the irregular lines. 






60 



50 



■to 



so 



10 







> 


X 






\ 
















"^ 




J 


7* 








\ 


\ 














/ 












"\ 












1 


























1 


I 


























\ 




























1! 


1 t ' 




" 1 1 T T 



Volumes 
Fig. 19. — Approximate Method of Evaluating Areas and Mean_ Effective Pressures of 
Indicator Cards and P.V. Diagrams. 



If the entire figure has irregular ends it may be necessary to subdivide one or 
both ends into strips in the other direction, as is done at the left-hand side of 
Fig. 19. The area of the entire figure will be equal to the summation of lengths 
of all such strips, multiplied by the common width. This total length may be 
obtained by marking on the edge of a strip of paper the successive lengths in 
such a way that the total length of the strip of paper when measured will be the 
total length of the strips. 

The mean height will be the total length of such strips divided by the num- 
ber of strip-widths in the length of the diagram. By a little practice the proper 
location of the ends of the strips can be made with reasonable accuracy, and 
consequently the results of this method will be very nearly correct if care is 
exercised. 

By dividing the diagram into equal parts, usually ten, and finding the length 
of the middle of each strip, an approximation to the mean height of each strip 



44 



ENGINEEKING THERMODYNAMICS 



will be obtained ; these added together and divided by the number will give 
the mean heightin inches from which the mean effective pressure may be found 
by multiplying by the scale as above, or the area in square inches by multiply- 
ing by the length in inches, which can be converted into work bjr multiplying 
by the foot-pounds per square inch of area as fixed by the scales. As the pres- 
sures usually vary most, near the ends of the diagram a closer approximation 
can be made by subdividing the end strips, as is done in Fig. 20, which repre- 
sents two steam engine indicator cards taken from opposite ends of the same 
cylinder and superimposed. The two diagrams are divided into ten equal 
spaces and then each end space is subdivided. The mean heights of the sub- 
divisions are measured and averaged to get the mean height of the whole end 
division, or average pressure in this case for the division. The average heights 
of divisions for diagram No. 1 are set down in a column on the left, while those 



S75- 
a 

CO 

ISO- 



'S 
a 

& 

.5 25- 



P< 















><^ 










> 



































10 



30 40 50 60 70 

Displacement in Per Cent of Stroke 



80 



90 



100 



Fig. 20. — Simpson's Method for Finding Mean Effective Pressure of Indicator Cards. 



for No. 2 are on the right; the sum of each column divided by ten and multiplied 
by the spring scale gives the whole m.e.p. The heights of No. 1 in inches 
marked off continuously on a slip of paper measured a total of 11.16 ins. and 
for No. 2, 10.79 ins., these quantities divided by 10 (number of strips) and multi- 
plied by the spring scale, 50, gives the m.e.p., as before. This method is 
often designated as Simpson's rule. 

The best and most commonly used method of area evaluation, whether for 
work or m.e.p. determination, is the planimeter, a well-known instrument 
specially designed for direct measurements of area. 

14. Indicated Horse-power. Work done by the fluid in a cylinder, because it 
is most often determined by indicator card measurements, measures the indicated 
horse-power, but the term is also applicable to work that would be done by the 
execution of a certain cycle of pressure volume changes carried out at a specified 
rate. The mean effective pressure in pounds per square inch, whether of an 
indicator card or ?F cycle, when multiplied by piston area in square inches, 



WORK AND POWER 45 

gives the average force acting on the piston for one stroke, whether the cycle 
required one, two or x strokes for its execution, and this mean force multiplied 
by the stroke in feet gives the foot-pounds of work done by the cycle. Therefore, 

Let m.e.p.=mean effective pressure in pounds per square inch for the cycle 

referred to one stroke; 

a = effective area of piston in square inches; 

L = length of stroke in feet; 

n = number of equal cycles completed per minute; 

N = number of revolutions per minute; 

# = mean piston speed = 2LN feet per minute; 

N 
z — number of revolutions to complete one cycle = — . Then will the 



indicated horse-power be given by, 



(m.e.p. )Lan 
33000 

(m.e.p. )LaN 
~ 330002 

(m.e.p.)a$ 
33000 X2z' 



TUP _ ym.c.p.;^u,», ,v 

I.H.P.- 3300Q (a) 



(&) 

(c) 



(17) 



When there are many working chambers, whether in opposite ends of the 
same cylinder or in separate cylinders, the indicated horse-power of each should 
be found and the sum taken for that of the machine. This is important not 
only because the effective areas are often unequal, as, for example, in opposite 
ends of a double-acting cylinder with a piston rod passing through one side 
only or with two piston rods Or one piston rod and one tail rod of unequal 
diameters, but also because unequal valve settings which are most common 
will cause different pressure volume changes in the various chambers. 

It is frequently useful to find the horse-power per pound mean effective pressure, 
which may be symbolized by K e , and its value given by 

„ Lan LaN 



33000 33000^ 

Using this constant, which may be tabulated for various values of n, stroke 
and bore, the indicated horse-power is given by two factors, one involving 
cylinder dimensions and cyclic speed or machine characteristics, and the other 
the resultant PV characteristic, of the fluid, symbolically, 

I,H.P.=X e (m.e.p.). 

These tables of horse-power per pound m.e.p. are usually based on piston speed 
rather than rate of completion of cycles and are, therefore, directly applicable 



46 ENGINEERING THERMODYNAMICS 

when z = i or n = 2N, which means that the two cycles are completed in one 
revolution, in which case, 

S = 2LN = Ln, 

and 

= aS 
e 33000' 

whence 

I.H.P. = 2f e (m.e.p.) = v 33Q qq (18) 

Table XI at the end of this chapter gives values of (H.P. per lb. m.e.p.) or 
K e for tabulated diameters of piston in inches and piston speeds in feet per 
minute. Tables are frequently given for what is called the engine constant, 
which is variously denned as either 

(a) „^ nnn , which must be multiplied by m.e.p. Xn to obtain H.P. , or 

(b) , which must be multiplied by m.e.p. XL Xn to obtain H.P. 

For an engine which completes two cycles per revolution, this is the same as 
multiplying by m.e.p. X^S. Before using such a table of engine constants it 
must be known whether it is computed as in (a) or in (b). 

Example. A 9 in. Xl2 in. double-acting steam engine runs at 250 R.P.M. and the 
mean effective pressure is 30 lbs. What is H.P. per pound m.e.p. and the I.H.P.? 

K Lan = 1X63.6X500 . 
e 33000 33000 ' 

I.H.P. =.9636x30 =28.908. 

Prob. 1. A pump has a piston speed of 250 ft. per minute; piston diameter is 24 ins. 
What is the H.P. per pound mean effective pressure? 

Prob. 2. A simple single-acting 2-cylinder engine has a piston 10 ins. in diameter with 
a 2-in. rod and a stroke of 15 ins. It runs with a mean effective pressure of 45 lbs. 
per square inch at a speed of 220 R.P.M. What is the H.P.? 

Prob. 3. A gas engine has one working stroke in every four. If the speed is 300 
R.P.M. what must be the m.e.p. to give 6 H.P. when the cylinder has a diameter of 6 ins. 
and a stroke of 12 ins.? 

Prob. 4. An air compressor is found to have a mean effective pressure of 50 lbs. If 
the cylinder is double acting and is 12 ins. diameter and 16 ins. stroke, what H.P. will 
be needed to drive it at 80 R.P.M.? 

Prob. 5. A gasoline engine has an engine constant (a) of .3. What must be the 
m.e.p. to give 25 H.P.? 

Prob. 6. A blowing engine has an m.e.p. of 10 lbs. Its horse-power is 500. What 
is the H.P. per pound m.e.p.? 



WORK AND POWER 47 

Prob. 7. Two engines of the same size and speed are so run that one gives twice the 
power of the other. How will the engine constants and m.e.p. vary? 

Prob. 8. From the diagrams following Section 9 what must have been the H.P. per 
pound m.e.p. to give 300 H.P. in each case? 

Prob. 9. How will the H.P. per pound m.e.p. vary in two engines if the speed of 
one is twice that of the other, if the stroke is twice, if the diameter of piston is twice? 

15. Effective Horse-power, Brake Horse-power, Friction Horse-power, 
Mechanical Efficiency, Efficiency of Transmission, Thermal Efficiency. Work 
is done and power developed primarily in the power cylinder of engines, and is 
transmitted through the mechanism with friction loss to some point at which 
it is utilized. There is frequently a whole train of transmission which may 
involve transformation of the energy into other forms, but always with some 
losses, including the mechanical friction. For example, a steam cylinder may 
drive the engine mechanism which in turn drives a dynamo, which transforms 
mechanical into electrical energy and this is transmitted to a distance over 
wires and used in motors to hoist a cage in a mine or to drive electric cars. 
There is mechanical work done at the end of the system and at a certain rate, 
so that there will be a certain useful or effective horse-power output for the 
system, which may be compared to the horse-power primarily developed in 
the power cylinders. A similar comparison may be made between the primary 
power or input and the power left after deducting losses to any intermediate 
point in the system. For example, the electrical energy per minute delivered 
to the motor, or motor input, is, of course, the output of the transmission line. 
Again, the electrical energy delivered to the line, or electrical transmission 
input, is the same as dynamo output, and mechanical energy delivered to the 
dynamo is identical with engine output. The comparison of these measure- 
ments of power usually takes one of two forms, and frequently both; first, 
a comparison by differences, and second, a comparison by ratios. The ratio 
of any horse-power measurement in the system to the I. H.P. of the power cylin- 
der is the efficiency of the power system up to that point, the difference between 
the two is the horse-power loss up to that point. It should be noted that, 
as both, the dynamo and motor transform energy from mechanical to electrical 
or vice versa, the engine mechanism transmits mechanical energy and the 
wires electrical energy, the system is made up of parts which have the 
function of (a) transmission without change of form, and (b) transformation 
of form. The ratio of output to input is always an efficiency, so that the efficiency 
of the power system is the product of all the efficiencies of transformation and of 
transfer or transmission, and the power loss of the system is the sum of trans- 
formation and transmission losses. Some of these efficiencies and losses have 
received names which are generally accepted and the meaning of which is gen- 
erally understood by all, but it is equally important to note that others have no 
names, simply because there are not names enough to go around. In dealing 
with efficiencies and power losses that have accepted names these names may 
with reason be used, but in other cases where names are differently under- 
stood in different places or where there is no name, accurate description must be 



ENGINEERING THERMODYNAMICS 

relied on. As a matter of fact controversy should be avoided by definition 
of the quantity considered, whether descriptive names be used or not. 

Effective horse-power is a general term applied to the output of a machine, 
or power system, determined by the form of energy output. Thus, for an engine 
it is the power that might be absorbed by a friction brake applied to the shaft, 
and in this case is universally called Brake Horse-power. The difference between 
brake and indicated horse-power of engines is the friction horse-power of the 
engine and the ratio of brake to indicated horse-power is the mechanical efficiency 
of the engine. For an engine, then, the effective horse-power or useful horse- 
power is the brake horse-power. When the power cylinders drive in one machine 
a pump or an air compressor, the friction horse-power of the machine is the 
difference between the indicated horse-power of the power cylinders and that 
for the pump or compressor cylinders, and the mechanical efficiency is the 
ratio of pump or air cylinder indicated porse-power to indicated horse-power 
of the power cylinders. Whether the indicated horse-power of the air or pump 
cylinders can be considered a measure of useful output or not is a matter of 
difference of opinion. From one point of view the machine may be as considered 
built for doing work on water or on air, in which case these horse-powers may 
properly be considered as useful output. On the other hand, the power pump 
is more often considered as a machine for moving water, in which case the 
useful work is the product of the weight of water moved into its head in feet, 
and includes all friction through ports, passages and perhaps even in pipes or 
conduits, which the indicated horse-power of the pump cylinder does not include, 
especially when leakage or other causes combine to make the pump cylinder 
displacement differ from the volume of water actually moved. With compressors 
the situation is still more complicated, as the air compressor may be considered 
useful only when its discharged compressed air has performed work in a rock 
drill, hoist or other form of an engine, in which case all sorts of measures of use- 
ful output of the compressor may be devised, even, for example, as the purely 
hypothetically possible work derivable from the subsequent admission and 
complete expansion of the compressed air in a separate air engine cylinder. 

Too accurate a definition, then, of output and input energy in machines and 
power systems is not possible for avoidance of misunderstanding, which may 
affect questions both of power losses and efficiency of transmission and trans- 
formation whether in a power system or single machine. It is interesting to 
note here that not only is the indicated work of the power cylinder always con- 
sidered the measure of power input for the system or machine, but, as in the 
other cases, it is itself an output or result of the action of heat on the vapor or 
gas and of the cycle of operations carried out. The ratio of the indicated power 
or cylinder work, to the heat energy both in foot-pound units, that was expended 
on the fluid is the thermal efficiency of the engine referred to indicated horse- 
power or the efficiency of heat transformation into work, the analysis of which 
forms the bulk of the subject matter of Chapter VI. Similarly, the ratio of 
any power measurement in the system to the equivalent of the heat supplied 
is the thermal efficiency of so much of the system as is included. 



WORE AND POWER 49 

£ 

Example. It has been found that when the indicated horse-power of an engine is 
250, a generator is giving out 700 amperes at 220 volts. At end of a transmis ion line is a 
motor using the output of the generator. This motor on test gave out 180 brake horse- 
power. Assuming no losses in the transmission line, what was the efficiency of the 
motor, of the generator, of the engine, and of the system? 

Motor efficiency =^^ =^^ =87.2%. 
J Input 220x700 

746 
Note: Volts X amperes = watts, and, watts -4-746 =H.P. 

Engine and generator efficiency =■ — — — =82.4%. 

250 

ISO 
Efficiency of system = —-=72% or 82.4x87.2=72%. 
ZoO 

Prob. 1. An engine is belted to a pump; the I.H.P. of the engine is 50, of the pump 
40, and the pump delivers 1200 gallons water per minute against 100-ft. head. What is 
the efficiency of each part and of the entire system. 

Prob. 2. An engine is geared to air compressor. Upon test, the efficiency of the engine 
alone, gearing alone and compressor alone were each 80 per cent. When the com- 
pressor H.P. was 100 what was that of the engine? 

Prob. 3- A water-wheel is run by the discharge from a pump. The B.H.P. of wheel 
is found to be 20 when the pump is delivering 45 gallons of water per minute at a 
head of 1000 lbs. per square inch. The water I.H.P. of the pump is 30 and the 
steam I.H.P. is 40. What are the efficiencies of each part of the system and the over-all 
efficiency? 

Prob. 4. Perry gives a rule for the brake horse-power of steam engines as being 
equal to .95 I.H.P. —10. On this basis find the mechanical efficiency of a 500-H.P. 
engine from 200 to 500 H.P. Show results by a curve with B.H.P. and per cent efficiency 
as coordinates. 

Prob. 5. Perry gives a rule for the efficiency of an hydraulic line as H - .71 —25 where 
H is the useful power of the pump and / is the indicated. Find / for values of H from 
100 to 300 and plot a curve of results. 

Prob. 6. An engine gives one I.H.P. for every 3 lbs. of coal per hour. One pound 
of coal contains 9,500,000 ft.-lbs. of energy. What is the thermal efficiency? 

Prob. 7. A gas engine has a mechanical efficiency of 70 per cent when delivering 
power to a generator which in turn has an efficiency of 90 per cent. If the engine uses 
15 cu.ft. of gas per indicated horse-power hour and the gas contains 700,000 ft.-lbs. per 
cubic foot, what is the net thermal efficiency of the system? 

16. Specific Displacement, Quantity of Fluid per Hour or per Minute per 

I.H.P. It has been shown that the work done in cylinders by pressure volume 
changes of the vapor or gas depends on the mean effective pressure and on the 
displacement, or that there is a relation between I.H.P. and displacement. 
The quantity of fluid used also depends on the displacement and may be expressed 



50 ENGINEERING THERMODYNAMICS 

in cubic feet per minute at either the low pressure or high pressure con- 
dition when the work is done between two definite pressure limits, or in terms 
of pounds per minute or hour, which involves the application of fluid densities 
to volumes and which eliminates the double expression for the two conditions 
of pressure. The displacement per hour per horse-power, termed the specific 
displacement, is the basis of computations on the steam consumption of steam 
engines, the horse-power per cubic feet of free air per minute for air compressors, 
the horse-power per ton refrigeration for refrigerating machines and the con- 
sumption of fuel per hour per horse-power for gas and oil engines. It is, 
therefore, a quantity of great importance in view of these applications. Apply- 
ing the symbols already defined to displacement in one direction of one side of 
a piston 

Displacement in cu.ft. per stroke = -kXyrr; 
Displacement in cu.ft. per minute = LXjjtXN; 

Displacement in cu.ft. per hour =60LX7TjXiV. 

T ,. , . (m.e.p.)Lan (m.e.p.)LaiV 

Indicated horse-power = 33QQQ = ^^ . 

Whence expressing displacement per hour per I.H.P. or specific displacement 
in one direction for one side of a piston by D s , 

D = mLX T£i XN = 60X3S000z = 13750^ -. - 

s ( m.e.p.)Laiy 144(m.e.p.) (m.e.p.)' ) 

33000^ 

From Eq. (19) it appears that the specific displacement is equal to zX 13,7 '50 
divided by the mean effective pressure in pounds per square inch. 

If two points, A and B, be so located on the indicator card, Fig. 21, as to 
have included between them a fluid transfer phase, either admission to, or 
expulsion from the cylinder, then calling 8« = pounds per cubic foot or density 
at point A, and 8 6 = pounds per cubic foot or density at point B, the weight 
of fluid present at A is, 

(D a +CQd a lbs., 

and weight of fluid present at B is 

(D b +Cl)db lbs., 

whence the weight that has changed places or passed in and out per stroke is, 
(D b +Cl)db-(D a +Cl)d a lbs. per stroke. 



WORK AND POWER 



51 



If both A and B lie on the same horizontal as A and B', d a = d b = d } the 
density of fluid at the pressure of measurement, whence the weight of fluid 
used per stroke, will be 

(Db'~Da)d, 

and the volume per stroke used at density d is 

ZV-DaCU.ft., 

which compared to the displacement is 

D b '-Da 

D ' 



















p 


















\ 
















1 






\ s 


\ 








\ 
















\ 














. B' 


\ 












B' 


^ R 


\ 


V 














\ 


s 










) 




\ 


s 










> 



































Fig. 21. — Determination of Consumption of Fluid per Hour per Indicated Horse-power from 

the Indicator Card. 

This is the fraction of the displacement representing the volume of fluid pass- 
ing through the machine at the selected pressure. Multiplying the specific 
displacement by this, there results, 



Cu.ft. of fluid per hr. at density (d) per I.H.P. 



13750 D^-Da 



(m.e.p.) D 



and 



Lbs. of fluid per hr. per I.H.P 



13750 (D h -D a 



(20) 



(m.e.p.) \ D 
More generally, that is, when A and B are not taken at the same pressures 

Lbs. fluid per hr. per I.H.P. = ^J^d \(D b +Cl)d b - (D a +Cl)d a ^ . (21) 

The particular forms which this may take when applied to special cases will be 
examined in the succeeding chapters. 



52 ENGINEERING THERMODYNAMICS 

Example. An air compressor whose cylinder is 18x24 ins. (18 ins. in diameter and 
stroke 2 ft.) runs at 60 R.P.M. and is double acting. The mean effective air pressure 
is 50 lbs. per square inch. What is the specific displacement? 

Cu.ft. per hour = ^777^ = 60 X2 X ^? X120 =25,600. 
144 144 

t tt -p _m.e.p.Law _50 X2 X254.5 X120 _ QO Q 
LH - F ' -"33^000 "3^000 9Z3; 

Cu.ft. per hour 25,600 



I.H.P. 92.3 



=277, 



or by the formula directly, 

13,750 __ 13,750 _ 275 
m.e.p. 50 

Prob. 1. What will be the cubic feet of free air per hour per horse-power delivered 
by a 56x72-in. blowing engine with 4 per cent clearance and mean effective pressure of 
10 lbs. per square inch? 

Prob. 2. An 18x22-in. ammonia compressor works with a mean effective pressure 
of 45 lbs. per square inch. What is the weight of NH 3 per I.H.P. hour if the speed is 
50 R.P.M. and compressor is double acting having a volumetric efficiency of 90 per 
cent? Use tabular NH 3 densities. 

Prob. 3. A steam engine whose cylinder is 9x12 ins. runs at a speed of 300 R.P.M. 
and is double acting. If the m.e.p. is 60 lbs. and the density of steam at end of the 
stroke is .03, how many pounds of steam are used per hour per indicated horse-power? 

17. Velocity Due to Free Expansion by PV Method. All the cases examined 
Jor the work done by PV cycles have been so far applied only to their execution 
in cylinders, but the work may be developed in nozzles accelerating the gas 
or vapor in free expansion, giving, as a consequence, a high velocity to the fluid. 
It was noted that for cylinders many combinations of phases might be found 
worthy of consideration as typical of possible actual conditions of practice, 
but this is not true of proper nozzle expansion, which has but one cycle, that of 
Fig. 22. That this is the cycle in question is seen from the following considera- 
tions. Consider a definite quantity of the gas or vapor approaching the nozzle 
from a source of supply which is capable of maintaining the pressure. It pushes 
forward that in front of it and work will be done, ABCD, equal to the admission 
of the same substance to a cylinder, so that its approach AB may be considered 
as a constant pressure, volume increasing phase for which the energy comes 
from the source of supply. This same substance expanding to the lower pres- 
sure will do the work CBEF; but there will be negative work equivalent to the 
pushing away or displacing of an equivalent quantity of fluid at the low pres- 
sure, or FEGD, making the work cycle A BEG, in which AG is the excess of 



WORK AND POWER 



53 



initial over back pressure or the effective working pressure, remaining constant 
during approach and lessening regularly during expansion to zero excess at E. 
The work done will be from Eq. (13), 



w=Pb v b+ M±[i-(^y]-p e v e . 



A 


B 














































































































































































G 


















E 
























F 





DC V 

Fig. 22. — Pressure-Volume Diagram for Nozzle Expansion Measuring the Acceleration 

Velocity and Horse-power of Jets. 



But 



PeV e 8 = PeVeVe*- 1 =PbV b * = P b VbVb*- 1 

8-1 



Whence 



s— 1 

Tr - ft7 »+r5[ 1 -(S)']- ftF - 



Pb 



-iVK 1 "®' 1 • • • • (22) 



54 ENGINEERING THEEMODYNAMICS 

Assuming the initial velocity to be zero, and the work of Eq. (22) to be done 
on 1 lb. ; the final or resultant velocity will be according to Eq. (6), 



or 



u = V2gW 

= >Mi ft M 1 -© S1 ] (23) 

M =8.0W^n[l-(g)~] (24) 

This velocity is in feet per second when pressures are in pounds per square foot 
and volumes in cubic feet of 1 lb. of substance, and is known as Zeuner's 
equation for the velocity of a gas or vapor expanding in a nozzle. It is 
generally assumed that such expansion, involving as it does very rapid motion 
of the fluid past the nozzle, is of the adiabatic sort, as there seems to be no 
time for heat exchange between fluid and walls. As already noted, the value 
of s for adiabatic expansion of vapors is not constant, making the correct 
solution of problems on vapor flow through orifices practically impossible by 
this method of pressure volume analysis, but as will be seen later the thermal 
method of solution is exact and comparatively easy. 

Note. A comparison of Eqs. (22) and (13) and the figures corresponding 
will show that the area under the process curve, which is the same as the work 
done during the compression or expansion, if multiplied by s will equal the area 
to the left of the process curve, which in turn represents, as in Fig. 21, for 
engines, the algebraic sum of admission, complete expansion, and exhaust work 
areas, or as in Fig. 9 for compressors, the algebraic sum of suction, compression 
and delivery work areas. This statement must not be thought to refer to the 
work area of such a cycle as that of Fig. 8, where expansion is incomplete, nor a 
case of over-expansion, Fig. 10. 

Example. In Fig. 22 assume the initial pressure at 100 lbs. per square inch absolute, 
back pressure at atmosphere, and expansion as being adiabatic. What will be the 
work per pound of steam and the velocity of the jet, if Vb is 4.36 cu.ft. and s = 1.3 for 
superheated steam? 



144xl00x4.36[l-(^) 1 - 3 ] 



.3 X 



= 27,206 X.608 = 16,541 ft.-lbs.; 



u = V2gW = 8.02 Vl6,541 
= 1028 ft. per second. 



WORK AND POWER 55 

Prob, 1. Taking the same pressure range as above, find W and u for adiabatic expan- 
sion of air, also for isothermal expansion. 

Prob. 2. How large must the effective opening of the suction valve be, in an air 
compressor 18x24 ins. to allow the cylinder to properly fill if the mean pressure-drop 
through the valve is 1 lb. per square inch and the compressor runs at 80 R.P.M.? 

Prob. 3. What must be the diameter of the inlet valve in a gas engine to fill a 
cylinder of \ cu.ft. capacity if the lift of the valve is f ins., allowing a pressure drop of 
1 lb. per square inch? Engine makes 150 working strokes per minute. 

Prob. 4. It has been found from experiment that the velocity of air issuing from a hole 
in plate orifice is 72 per cent of what would be expected from calculation as above when 
the absolute pressure ratio is 2 to 1, and 65 per cent when the absolute pressure ratio 
is 1 1 to 1. What will be the actual velocity for air flowing from a tank to atmosphere 
for these pressure ratios? 

Prob. 5. C0 2 stored in a tank is allowed to escape through an orifice into the air. 
What will be the maximum velocity of the jet if the pressure on the tank be 100 lbs. 
per square inch gage? 

Note: 1 lb. C0 2 at pressure of 100 lbs. per square inch gage occupies 1.15 cu.ft. 

Prob. 6. If ammonia gas and hydrogen were allowed to expand from the same pres- 
sure, how would their maximum velocities compare? Vol. of 1 lb. of NH 3 at 50 lbs. per 
square inch gage is 4.5 cu.ft. Vol. of 1 lb. of H at same pressure is 77.5 cu.ft. 

18. Weight of Flow through Nozzles. Applying an area factor to the 
velocity equation will give an expression for cubic flow per second which 
becomes weight per second by introducing the factor, density. 

Let the area of an orifice at the point of maximum velocity, u, be A sq.ft., 
then will the cubic feet per second efflux be Au. Assume the point of maxi- 
mum velocity, having area A, to be that part of the nozzle where the pressure 
has fallen to P e , Fig. 22, and the gas or vapor to have the density B e pounds 
per cubic foot. Then will the nozzle flow in pounds per second be 

W = uAde. 

But the weight per cubic foot is the reciprocal of the cubic feet per pound, 
V e , which it has already been assumed, is the final volume, of one pound of the 
fluid. Hence, 

uA 

r e • 

This may be put in terms of initial gas or vapor conditions for, 



V c =V b 



Whence 



w = 



uA 



•<w 



= uA(Pe\ 



56 ENGINEERING THERMODYNAMICS 

Substituting in this the value of u from Eq. (24), 

HW^HH9f] _ 

This weight will be a maximum for a certain value of the pressure ratio, depend- 
ing on the value of s only, and this value can be found by placing the first dif- 
ferential coefficient of w with respect to ( -^ ) equal to 



!ffl 



zero. 



(ft) 



= 0. 



To accomplish this, rearrange Eq. (25) as follows: 

2 s + 1 , 1 

A- 8 Ph\ (Pe\ 
W 



-A^-MW -©T- 



But as the other factors do not enter to effect the result so long as P& does not 
vary, w is a maximum when the bracket 



+ i 



is a maximum or when 



iw~'-cmf"=« 



or 

i i 



or 

But as (^ j cannot be equal to zero in practice, then 

2 (sr- (s+i) =°' , 

which gives the condition that w is a maximum when 



(ft) 



8 + 1 

2 ' 



WORK AND POWER 57 

or maximum flow for given initial pressure occurs when 

m-i^r « 

For air expanding adiabatically s = 1.407. Maximum flow occurs when 

P 

^ = .528 and for most common values of s it will be between .50 and .60 
Pb 

This result is quite remarkable and is verified by experiment reasonably closely. 

It shows that, contrary to expectation, the weight of efflux from nozzles will not 

continuously and regularly increase with increasing differences in pressure, but 

for a given initial pressure the weight discharged per second will have reached 

its limit when the final pressure has been diminished to a certain fraction of the 

initial, and any further decrease of the discharge pressure will not increase the 

flow through an orifice of a given area. 

The subject of flow in nozzles will be treated more completely in 

Chapter VI. 

Prob. 1. For the following substances under adiabatic expansion determine the 
pressure ratio for maximum flow and find the rate of flow per square inch of orifice under 
this condition when flow is into a vacuum of 10 ins. of mercury with standard barometer: 

(a) Carbon dioxide. 

(b) Nitrogen. 

(c) Hydrogen. 

(d) Ammonia. 

(e) Dry steam accordingto saturation law. 

19. Horse-power of Nozzles and Jets. Although, strictly speaking, nozzles 
can have no horse-power, the term is applied to the nozzle containing the 
orifice through which flow occurs and in which a certain amount of work is 
done per minute in giving to a jet of gas or vapor initially at rest a certain 
final velocity, and amount of kinetic energy. The foot-pounds of work per 
pound of fluid multiplied by the pounds flowing per second will give the foot- 
pounds of work developed per second within the nozzle, and this divided by 
550 will give the horse-power developed by the jet, or the nozzle horse-power. 
Accordingly, 

VT> ,., Wxw W^A/P e \T 

H.P.ofjet = w — X^) X. (a) 



WA (^WWW . . (6) 



550 7 . 

i 



=-<(ri^4'-(8n 



\{c) 



(27) 



58 ENGINEERING THERMODYNAMICS 

where the expression in the bracket is the work done per pound of substance. 
The pressures are expressed in pounds per square foot, areas in square feet 
and volumes in cubic feet. 

Example. A steam turbine operates on wet steam at 100 lbs. per square inch abso- 
lute pressure which is expanded adiabatically to atmospheric pressure. What must be 
the area of the nozzles if the turbine is to develop 50 H.P. ideally? 

Note : 1 cu.ft. of steam at 100 lbs. = .23 lb. 

s 

/ 2 \«-i 
By Eq. (26), maximum flow occurs when the pressure ratio is (-77) , or, for 

l.ii 
/ 2 \ .11 
this case when the pressure is 100 -^ (^tt) =58 lbs. per square inch absolute. As 

the back pressure is one atmosphere, the flow will not be greater than for the above 
critical pressure. Substituting it in Eq. (25) will give the flow weight w, and using the 
actual back pressure in Eq. (22) will give the work W. 

.11 

_ _ , s „ T /1.11\ / 14400 \ [■ / 2116 \ 1.11" 
B y E q .(22),W-(— ) x (_)|^(_) 

= 110000 ft.-lbs. per pound of steam. 

J_ ( 1.11 14400 T — 

ByEq.(25), w=8.02AX-23X(.58)i.n j— X-— l-(.58)i.n 

= 198A lbs. of steam per second. 
By Eq. (27a), 

ttt> Wxw m 110000 X198A 



Whence 



550 550 



'-&—«• 



Prob. 1. What will be the horse-power per square inch of nozzle for a turbine using 
hot gases if expansion follows law PVs=k, when s = 1.37, the gases being at a pressure of 
200 lbs. per square inch absolute and expanding to atmosphere. 

Let the volume per pound at the high pressure be 2 cu.ft. 

Prob. 2. What will be the horse-power per square inch of nozzle for the problems 
of Section 17? 

Prob. 3. Suppose steam to expand according to law PVs=k, where s =1.111, from 
atmosphere to a pressure of 2 lbs. per sq. inch absolute. How will the area of the ori- 
fice compare with that of the example to give the same horse-power? 

Note: V b =2QA. 

Prob. 4. Suppose steam to be superheated in the case of the example and of the last 
problem, how will this affect the area of nozzle? 

Note: Let Vb =5 and 32 respectively. 

Prob. 5. How much work is done per inch of orifice if initial pressure is 100 lbs. 
absolute on one side and final 10 lbs. absolute on other side of a valve through which 
air is escaping? 



WORK AND POWER 59 



GENERAL PROBLEMS ON CHAPTER I 

1. An air compressor is required to compress 500 cu.ft. of free air per minute to a 
pressure of 100 lbs. per square inch gage; the compressor is direct connected to a steam 
engine. The mechanical efficiency of the machine is 80 per cent. What will be the 
steam horse-power if compression is (a) isothermal; (6) adiabatic? 

2. A mine hoist weighing 10 tons is raised 4500 ft. in one minute. In the first 20 
seconds it is accelerated from rest to a speed of 6000 ft. per minute; during the next 30 
seconds speed is constant at this value, and during last 10 seconds it is brought to rest. 
What will be (a) work of acceleration for each period; (b) work of lift for each period; 
(c) total work supplied by engine; (d) horse-power during constant velocity period? 

3. The engine driving the above hoist is driven by compressed air. If air is supplied at 
a pressure of 150 lbs. per square inch gage and is admitted for three-quarters of the 
stroke, then expanded adiabatically for the remainder of the stroke and exhausted to the 
atmosphere find (a) what must the piston displacement be to lift the hoist, the work of 
acceleration being neglected? (b) To what value could the air pressure be reduced if 
air were admitted full stroke? 

4. It is proposed to substitute an electric motor for the above engine, installing a 
water-power electric plant at a considerable distance. The type of wheel chosen is one in 
which a jet of water issuing from a nozzle strikes against a series of revolving buckets. 
The available head at the nozzle is 1000 ft. Assuming the efficiency of the motor to be 
85 per cent, transmission 80 per cent, generators 90 per cent, and water-wheels 60 per 
cent, what will be the cubic feet of water per minute? 

5. A steam turbine consists of a series of moving vanes upon which steam jets issuing 
from nozzles impinge. It is assumed that for best results the speed of the vanes should be 
half that of the jets. The steam expands from 100 lbs. per square inch gage to 5 lbs. 
per square inch absolute, (a) What must be the best speed of vanes for wet steam where 
s = 1.111? (6) If 55 per cent of the work in steam is delivered by the wheel what must 
be the area of nozzle per 100 H.P., and weight of steam per hour 100 H.P? 

Note: 7&=3.82. 

6. It has been found that a trolley car uses a current of 45 amperes at 550 volts when 
running 12 miles per hour. If the motor is 80 per cent efficient, what is tractive effort? 

Note: Volts X amperes = watts, and watts -k746 =H.P. 

7. An engine whose cylinder is 18x24 ins. (diameter = 18 ins., stroke =24 ins.) is 
double acting and runs at a speed of 125 R.P.M. Steam is admitted for one-quarter 
stroke at a pressure of 125 lbs. per square inch gage, allowed to expand for the rest of 
the stroke and then exhausted into a vacuum of 28.25 ins. of Hg. (a) Draw a P7 
diagram of the cycle and find the m.e.p. from the diagram and by calculation, and then 
find the horse-power. (6) Consider steam to be admitted one-half stroke without 
other change. How will the horse-power vary? (c) What will be the horse-power for 
one-quarter admission if the exhaust pressure is 15 lbs. per square inch absolute? (d) 
What will be the horse-power if the steam pressure be made 150 lbs. per square inch 
absolute, other conditions as in (a)? (e) Suppose all conditions as in (a) but the 
speed lowered to 75 R.P.M. What will be the horse-power? 

8. Assuming that 50 per cent of the work in the jet is transformed to useful work, 
what must be the total area of the nozzles of a steam turbine to develop the same horse- 



60 ENGINEERING THERMODYNAMICS 

power as the engine in problem (7a), the pressure range being the same and s being 1.3? 
76=3.18. 

9. Water is being pumped from the bottom of a shaft 700 ft. deep at the rate of 1000 
gallons per minute by an electrically driven pump. Efficiency of pump is 70 per cent, 
motor 90 per cent, transmission line 95 per cent, generator 85 per cent, and mechanical 
efficiency of engine 80 per cent. What will be the indicated horse-power of the engine? 
If the above installation were replaced with an air-driven pump of 65 per cent efficiency, 
efficiency of transmission being 100 per cent, and that of the compressor and engine 
80 per cent, what would be the horse-power of this engine? 

10. Show by a PV diagram, assuming any convenient scales, that the quantity of air 
discharged by a compressor and the horse-power, both decrease as the altitude increases, and 
that the horse-power per cubic foot of air delivered increases under the same condition. 

11. A centrifugal pump is driven by a steam engine directly connected to it. The 
pump is forcing 1000 gallons of water per minute against a head of 250 ft. and runs at a 
speed of 450 R.P.M. The engine is double acting and its stroke equals the diameter of 
the cylinder. Steam of 100 lbs. per square inch gage is admitted for half stroke, allowed 
to expand the rest of the stroke so that s = 1, and is then exhausted to atmosphere. What 
must be the size of the engine if the pump efficiency is 65 per cent and the engine 
efficiency 75 per cent? 

12. (a) What will be the pounds of steam used by this engine per hour per horse- 
power? (b) If the steam were admitted but one-quarter of the stroke and the initial 
pressure raised sufficiently to maintain the same horse-power, what would be the new 
initial pressure and the new value of the steam used per horse-power per hour? 

Note : Weight of steam per cubic foot for (a) is .261 ; for (b) is .365. 

13. If it were possible to procure a condenser for the above engine so that the exhaust 
pressure could be reduced to 2 lbs. per square inch absolute, (a) how much would the 
power be increased for each of the two initial pressures already given? (6) How would 
the steam consumption change? 

14. A motor-fire engine requires a tractive force of 1300 lbs. to drive it 30 miles per 
hour, its rated speed. The efficiency of engine and transmission is 80 per cent. When 
the same engine is used to actuate the pumps 70 per cent of its power is expended on 
the water. What will be the rating of the engine in gallons per minute when pumping 
against a pressure of 200 lbs. per square inch? 

15. A compressor when compressing air at' sea level from atmosphere to 100 lbs. per 
square inch absolute, expends work on the air at the rate of 200 H.P., the air being com- 
pressed adiabatically. (a) How many cubic feet of free air are being taken into the 
compressor per minute and how many cubic feet of high pressure air discharged? 
Compressor is moved to altitude of 8000 ft. (6) What will be the horse-power if the same 
amount of air is taken in and how many cubic feet per minute will be discharged? 

(c) What will be the horse-power if the same number of cubic feet are discharged as in 
case (a) and what will be the number of cubic feet of low pressure air drawn in? 

(d) Should superheated ammonia be substituted for air at sea level, what would be the 
necessary horse-power? 

16. A car weighing 40 short tons is at rest and is struck by a train of four cars, each 
of the same weight as the first, (a) Upon impact the single car is coupled to the train and 
all move off at a certain velocity. If the original velocity of the train was 3 miles 
per hour, what will it be after attachment of the extra car? (6) If instead of coupling, 
the extra car after impact moved away from the train at twice the speed the train was 
then moving, what would be the speed of train? 



WORK AND POWER 61 

17. To drag a block of stone along the ground requires a pull of 1000 lbs. If it be 
placed on rollers the pull will be reduced to 300 lbs., while if it be placed on a wagon with 
well-made wheels, the pull will be but 200 lbs. Show by diagram how the work required 
to move it 1000 ft. will vary. 

18. 100 cu. ft. of air, at atmospheric pressure, is compressed to six times its 
original pressure, (a) What will be the difference in horse-power to do this in 45 seconds 
isothermally and adiabatically at an elevation of 8000 ft. (b) What will be the final vol- 
umes? (c) What will be the difference in horse-power at sea level? (d) What will be the 
final volumes? 

19. An engine operating a hoist is run by compressed air at 80 lbs. per square inch 
gage. The air is admitted half stroke, then expanded for the rest of the stroke so that 
s = 1.3 and then exhausted to atmosphere. The engine must be powerful enough to lift 
a ton cage and bring it up to a speed of 2000 ft. per minute in 15 seconds. What will be 
the necessary displacement per minute? 

20. Construct P V diagrams for Probs. 1, 11, 13 and 15, showing by them that the 
work of admission, compression or expansion, and discharge or exhaust, is equal to that 
found algebraically. 

' 21. The elongation of wrought iron under a force F is equal to the force times the 
length of the piece divided by 25,000,000 times the cross-section of metal in the piece. 
A 4i in. pipe has 3.75 sq. ins. of metal; a line of this pipe 1 mile long is running full of 
water with a velocity of 600 ft. per minute. This is stopped in 1 second by closing a 
valve. Assuming pipe did not burst, what would be the elongation? 

22. Two steam turbines having nozzles of equal throat areas are operating on a steam 
pressure of 150 lbs. per square inch gage. One is allowing steam to expand to atmosphere 
the other to 2 lbs. per square inch absolute, both cases having an exponent for expansion 
of 1.11. Find the relation of the horse-power in the two cases. 

23. The power from a hydro-electric plant is transmitted some distance and then 
used to drive motors of various sizes. At the time of greatest demand for current it has 
been found that 1000 horse-power is given out by the motors. Taking the average 
efficiency of the motors as 70 per cent, transmission efficiency as 85 per cent, generator 
efficiency as 85 per cent, and water-wheel efficiency as 70 per cent, how many cubic feet 
of water per second will the plant require if the fall. is 80 ft.? 

24. A small engine used for hoisting work is run by compressed air. Air is admitted 
for three-quarters of the stroke and then allowed to expand for the rest of the stroke in 
such a way that s = lA and finally exhausted to atmosphere. For the first part of the 
hoisting,full pressure (80 lbs. per square inch gage) is applied, but after theload has been 
accelerated the pressure is reduced to 30 lbs. per square inch gage. If the engine has 
two cylinders each 9 Xl2 in., is double acting and runs at 200 R.P.M., what will be the 
(a) horse-power in each case, (b) the specific displacement? 

25. A steam engire has a cylinder 12x18 in., is double acting, and runs at 150 
R.P.M. (a) What is the engine constant, and (b) horse-power per pound m.e.p.? 

26. A water-power site has available at all times 3500 cu.ft. of water per min- 
ute at a 100-ft. fall. Turbines of 70 per cent efficiency are installed which take the place 
of two double-acting steam engines whose mechanical efficiencies were 85 per cent. 
The speed of the engines was 150 R.P.M. , m.e.p. 100 lbs. per square inch, and stroke was 
twice the diameter. What was the size of each engine? 

27. Assuming the frictional losses in a compressor to have been 15 per cent, how many 
cu.ft. of gas per minute could a compressor operated by the above engines compress 
from atmosphere to 80 lbs. per square inch gage if s = 1.35? 



62 



ENGINEERING THERMODYNAMICS 



Table I 
CONVERSION TABLE OF UNITS OF DISTANCE 



Meters. 1 


Kilometers. 


Inches. 


Feet. 


Statute Miles. 


Nautical Miles. 


1 

1000 

0.0254 
0.304801 

1609.35 

1853.27 


0.001 

1 

0.0000254 
0.0003048 
1 .60935 
1.85327 


39.37 
39370.1 
1 

12 
63360 
72963.2 


3.28083 
3280.83 

0.083333 

1 
5280 
6080.27 


0.000621370 

0.62137 

0.0000157828 

0.000189394 

1. 

1.15157 


0.000539587 

0.539587 

0.0000137055 

0.000164466 

0.868382 

1. 



In accordance with U. S. Standards (see Smithsonian Tables). 



Table II 
CONVERSION TABLE OF UNITS OF SURFACE 



Sq. Meters. 


Sq. Inches. 


Sq. Feet. 


Sq. Yards. 


Acres. 


Sq. Miles. 


1 

.000645 
.0929 
.8361 
4046.87 


1550.00 

1 

144 

1296 


10.76387 
.00694 
1 
9 
43560 
27878400 


1 . 19599 

.111 
1 

4840 
3097600 


.000247 

.000206 

1 
640 


001562 


2589999 




1 









Table III 
CONVERSION TABLE OF UNITS OF VOLUME 



Cu. Meters. 


Cu. Inches. 


Cu. Feet. 


Cu. Yards. 


Lities 
(1000 Cu. Cm.) 


Gallons (U.S.) 


1 


61023.4 

1 

1728 

46656 

61.023 

231 


35.3145 
.000578 
1 

27 

.035314 
. 13368 


1.3079 


1000 

.016387 
28.317 


264.170 
00433 


.028317 
.76456 


.03704 
1 
.001308 
.004951 


7 . 4805 
201 974 


.001 
.003785 


1 

3.7854 


.26417 
1 



TABLES 



63 



Table IV. 
CONVERSION TABLE OF UNITS OF WEIGHT AND FORCE 



Kilogrammes. 


Metric Tons. 


Pounds. 


U. S. or Short Tons. 


British or Long Tons. 


1. 

1000. 

0.453593 
907.186 
1016.05 


0.001 
1. 

0.000453593 

0.907186 

1.01605 


2.20462 
2204.62 

1. 
2000. 
2240. 


0.00110231 

1.10231 

0.0005 

1. 

1 . 12000 


0.000984205 

0.984205 

0.000446429 

0.892957 

1. 



Table V 
CONVERSION TABLE OF UNITS OF PRESSURE 



Pounds per 
Square Foot. 



Pounds per 
Square Inch. 



Inches of 

Mercury at 

32° F. 



Atmospheres 
(Standard at 
Sea Level). 



One lb. per sq. ft 

One lb. per sq. in 

One ounce per sq. in 

One atmosphere (standard at sea 
level) 

One kilogramme per square meter . . 

One gramme per square millimeter . 

One kilogramme per square centi- 
meter 

FLUID PRESSURES 

One ft. of water at 39.1° F. (max. 
dens.) 

One ft. of water at 62° F 

One in. of water at 62° F 

One in. of mercury at 32° F. (stand- 
ard) l 

One centimeter of mercury at 0° C. . 

One ft. of air at 32° F., one atmos. 
press 

One ft. of air, 62° F 



1 
144. 
9. 

2116.1 
20.4817 
204.817 

2048.17 



62.425 

62.355 

5.196 

70.7290 

27.8461 



0.08071 
0.07607 



0.006944 

1. 

0.0625 

14.696 
0.142234 
1.42234 

14.2234 



0.43350 
0.43302 
0.036085 

0.491174 
0.193376 

0.0005604 
0.0005282 



0.014139 

2.03594 

0.127246 

29.924 
. 289579 
2.89579 

28.9579 



0.88225 
0.88080 
0.07340 

1. 
0.393701 

0.0011412 
0.0010755 



0.0004724 

0.06802 

0.004252 

1. 

0.009678 

0.09678 

0.9678 



0.029492 
0.029460 
0.002455 

0.033416 
0.013158 

0.00003813 
0.00003594 



1 Pressures Measured by the Mercury Column. For temperatures other than 32° F., the density 
of mercury, pounds per cubic inch, and hence the pressure, pounds per square inch, due to a column of mercury 
1 inch high, is given with sufficient accuracy by the following formula: 

p = 0.4912-(i-32) X0.0001. 

The mercurial barometer is commonly made with a brass scale which has its standard or correct length 
at 62° F, and a linear coefficient of expansion of about 0.000001 for each degree Fahrenheit. Hence, to 
correct the standard of mercury at 32° F., the corrected reading will be 

H** = H t -H tx nooo , 



where H^ is the observed height at a temperature of t° F. 



64 



ENGINEERING THERMODYNAMICS 



Table VI 
CONVERSION TABLE OF UNITS OF WORK l 



Kilogrammeters. 


Foot-pounds. 


Foot Tons (Short Tons). 


Foot Tons (Long Tons). 


1. 

0.138255 
276.510 
309.691 


7.23300 

1. 
2000. 
2240. 


0.00361650 

0.000500 

1. 

1 . 12000 


0.00322902 
0.000446429 
0.892857 
1. 



!See 



more complete table of Units of Work and Energy in Chapter IV on Work and Heat. 



Table VII 
CONVERSION TABLE OF UNITS OF POWER 



Foot-pounds per 
Second. 


Foot-pounds per 
Minute. 


Horse-power. 


Cheval-Vapeur. 


Kilogrammeters per 
Minute. 


1. 

0.0166667 
550.000 
542.475 

0.120550 


60. 

1. 
33000. 
32548.5 

7.23327 


0.00181818 

0.000030303 

1. 

0.986319 

0.000219182 


0.00184340 

0.0000307241 

1.01387 

1. 

0.000222222 


8.29531 
0.138252 

4562.42 

4500.00 
1. 



Table VIII 
UNITS OF VELOCITY 



Feet per Minute. 



Feet per Second. 



One foot per second 

One foot per minute 

One statute mile per hour 

One nautical mile per hour = 1 knot 

One kilometer per hour 

One meter per minute 

One centimeter per second 



60. 
1. 

88. 

101.338 
54.6806 
3.28084 
2.00848 



0.016667 

1.4667 

1.6890 

0.911344 

0.054581 

0.032808 



TABLES 



65 



Table IX 
TABLE OF BAROMETRIC HEIGHTS, ALTITUDES, AND PRESSURES 

(Adapted from Smithsonian Tables) 

Barometric heights are given in inches and millimeters of mercury at its standard density 
(32° F.). 

Altitudes are heights above mean sea level in feet, at which this barometric height is 
standard. (See Smithsonian Tables for corrections for latitude and temperature.) 

Pressures given are the equivalent of the barometric height in lbs. per sq. in. and per 
sq. ft. 



Standard Barometer. 


Altitude, Feet above 
Sea Level. 


Pressure, P 


ounds per 










Inches. 


Centimeters. 




Square Inch. 


Square Foot. 


17.0 


43.18 


15379 


8.350 


1202.3 


17.2 


43.69 


15061 


8.448 


1216.6 


17.4 


44.20 


14746 


8.546 


1230.7 


17.6 


44.70 


14435 


8.645 


1244.8 


17.8 


45.21 


14128 


8.742 


1259.0 


18.0 


45.72 


13824 


8.840 


1273.2 


18.2 


46.23 


13523 


8.940 


1287.3 


18.4 


46.73 


13226 


9.038 


1301.4 


18.6 


47.24 


12931 


9.136 


1315.6 


18.8 


47.75 


12640 


9.234 


1329.7 


19.0 


48.26 


12352 


9.332 


1343.8 


19.2 


48.77 


12068 


9.430 


1357.9 


19.4 


49.28 


11786 


9.529 


1372.1 


19.6 


49.78 


11507 


9.627 


1386.3 


19.8 


50.29 


11230 


9.726 


1400.4 


20.0 


50.80 


10957 


9.825 


1414.6 


20.2 


51.31 


10686 


9.922 


1428.7 


20.4 


51.82 


10418 


10.020 


1442.9 


20.6 


52.32 


10153 


10.118 


1457.0 


20.8 


52.83 


9890 


10.217 


1471.2 


21.0 


53.34 


9629 


10.315 


1485.3 


21.2 


53.85 


9372 


10.414 


1499.4 


21.4 


54.36 


9116 


10.511 


1513.6 


21.6 


54.87 


8863 


10.609 


1527.7 


21.8 


55.37 


8612 


10.707 


1541.8 


22.0 


55.88 


8364 


10.806 


1556.0 


22.2 


56.39 


8118 


10.904 


1570.1 


22.4 


56.90 


7874 


11.002 


1584.3 


22.6 


57.40 


7632 


11.100 


1598.4 


22.8 


57.91 


7392 


11.198 


1612.6 


23.0 


58.42 


7155 


11.297 


1626.7 


23.2 


58.92 


6919 


11.395 


1640.8 


23.4 


59.44 


6686 


11.493 


1655.0 


23.6 


59.95 


6454 


11.592 


1669.3 


23.8 


60.45 


6225 


11.690 


1683.3 


24.0 


60.96 


5997 


11.788 


1697.4 


24.2 


61.47 


5771 


11.886 


1711.6 


24.4 


61.98 


5547 


11.984 


1725.7 


24.6 


62.48 


5325 


12.083 


1739.9 


24.8 


62.99 


5105 


12.182 


1754.0 


25.0 


63.50 


4886 


12.280 


1768.2 


25.2 


64.01 


4670 


12.377 


1782.3 


25.4 


64.52 


4455 


12.475 


1796.5 


25.6 


65.02 


4241 


12.573 


1810.7 


25.8 


65.53 


4030 


12.671 


1824.8 




i 


i 


7 ' 





66 



ENGINEERING THERMODYNAMICS 



Table IX — Continued 



Standard Barometer. 


Altitude, Feet above 
Sea Level. 


Pressure, Pounds per 


Inches. 


Centimeters. 


Square Inch. 


Square Foot. 


26.0 
26.1 
26.2 
26.3 
26.4 


65.04 
66.30 
66.55 
66.80 
67.06 


3820 
3715 
3611 
3508 
3404 


12.770 

12.819 
12.868 
12.918 
12.967 


1838.9 
1846.0 
1853.1 
1860.2 
1867.3 


26.5 
26.6 
26.7 
26.8 
26.9 


67.31 
67.57 
67.82 
68.08 
68.33 


3301 
3199 
3097 
2995 
2894 


13.016 
13.065 
13.113 
13 163 
13.212 


1874.3 
1881.4 
1888.5 
1895.5 
1902.6 


27.0 
27.1 

27.2 
27.3 
27.4 


68.58 
68.84 
69.09 
69.34 
69.60 


2793 
2692 
2592 
2493 
2393 


13.261 
13.310 
13.359 
13.408 
13.457 


1909.7 
1916.7 
1923.8 
1930.9 
1938.0 


27.5 
27.6 

27.7 
27.8 
27.9 


69.85 
70.10 
70.35 
70.61 

70.87 


2294 
2195 
2097 
1999 
1901 


13.507 
13.556 
13.605 
13.654 
13.704 


1945.1 
1952.1 
1959.2 
1966.3 
1973.3 


28.0 

28.1 
28.2 
28.3 
28.4 


71.12 
71.38 
71.63 

71.88 
72.14 


1804 
1707 
1610 
1514 
1418 


13.753 

13.802 
13.850 
13.899 
13.948 


1980.4 
1987.5 
1994.5 
2001.6 

2008.7 


28.5 
28.6 
28.7 
28.8 
28.9 


72.39 
72.64 
72.90 
73.15 
73.40 


1322 
1227 
1132 
1038 
943 


13.998 
14.047 
14.096 
14.145 
14.194 


2015.7 
2022.8 
2030.0 
2037.0 
2044.1 


29.0 
29.1 
29.2 
29.3 
29.4 


73.66 
73.92 
74.16 
74.42 
74.68 


849 
756 
663 
570 
477 


14.243 
14.293 
14.342 
14.392 
14.441 


2051 . 2 
2058 . 2 
2065.3 
2072.4 
2079.4 


29.5 
29.6 
29.7 
29.8 
29.9 


74.94 
(75.18 
75.44 
75.69 
75.95 


384 
292 
261 
109 

+18 


14.490 
14.539 

14.588 
14.637 
14.686 


2086.5 
2093.6 
2100.7 
2107.7 
2114.7 


29.92 


76.00 





14.696 


2116.1 


30.0 
30.1 
30.2 
30.3 
30.4 


76.20 
76.46 
76.71 
76.96 

77.22 


- 73 
-163 
-253 
-343 
-433 


14.734 
14.783 
14.833 
14.882 
14.931 


2121.7 
2128.8 
2135.9 
2143.0 
2150.1 


30.5 
30.6 
30.7 
30.8 
30.9 


77.47 
77.72 
77.98 
78.23 
78.48 


-522 
-611 
-700 

-788 
-877 


14.980 
15.030 
15.078 
15.127 
15.176 


2157.2 
2164.2 
2171.3 
2178.4 
2185.5 


31.0 


78.74 


-965 


15.226 


2192.6 



- 



TABLES 



67 



Table X 

VALUES OF s IN THE EQUATION PV S = CONSTANT FOR VARIOUS SUBSTANCES 

AND CONDITIONS 



Substance. 




s 


Remarks or Authority. 


All gases 


Isothermal 
Constant pressure 

Isothermal 

Constant volume 

Adiabatic 

Compressed in cylinder 

Adiabatic, wet 

Adiabatic, superheated 

Adiabatic 

Adiabatic 

Adiabatic 

Adiabatic 
Adiabatic 

Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 

Adiabatic 

Adiabatic 

Adiabatic 

Expanding in cylinder 

Saturation law 


« ) 

00 J 
1.4066 
1.4 
1.1 
1.3 
1 . 293 
1.300 
1.403 

1.200 
1.323 

1.106 
1.029 
1.410 
1.276 
1.316 
1.410 
1.291 
1.24 
1.26 
1.300 
Variable 

1.111 

1 + 14X% moist. 

1.035 + .1X% moist. 

1. 

1.0646 




All gases and vapors . . 
All saturated vapors . . 
All gases and vapors . . 
Air 


Accepted thermody- 
namic law 

Smithsonian Tables 


Air ... 


Experience 


Ammonia (NH 3 )' 

Ammonia (NH 3 ) 

Bromine 


Average 

Thermodynamics 
Strecker 


Carbon dioxide (C0 2 ) 
Carbon monoxide (CO) 
Carbon disulphide 

(CS2) • 


Rontgen, Wullner 
Cazin, Wullner 

Beyne 
Strecker 

Beyne, Wullner 

Muller 

Cazin 

Muller 

Muller 

Cazin 

Wullner 

Pintsch Co. 


Chlorine (CI) 

Chloroform 

(CCUCH(OH) 2 ).... 
Ether (C2H5OC2H5)... 

Hydrogen (H 2 ) 

Hydrogen sulph . (H 2 S) 

Methane (CH 4 ) 

Nitrogen (N 2 ) 

Nitrous oxide (N0 2 ) . . 
Pintsch gas 


Sulphide diox (S0 2 ) ... 
Steam, superheated . .. 
Steam, wet 


Cazin, Muller 
Smithsonian Tables 
(From less than 1 to 


Steam, wet 


more than 1.2) 
Rankine 


Steam, wet .... 


Perry 

Gray 

Average from practice 

R egnault 


Steam, wet 


Steam, wet 


Steam, dry 







68 



ENGINEERING THERMODYNAMICS 



Table XI 
HORSE-POWER PER POUND MEAN EFFECTIVE PRESSURE. 

aS 



VALUE OF K e = 



33000 



Diameter 
of 


Speed of Piston in Feet per Min 


ute. 






Cylinder, 
Inches. 


100 


200 


300 


400 


500 


600 


700 


800 


900 


4 


0.0381 


0.0762 


0.1142 


0.1523 


0.1904 


0.2285 


0.2666 


0.3046 


0.3427 


4-i 


0.0482 


0.0964 


0.1446 


0.1928 


0.2410 


0.2892 


0.3374 


0.3856 


0.4338 


5 


0.0592 


0.1190 


0.1785 


0.2380 


0.2975 


0.3570 


0.4165 


0.4760 


0.5355 


51 


0.0720 


0.1440 


0.2160 


0.2880 


0.3600 


0.4320 


0.5040 


0.5760 


0.6480 


6 


0.0857 


0.1714 


0.2570 


0.3427 


0.4284 


0.5141 


0.5998 


0.6854 


0.7711 


6i 


0.1006 


0.2011 


0.3017 


0.4022 


0.5028 


0.6033 


0.7039 


0.8044 


0.9050 


7 


0.1166 


0.2332 


0.3499 


0.4665 


0.5831 


0.6997 


0.8163 


0.9330 


1 . 0490 


7* 


0.1339 


0.2678 


0.4016 


0.5355 


0.6694 


0.8033 


0.9371 


1.0710 


1 . 2049 


8 


0.1523 


0.3046 


0.4570 


0.6093 


0.7616 


0.9139 


1.0662 


1.2186 


1 . 3709 


81 


0.1720 


0.2439 


0.5159 


0.6878 


0.8598 


1.0317 


1.2037 


1.3756 


1.5476 


9 


0.1928 


0.3856 


0.5783 


0.7711 


0.9639 


1 . 1567 


1 . 3495 


1 . 5422 


1.7350 ' 


91 


0.2148 


0.4296 


0.6444 


0.8592 


1.0740 


1.2888 


1.5036 


1 . 7184 


1.9532 


10 


0.2380 


0.4760 


0.7140 


0.9520 


1.1900 


1.4280 


1.6660 


1 . 9040 


2.1420 


11 


0.2880 


0.5760 


0.8639 


1.1519 


1.4399 


1 . 7279 


2.0159 


2.3038 


2.5818 


12 


0.3427 


0.6854 


1.0282 


1.3709 


1.7136 


2 . 0563 


2.3990 


2.7418 


3 . 0845 


13 


0.4022 


0.8044 


1.2067 


1.6089 


2.0111 


2.4133 


2.8155 


3.2178 


3 . 6200 


14 


0.4665 


0.9330 


1.3994 


1.8659 


2.3324 


2.7989 


3 . 2654 


3.7318 


4.1983 


15 


0.5355 


1.0710 


1.6065 


2 . 1420 


2.6775 


3.2130 


3 . 7485 


4.2840 


4.8195 


.16 


0.6093 


1.2186 


1.8278 


2.4371 


3.0464 


3.6557 


4.2650 


4.8742 


5.4835 


17 


0.6878 


1.2756 


1.9635 


2.6513 


3.3391 


4.0269 


4.6147 


5.4026 


6.1904 


18 


0.7711 


1.5422 


2.3134 


3.0845 


3.8556 


4.6267 


5.3987 


6.1690 


6.4901 


19 


0.8592 


1.7184 


2.5775 


3.4367 


4.2858 


5.1551 


6.0143 


6.8734 


7.7326 


20 


0.9520 


1 . 9040 


2.8560 


3.8080 


4.7600 


5.7120 


6.6640 


7.6160 


8 . 5680 


21 


1.0496 


2 . 0992 


3.1488 


4.1983 


5.2475 


6.2975 


7.3471 


8.3966 


9.4462 


22 


1.1519 


2.3038 


3.4558 


4.6077 


5.7596 


6.9115 


8.0643 


9.2154 


10.367 


23 


1.2590 


2.5180 


3.7771 


5.0361 


6.2951 


7.5541 


8.8131 


10.072 


11.331 


24 


1.3709 


2.7418 


4.1126 


5.4835 


6.8544 


8.2253 


9 . 5962 


10.967 


12.338 


25 


1.4875 


2.9750 


4.4625 


5.9500 


7.4375 


8.925a 


10.413 


11.900 


13.388 


26 


1.6089 


3.2178 


4.8266 


6.4355 


8.0444 


9.6534 


11.262 


12.871 


14.480 


27 


1.7350 


3.4700 


5.2051 


6.9401 


8.6751 


10.410 


12.145 


13.880 


15.615 


28 


1.8659 


3.7318 


5.5978 


7.4637 


9.3296 


11.196 


13.061 


14.927 


16.793 


29 


2.0016 


4.0032 


6.0047 


8.0063 


10.008 


12.009 


14.011 


16.013 


18.014 


30 


2.1420 


4.2840 


6.4260 


8.5680 


10.710 


12.852 


14.994 


17.136 


19.278 


31 


2.2872 


4.5744 


6.8615 


9.1487 


11.436 


13.723 


16.010 


18.287 


20.585 


32 


2.4371 


4.8742 


7.3114 


9.7485 


12.186 


14.623 


17.060 


19.497 


21.934 


33 


2.5918 


5.1836 


7.7755 


10.367 


12 . 959 


15.551 


18.143 


20.735 


23 . 326 


34 


2.7513 


5.5026 


8.2538 


11.005 


13.756 


16.508 


19.259 


22.010 


24.762 


35 


2.9155 


5.8310 


8.7465 


11.662 


14.578 


17.493 


20.409 


23.224 


26.240 


36 


3 . 0845 


6 . 1690 


9.2534 


12.338 


15.422 


18.507 


21.591 


24.676 


27.760 


37 


3.2582 


6.5164 


9.7747 


13 . 033 


16.291 


19.549 


22.808 


26.066 


29.324 


38 


3.4367 


6.8734 


10.310 


13.747 


17.184 


20.620 


24.057 


27.494 


30.930 


39 


3 . 6200 


7.2400 


10.860 


14.480 


18.100 


21.720 


25.340 


28.960 


32 . 580 



TABLES 
Table XI — Continued 



69 



Diameter 

of 

Cylinder, 

Inches. 


Speed of Piston in Feet per Minute. 


100 


200 


300 


400 


500 


600 


700 


800 


900 


40 


3.8080 


7.6160 


11.424 


15.232 


19.040 


22.848 


26.656 


30.464 


34.272 


41 


4.0008 


8.0016 


12.002 


16.003 


20.004 


24.005 


28.005 


32.006 


36.007 


42 


4.1983 


8.3866 


12.585 


16.783 


20.982 


25.180 


29.378 


33.577 


37.775 


43 


4.4006 


8.8012 


13.202 


17.602 


22.003 


26.404 


30.804 


35 . 205 


39.606 


44 


4.6077 


9.2154 


13.823 


18.431 


23 . 038 


27.646 


32.254 


36.861 


41.469 


45 


4.8195 


9.6390 


14.459 


19.278 


24.098 


28.917 


33.737 


38 . 556 


43.376 


46 


5.0361 


10.072 


15.108 


20.144 


25.180 


30.216 


35.253 


40.289 


45.325 


47 


5.2574 


10.515 


15.772 


21.030 


26.287 


31.545 


36.802 


42.059 


47.317 


48 


5.3845 


10.967 


16.451 


21.934 


27.418 


32.901 


38.385 


43.868 


49.352 


49 


5.7144 


11.429 


17.143 


22.858 


28.572 


34.286 


40.001 


45.715 


51.429 


50 


5.9500 11.900 


17.850 


23.800 


29 . 750 


35.700 


41.650 


47.600 


53.550 


51 


6.1904! 12.381 


18.571 


24.762 


30.952 


37.142 


43.333 


49.523 


55.713 


52 


6.4355 12.871 


19.307 


25.742 


32.178 


38.613 


45.049 


51.484 


57.920 


53 


6.6854 13.371 


20.056 


26.742 


33.427 


40.113 


46.798 


53.483 


60.169 


54 


6.9401 


13.880 


20.820 


27.760 


34.700 


41.640 


48.581 


55.521 


62.461 


55 


7.1995 


14.399 


21.599 


28.798 


35.998 


43.197 


50.397 


57.596 


64.796 


56 


7.4637 14.927 


22.391 


29.855 


37.318 


44.782 


52.246 


59.709 


67.173 


57 


7.7326 


15.465 


23.198 


30.930 


38.663 


46.396 


54.128 


61.861 


69.594 


58 


8.0063 


16.013 


24.019 


32.025 


40.032 


48.038 


56.044 


64.051 


72.057 


59 


8.2849 


16 . 570 


24.854 


33.139 


41.424 


49.709 


57.993 


66.278 


74.563 


60 


8.5680 


17. 136 ! 25.704 

i 


34.272 


42.840 


51.408 


59.976 


68.544 


77.112 



70 ENGINEERING THERMODYNAMICS 



GENERAL FORMULA RELATING TO PRESSURE- VOLUME 
CALCULATIONS OF WORK AND POWER 

Work = W= Force X Distance =FxL; 

=Pressure xArea xDistance =P XA XL; 
= Pressure X Volume change =Px(V 2 —V 1 ). 
Force of acceleration = mass X acceleration. 

9 dx 
Work of acceleration =5" mass X difference of (velocity) 2 , 

- Velocity due to work of acceleration = square root of the sum of (initial velocity) 2 
plus 2<7Xwork per pound of substance accelerated. 



/ 2 , 64.32IT 



u 2 = \u l 2 + 



w 
or if initial velocity is zero, 



= V64.32— =8.02^ — . 



u 2 

w 



Pressure Volume Relation for Expansion or Compression 
PV S =Pi7! s = P 2 V 2 S =K, a constant. (See Table VIII for values of s.) 

log (y) 
(Note graphical method for finding s when variable, see text.) 



FORMULA RELATING TO PRESSURE- VOLUME CALCULATIONS 71 



Work done during a pressure volume change represented by the equation (PV 8 =K) 
between points represented by 1 and 2 in figure =area under curve, Wi. 






dV 
V s ' 



If 5 = 1, 



T^P^log*^ 



= P 2 V 2 loge 
-PlF, loge 

=P 2 F 2 log e 



f\ 
Pa 

P, 
P 2 J 



\ s = l. 



If s is not equal to 1, 



\s^l. 



".^■[(fr-] i 
■si-©"] 

Work of admission, complete expansion and exhaust for engines = area to left of 
curve (see figure) W 2 . Same for admission, compression and expulsion for com- 
pressors. Both cases without clearance. 

(When s = 1, W 2 is same as area under curve , W x (see above). 
Whens^l ^=^ 1 P 2 F ? [(^- 2 ) S ~ 1 -l] 

-^■[©"-'] 

-r^r.[.-.©-"] , 

Clearance, expressed as a fraction of displacement =c, as a volume = CI. 



sxW,. 



72 ENGINEERING THERMODYNAMICS 

If P a V a =P b V b s , 

Db_Da(Pa\l 

D D \P b J 



CI 
c = 



D I 



© 



Indicated horse-power = (mean effective pressure, pounds per square inch X effective 
area of piston, square inches Xlength of stroke, feetXnumber of working cycles per- 
formed per minute) divided by 33,000. 

t tt t> (m.e.o.'Lan 
IH>R= 33 000 • 

Specific displacement = displacement in one direction for one side of a piston, in 

cuit. per hour per H.P. =D S = 13,750,--?-— , 

(m.e.p.) 

where z is the number of strokes required o compl e one cycle. 

Velocity of a jet due to its own expansion =the square root of the product of 2g Xwork 
done by admission, complete expansion and exhaust of 1 lb. of the substance. 

„ = y/2gw, ; =8.02 1 7 ^ T P,r i [l - (g) ~ ] \ . 

Weight of flow through nozzle or orifices, pounds per second =w = (velocity, feet per 
second X area, square feet, of orifice) -v- (volume per pound of substance at section where 
area is measured). 



w 



^KffirkT^HftHP 



Maximum discharge w for a given initial pressure occurs when 

Pi / 2 



P2 \s+l) 



CHAPTER II 

WORK OF COMPRESSORS, HORSE-POWER AND CAPACITY OF AIR, GAS AND 
VAPOR COMPRESSORS, BLOWING ENGINES AND DRY VACUUM PUMPS. 

1. General Description of Structure and Processes. There is quite a 
large class of machines designed to receive a cylinder full of some gas at one 
constant pressure and after the doing of work on the gas through decreasing 
volumes and rising pressures, to discharge the lesser volume of gas against a 
constant higher pressure. These machines are in practice grouped into sub- 
classes, each having some specific distinguishing characteristic. For example, 
blowing engines take in air at atmospheric pressure or as nearly so as the 
valve and port resistance will permit, and after compression deliver the air 
at a pressure of about three atmospheres absolute for use in blast furnaces. 
These blowing engines are usually very large, work at low but variable speeds, 
but always deliver against comparatively low pressures; they, therefore, have 
the characteristics of large but variable capacity and low pressures. A great 
variety of valves and driving gears are used, generally mechanically moved 
suction and automatic spring closed discharge valves, but all valves may be 
automatic. The compressor cylinder is often termed the blowing tub and the 
compressed or blast air frequently is spoken of as wind by furnace men. They 
are all direct-connected machines, an engine forming with the compressor one 
machine. The engine formerly was always of the steam type, but now a change 
is being affected to permit the direct internal combustion of the blast furnace 
waste gases in the cylinders of gas engines. These gas-driven blowing engines, 
showing approximately twice the economy of steam-driven machines, will in 
time probably entirely displace steam in steel plants, and this change will take 
place in proportion to the successful reduction of cost of repairs, increase of 
reliability and life of the gas-driven blowing engines to equal the steam-driven. 
Some low-pressure blowers are built on the rotary plan without reciprocating 
pistons, some form of rotating piston being substituted, and these, by reason 
of greater leakage possibilities, are adapted only to such low pressures as 5 lbs. 
per square inch above atmosphere or thereabouts. These blowers are coming 
into favor for blasting gas producers, in which air is forced through thick coal 
beds either by driving the air or by drawing on the gas produced beyond the 
bed. They are also used for forcing illuminating gas in cities through pipes 
otherwise too small, especially when the distances are long. In general very 
low pressures and large capacities are the characteristics of the service whether 
the work be that of blowing or exhausting or both. For still lower pressures, 
measurable by water or mercury columns, fans are used of the disk or propeller 

73 



74 ENGINEERING THERMODYNAMICS 

or centrifugal type. These fans are most used for ventilation of buildings and 
mines, but a modification, based on the principles of the steam turbine reversed, 
and termed turbo-compressors, is being rapidly adapted to such higher pres- 
sures as have heretofore required piston compressors. 

When high-pressure air is required for driving rock drills in mines and for 
hoisting engines, for tools, as metal drills, riveters, chipping chisels, for car 
air brakes, the compressors used to provide the air are termed simply air 
compressors. These compressors usually take in atmospheric air and compress 
it to the desired pressure, the capacity required being usually adjustable; 
they have valves of the automatic type throughout commonly, but in large 
sizes frequently are fitted with mechanically operated suction valves to 
decrease the resistance to entrance of air and so increase economy, a com- 
plication not warranted in small machines. When the pressures of delivery 
are quite high the compression is done in stages in successive cylinders, the 
discharge from the first or low-pressure cylinder being delivered through a 
water cooler or intercooler to the second cylinder and occasionally to a third 
in turn. This staging with intermediate or intercooling results in better 
economy, as will be seen later in detail, and permits the attainment of the 
desired quantity of cool compressed air for subsequent use with the expenditure 
of less work, the extra complication and cost being warranted only when 
machines are large and final pressures high. 

In the operation of large steam condensers, non-condensible gases will 
collect and spoil the vacuum, which can be maintained only by the continuous 
removal of these gases, consisting of air, carbon dioxide and gases of animal 
and vegetable decomposition originally present in the water. W^hen these gases 
are separately removed the machine used is a special form of compressor termed 
a dry vacuum pump which, therefore, receives a charge at the absolute pres- 
sure corresponding to the vacuum, or as nearly so as the entrance resistance 
permits, and after compression discharges into the atmosphere at a pressure 
in the cylinder above atmosphere equivalent to discharge resistance. Natural- 
gas wells near exhaustion can sometimes be made to flow freely by the applica- 
tion of a compressor capable of drawing a charge at a pressure below atmosphere, 
but whether the charge be received below atmospheric pressure or above as 
in normal wells, the compressor will permit the delivery of the gas to distant 
cities or points of consumption even 250 miles away through smaller pipes 
than would be otherwise possible. Natural-gas compressors, some steam- and 
some gas-engine driven, are in use for both these purposes, compressing natural 
gas from whatever pressure may exist at the well to whatever is desired at the 
beginning of the pipe line. 

In the preparation of liquid ammonia or carbonic acid gas for the market, 
as such, or in the operation of refrigerating machinery, wet or dry vapor is com- 
pressed into a condenser to permit liquefaction by the combined effect of high 
pressure and cooling. One form of refrigerating machine merely compresses 
air, subsequently expanding it after preliminary cooling by water, so that 
after expansion is complete it will become extremely cold. 



WORK OF COMPRESSORS 75 

All these compressing machines have, as a primary purpose, either the 
removal of a quantity of low-pressure gases from a given place, or the delivery 
of a quantity of higher-pressure gas to another place or both, but all include 
compression as an intermediate step between constant-pressure admission and 
constant-pressure discharge as nearly as structure may permit. They will all 
involve the same sort of physical operations and can be analyzed by the same 
principles except the wet-vapor or wet-gas compressors, in which condensation 
or evaporation may complicate the process and introduce elements that can 
be treated only by thermal analysis later. Safe compressors cannot be built 
with zero cylinder clearance, hence at the end of delivery there will remain in 
the clearance space a volume of high-pressure gases equal to the volume of the 
clearance space. On the return stroke this clearance volume will expand until 
the pressure is low enough to permit suction, so that the new charge cannot 
enter the cylinder until some portion of the stroke has been covered to permit 
this re-expansion of clearance gases. . 

It is quite impossible to study here all the effects or influences of structure 
as indicated by the compressor indicator cards, but a quite satisfactory treat- 
ment can be given by the establishment of reference diagrams as standards of 
comparison and noting the nature of the differences between the actual cases and 
the standard reference diagram. These standard reference diagrams will really 
be pressure-volume diagrams, the phases of which correspond to certain hypoth- 
eses capable of mathematical expression, such as constant pressure, constant 
volume, expansion, and compression, according to some law, or with some 
definite value of s fixing either the heat-exchange character of the process or 
the substance, as already explained. 

2. Standard Reference Diagrams or PV Cycles for Compressors and Methods 
of Analysis of Compressor Work and Capacity. All the standard reference 
diagrams will include constant-pressure lines corresponding to delivery and 
supply at pressures assumed equal to whatever exists outside the v cylinder on 
either delivery or suction side, that is, assuming no loss of pressure on delivery 
or suction. The compression may be single or multi-stage with various 
amounts of cooling in the intercooler, but in multi-stage compression 
the standard reference diagram will be assumed to involve intermediate 
cooling of the gases to their original temperature, so that the gases 
entering all cylinders will be assumed to have the same temperature and 
to maintain it constant during admission. Another difference entering into 
the classification of standard reference diagrams is the laws of compression 
as defined by the exponent s. Integration of the differential work expres- 
sion will take a logarithmic form for s = l, and an exponential form for all 
other values, thus giving two possible reference compression curves and two 
sets of work equations. 

(a) The isothermal for which s=l, no matter what the gas, and which is 
the consequence of assuming that all the heat liberated by compression is con- 
tinuously carried away as fast as set free, so that the temperature cannot rise 
at all. 



76 ENGINEERING THERMODYNAMICS 

(b) The exponential for which s has a value greater than one, generally 
different for every gas, vapor or gas-vapor mixture, but constant for any one 
gas, and also for dry vapors that remain dry for the whole process. Wet vapors 
having variable values of s cannot be treated by the simple pressure-volume 
analysis that suffices for the gases, but must be analyzed thermally. The 
adiabatic value of s is a consequence of assuming no heat exchange at all 
between the gas and anything else and is a special case of the general exponen- 
tial class. 

Just why these two assumptions of thermal condition should result in the 
specified values of s will be taken up under the thermal analysis part of this 
work, and is of no interest at this time. 

As a consequence of these phase possibilities there may be established eight 
standard reference diagrams or pressure-volume cycles defined by their phases, 
as shown in Fig. 23, four for single-stage compression and two each for two and 
three stages. These might be extended by adding two more for four stages 
and so on, but as it seldom is desirable, all things being considered, to go beyond 
three, the analysis will stop with the eight cycles or reference diagrams shown. 

Single-stage Compression Reference Cycles or PV Diagrams 

Cycle 1. Single-stage Isothermal Compression without Clearance. 
Phase (a) Constant pressure supply. 

" (6) Isothermal compression. 

" (c) Constant pressure delivery. 

" (d) Constant zero-volume pressure drop. 
Cycle 2. Single-stage Isothermal Compression with Clearance. 
Phase (a) Constant pressure supply. 

" (b) Isothermal compression. 

" (c) Constant pressure delivery. 

" (d) Isothermal re-expansion. 
Cycle 3. Single-stage Exponential Compression without Clearance. 
Phase (a) Constant pressure supply. 

" (b) Exponential compression. 

" *(c) Constant pressure delivery. 

" (d) Constant zero-volume pressure drop. 
Cycle 4. Single-stage Exponential Compression with Clearance. 
Phase (a) Constant pressure supply. 

" (6) Exponential compression. 

" (c) Constant pressure delivery. 

" (d) Exponential re-expansion. 



WORK OF COMPRESSORS 



77 



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78 ENGINEEEING THERMODYNAMICS 



Multi-stage Compression 

The phases making up multi-stage compression cycles may be considered in 
two ways, first, as referred to each cylinder and intercooler separately, or 
second, as referred to the pressure volume changes of the gases themselves 
regardless of whether the changes take place in cylinders or intercoolers. 

For example, if 10 cu.-ft. of hot compressed air be delivered from the first 
cylinder of 50 cu.-ft. displacement, the phase referred to this cylinder is a con- 
stant-pressure decreasing volume, delivery line whose length is \ of the whole 
diagram, exactly as in single-stage compression. If this 10 cu.-ft. of air deliv- 
ered to an intercooler became 8 ft. at the same constant pressure as the first 
cylinder delivery, the phase would be indicated by a constant-pressure volume 
reduction line 2 cu.-ft long to scale, or referred to the original volume of air 
admitted to the first cylinder, a line jj of its length. Finally, admitting this 
8 cu.-ft of cool air to the second cylinder and compressing it to } of its vol- 
ume would result in a final delivery line at constant pressure of a length of -J of 
the length of the second cylinder diagram, but as this represents only 8 cu.-ft., 
the final delivery will represent only ^X8 = 1.6 cu.-ft. This 1.6 cu.-ft. will, 
when referred to the original 50 cu.-ft. admitted to the first cylinder, be repre- 

1.6 
sented by a constant-pressure line, -rr=.032, of the whole diagram length, 

5U 

which in volume is equivalent to \ of the length of the second cylinder 
diagram. It should be noted also that three volume change operations take 
place at the intermediate pressure; first, first cylinder delivery; second, 
volume decrease due to intercooling; third, second cjdinder admission, the 
net effect of which referred to actual gas volumes, regardless of place where 
the changes happen, is represented by the volume decrease due to inter- 
cooling only. A diagram of volumes and pressures representing the resultant 
of all the gas processes is called in practice the combined PV diagram for the 
two cylinders, or when plotted from actual indicator cards with due regard 
for the different clearances of each cylinder the combined indicator diagrams. 
It is proper in the study of the whole process of compression to consider the 
cycle consisting of phases referred to true gas volumes rather than phases 
referring to separate cylinder processes, which is equivalent to imagining the 
whole cycle carried out in one cylinder. 

Intercooling effects measured by the amount of decrease of volume at 
constant pressure will, of course, depend on the amount of cooling or reduc- 
tion of temperature, but in establishing a standard reference diagram some 
definite amount capable of algebraic description must be assumed as an 
intercooling hypothesis. 

It has already been shown, Fig. 6, Chapter I, that from any original state 
of pressure and volume the exponential and isothermal could be drawn, diverging 
an amount depending on the difference between the defining exponent, s. 
If, after reaching a given state on the exponential curve, the gas be cooled at 



WORK OF COMPRESSORS 79 

constant pressure to its original temperature, the point indicating its condition 
will lie by definition on the other curve or isothermal and the cooling process 
is represented by a horizontal joining the two curves. Such intercooling as 
this will be defined as perfect intercooling, for want of a better name, and its 
pressure-volume effects can be treated by the curve intersections. It is now 
possible to set down the phases for the standard reference diagrams of multi- 
stage compression, if in addition to the above it be admitted, as will be proved 
later, that there is a best or most economical receiver pressure. 



Two-stage Compression Reference Cycles or PV Diagrams 

Cycle 5. Two-stage Exponential Compression without Clearance, Perfect 
Intercooling at Best Receiver Pressure. 
Phase (a) Constant pressure supply. 

(6) Exponential compression to best receiver pressure. 

(c) Constant pressure perfect intercooling of delivered gas. 

(d) Exponential compression from best receiver pressure. 

(e) Constant pressure delivery. 
(/) Constant zero-volume pressure drop. 

Cycle 6. Two-stage Exponential Compression with Clearance, Perfect 
Intercooling at Best Receiver Pressure. 
Phase (a) Constant pressure supply. 

(b) Exponential compression to best receiver pressure. 

(c) Constant pressure perfect intercooling of delivered gas. 

(d) Exponential re-expansion of first stage clearance. 

(e) Exponential compression from best receiver pressure. 
(/) Constant pressure delivery. 
(g) Exponential re-expansion of second stage clearance. 



Three-stage Compression Reference Cycles or PV Diagrams - 

Cycle 7. Three-stage Exponential Compression, without Clearance, Per- 
fect Intercooling at Best Two Receiver Pressures. 
Phase (a) Constant pressure supply. 
' ' (b) Exponential compression to first receiver pressure. 
11 (c) Perfect intercooling at best first receiver presssure, 
li (d) Exponential compression from best first to best second 

receiver pressure. 
1 ' (e) Perfect intercooling at best second receiver pressure. 
1 ' (/) Exponential compression from best second receiver pressure. 
' ' (g) Constant pressure delivery. 
" (h) Constant zero-volume pressure drop. 



80 ENGINEERING THERMODYNAMICS 

Cycle 8. Three-stage Adiabatic Compression with Clearance, Perfect 
Intercooling at Best Two Receiver Pressures. 
Phase (a) Constant pressure admission. 
' ' (b) Exponential compression to best first receiver pressure. 
1 ' (c) Perfect cooling of delivered gas at best first receiver pressure. 
1 l id) Exponential re-expansion of first stage clearance. 
" (e) Exponential compression from best first to best second 

receiver pressure. 
1 1 (/) Perfect intercooling of delivered gas at best second receiver 

pressure. 
11 (g) Exponential re-expansion of second stage clearance. 
11 (h) Exponential compression from best second receiver pressure. 
1 ' (i) Constant pressure delivery. 
' ' (j) Exponential re-expansion of third stage clearance. 

It should be noted that cycles 6 and 8 may be sub-divided into any number 
of cases, of which some of the most characteristic are shown: (a) where the 
clearance volume in each cylinder bears the same ratio to the displacement 
of that cylinder, and commonly called equal clearances; (6) where the clearances 
are such that the volume after re-expansion in the higher-pressure cylinder 
is equal to the volume of clearance in the next lower-pressure cylinder, causing 
the combined diagram to have a continuous re-expansion line, a case which 
may be called proportionate clearance; and (c) the general case in which there 
is no particular relation between clearances in the several cylinders. 

By means of these definitions or their mathematical equivalents in symbols 
it will be possible to calculate work as a function of pressures and volumes 
and by various transformations of a general expresssion for work of a reference 
cycle to calculate the horse-power corresponding to the removal of a given 
volume of gas per minute from the low-pressure supply or to the delivery of 
another volume per minute to the high-pressure receiver or per unit weight 
per minute. It will also be possible to calculate the necessary cylinder size 
or displacement per unit of gas handled, and the horse-power necessary to 
drive the compressing piston at a specified rate and further to calculate the 
work and horse-power of cylinders of given size and speed. In order that these 
calculations of a numerical sort may be quickly made, which is quite necessary 
if they are to be useful, the formulas must be definite and of proper form, the 
form being considered proper when little or no algebraic transformation is 
necessary before numerical work is possible. While special expressions for 
each case are necessary to facilitate numerical work, it is equally important, 
if not more so, to make clear the broad general principles or methods of attack, 
because it is quite impossible to set down every case or even to conceive at the 
time of writing of all different cases that must in future arise. The treat- 
ment, then, must be a combination of general and special, the general methods 
being applied successively, to make them clear and as a matter of drill, not 
to every possible case, but only to certain characteristic or type forms 



WORK OF COMPRESSORS 81 

of cases, such as are here set down as standard reference diagrams. 
Individual cases may be judged by comparison with these and certain factors 
of relation established which, being ratios, may be and are called efficiencies. 
Thus, if a single-stage compressor should require two horse-power per cubic 
foot of free air compressed per minute, and Cycle I should for the same 
pressure limits require only one horse-power for its execution, then the efficiency 
of the real compression would be 50 per cent referred to Cycle I, and similar 
factors or efficiencies for other compressors similarly obtained; a com- 
parison of the factors will yield information for a judgment of the two 
compressors. 

In what follows on the work and gas capacity of compressors two methods 
of attack will be used. 

1. General pressure-volume anal)- sis in terms of gas pressures and volumes 
resulting in the evaluation of work per cubic foot of low- or high- pressure 
gaseous substance. 

2. Transformation of results of (1) to yield volumetric efficiencies, mean 
effective pressures, work, horse-power, and capacity in terms of dimensions 
of cylinders and clearances. 

3. Single-stage Compressor, No Clearance, Isothermal Compression 
(Cycle 1), Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
and Volumes. The standard reference diagram is represented by Fig. 24, on 
which the process (A to B) represents admission or supply at constant pressure; 
(B to C) compression at constant temperature; (C to D) delivery at constant 
pressure; and (D to A) zero-volume. 

Let V b = The number of cubic feet of low pressure gas in the cylinder after 

admission, represented to scale on the diagram by AB and equal 

to the volume at B; 
V c = volume in cubic feet of the gas in cylinder when discharge begins, 

represented by DC, which is the volume at C; 
Pb = absolute pressure in pounds per square foot, at which supply 

enters cylinder = (Sup.Pr.) = pressure at B; 
p b = P b -T- 144 = absolute supply pressure in pounds per square inch = 

(sup.pr.); 
P c = absolute pressure in pounds per square foot, at which delivery 

occurs = (Del. Pr.) = pressure at C; 
Pc = Pc-+- 144 = absolute delivery pressure in pounds per square inch 

= (del.pr.); 

Pc 

" R p = ~ = ratio of delivery pressure to supply pressure; 

W = foot-pounds work done for the cycle; 

(H. P. Cap.) = volume of gas delivered in cubic feet per cycle, at 

temperature same as that of supply; 
(L. P. Cap.) = volume of gas drawn into cylinder, cubic feet per cycle. 

For this no clearance case (L. P. Cap.) = Vb. 



82 



ENGINEERING THERMODYNAMICS 



Referring to Fig. 24, the work for the cycle is the sum of compression and 
delivery work, less admission work, or by areas 



Net work ABCD = compression work EBCG+ delivery work GCDF 

— admission work EBAF. 
Algebraically this is equivalent to 



Volumes in Cubic Feet 



L.P. Cap 



w=p b v b io ge y 

(H.P. Cap) 


+P c V c -P b V b . 










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Fig. 24. — One-stage Compressor Cycle 1, No Clearance, Isothermal. 
But since P c V c = PbVb the expression becomes 



W = P b V b \og e 



P b ' 



(28) 



which is the work for the execution of the cycle when pressures and volumes 
are in pounds per square foot, and cubic feet. The equivalent expression 
for pounds per square inch and cubic feet is 



W 



144 p b V b log e ^ 
p b 



(29) 



WORK OF COMPRESSORS 83 

Since, when there is no clearance the volume taken into the cylinder for 
each cycle (L. P. Cap.) is equal to the volume at B, Vb, the expression Eq. (29) 
may be stated thus, symbolic form. 

TT=144(sup.pr.)(L. P. Cap.) logo R P (30) 

The work per cubic foot of low pressure gas, foot-pounds, will be the above 
expression divided by (L. P. Cap.), or 

! W 

(L. P. Cap.) = 144 ( su P-P r -) lo & r p ( 31 ) 

The work per cubic foot of high-pressure gas delivered will be 

W 

= 144 (sup.pr.) R v log e R p , .... (32) 





(H. P. Cap.) _x " v 


since 


P b V b = P c V c 


or 


v b =v£, 



which expressed symbolically is 

(L.P.Cap.) = (H.P.Cap.)X# P (33) 

Expressions (31) and (32) for the isothermal compressor are especialfy useful 
as standards of comparison for the economy of the compressors using methods 
other than isothermal. It will be found that the work per cubic foot of either 
low pressure or cooled high-pressure gas is less by the isothermal process than 
by any other process discussed later, and that it is the limiting case for the 
economy of multi-stage compressors with a great number of stages. The fact 
that this process^ of isothermal compression is seldom if ever approached in 
practice does not make it any the less a suitable basis for comparison. 

Example 1. Method of calculating Diagram Fig. 24. 
Assumed Data. 

P a =Pb = 2116 lbs. per square foot. V a = Va =0. 

Pc =Pa = 18,000 lbs. per square foot. Capacity =5 cu.ft. 

s = l. 

To obtain point C, 



PbV^PcVc Or Vc=wV b = ,onnn =.59. 



Pb 5X2116 
P c Vb ~ 18,000 



.59, P c = 18,000. 



84 ENGINEERING THERMODYNAMICS 

Intermediate points B to C are obtained by assuming various pressures and finding 
corresponding volumes as for V c . 

Example 2. To compress and deliver 5 cu.ft. of air from atmospheric pressure (2117 
lbs. per square foot) to 8.5 atmospheres (18,000 lbs. per square foot) isothermally 
without clearance, how much work is necessary? 

P 6 =2116 P c = 18,000 

F 6 =5 

y c =k= 8 - 5 •• Fc= - 588 

Work of admission =P b Vb =2116 X5 = 10,585 ft.-lbs. 

p 
Work of compression = P b V b log e ~ =10,585 Xloge 8.5 =22,600 ft.lbs. 

■Lb 

Work of delivery =P C V C = 10,585 ft.-lbs. 

Total work = 10,585+22,600 - 10,585 =22,600 ft.-lbs. 

Or by the general formula, 

W = (sup.pr.)(L.P.Cap.) log e fl, =2116x5 Xlog e 8.5 =2116x5x2.14=22,652 ft.-lbs. 

Prob. 1. How many cubic feet of free air may be compressed and delivered per 
minute from 14 lbs. absolute to 80 lbs. per square inc TI , absolute per horse-power in a 
compressor with zero clearance if compression is isothermal? 

Prob. 2. Gas is being forced through mains at the rate of 10,000 cu.ft. per minute 
under a pressure of 5 lbs. per square inch above atmosphere. The gas is taken into the 
compressor at atmospheric pressure and compression is isothermal. What horse- 
power will be needed at sea level and at an elevation of 5000 feet? 

Prob. 3. Natural gas is drawn from a well, compressed isothermally and forced through 
a main at the rate of 200,000 cu.ft. per hour measured at the pressure on the suction 
side. What steam horse-power will be required to operate the compressor if the 
mechanical efficiency be 80 per cent? Suction pressure is 8 lbs. per square inch 
absolute, delivery pressure 60 lbs. per square inch absolute. 

Prob. 4. A vacuum cleaning pump is required to maintain a pressure of 14 lbs. per 
square inch absolute, move 500 cu.ft. of free air per minute and discharge it against 
an atmospheric pressure of 15 lbs. per square inch absolute. What horse-power will 
be required (isothermal)? 

Prob. 5. A blower furnishes 45 cu.ft. of air a minute at a pressure of 5 ins. of mercury 
above atmosphere. Assuming compression to be isothermal and supply pressure to be 
atmospheric, what horse-power will be needed? 

Prob. 6. A compressor has a piston displacement of 3 cu.ft. At what speed can it be 
run if air be compressed isothermally from 1 to 10 atmospheres and the horse-power 
supplied is 100? 

Prob. 7. A tank of 1000 cu.ft. capacity contains air at atmospheric pressure. A 
compressor taking air from atmosphere compresses it isothermally and discharges it into 
the tank until the pressure reaches 100 lbs. per square inch gage. What horse-power 
will be required to fill tank at this pressure in ten minutes? 

Prob. 8. A compressor receives air at atmosphere and compresses it isothermally to 
five atmospheres above atmosphere. It takes in 1000 cu.ft. of free air per minute. 
How much would the capacity increase if the discharge pressure dropped to 3 atmos- 
pheres and the horse-power remained the same? 



WORK OF COMPRESSORS 



85 



Prob. 9. Suppose that the pressure in the above problem were raised to 8 atmos- 
pheres. How much would the capacity decrease if the horse-power remained the same 
and how much more power would be required to keep the capacity the same? 

Prob. 10. Ity means of suitable apparatus, the water from the side of a waterfall is 
diverted to a vertical shaft, and in falling 126 ft. compresses air from atmospheric pressure 
to a value equal to 90 per cent of the head of the water. To deliver 1000 cu.ft. of com- 
pressed air per hour, how much water is required if the work of falling water is 80 per 
cent useful in compressing the air? 

4. Single-stage Compressor with Clearance, Isothermal Compression, 
(Cycle 2) . Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
and Volumes. 



l&iUO 



14400 



& 
CO 

hi 10800 



a 7200 



Cc D 





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Fig. 25. — One-Stage Compressor Cycle 2, Clearance, Isothermal. 

Referring to Fig. 25, the work of the cycle is, by areas. 
Net work area = EBCG+GCDF 



■HADF-EBAH 
Area ABCD. 



It is easily seen that this area is also equal to (JBCL)-(JADL), both 
of which are areas of the form evaluated in the preceding section. Accordingly 



Net work area = JBCL-JADL, 



86 ENGINEERING* THERMODYNAMICS 

Algebraically, 

W = P b V b \og e ^-P a V a \og e ^ 

*b -t a 

-P»<y.-VJlog,j? (34) 

which is the general expression for the work of the cycle in foot-pounds when 
pressures are in pounds per square foot, and volumes in cubic feet. Substituting 
the symbolic equivalents and using pressures in pounds per square inch, there 
results, since (V b — 7 a ) = (L. P. Cap.), 

Work = 144 (sup.pr.)(L. P. Cap.) log e R P) .... (35) 

which is identical with Eq. (30), showing that for a given low-pressure capacity 
the work of isothermal compressors is independent of clearance. The value of 
the low-pressure capacity (V b — V a ) may not be known directly, but may 
be found if the volume before compression, V b , the clearance volume befor 9 
re-expansion, Va, and the ratio of delivery to supply pressure, R P) are known, 
thus 

V a =V d R p , 
from which 

(L. P. Cw.) = (V b -V d R p ) (36) 

From Eq. (35) the work per cubic foot of low-pressure gas is, in foot-pounds, 

W 
(L P Cap.) = 144 ( su P-P r -) lo & R p ( 37 ) 

and the work per cubic foot of high-pressure gas delivered, ft.-lbs. 

W 
r^-p-Q— ■y=144(sup.pr.)i2plog B fi„ (38) 

By comparison, Eqs. (37) and (38) are found to be identical with (31) and 
(32) respectively, since clearance has, as found above, no effect on the work done 
for a given volume of gas admitted, however much it may affect the work of the 
cycle between given volume limits or work per unit of displacement. 

It is interesting to note that the work areas of Figs. 24 and 25 are equal 
when plotted on equal admission lines AB or delivery lines CD and any 
horizontal intercept xy will be equal in length on both if drawn at the same 
pressure. 

In what precedes, it has been assumed that AB represents admission volume 
and CD represents delivery volume which is true for these established cycles 



WORK OF COMPRESSORS 87 

of reference, but it is well to repeat that for real compressors these are only 
apparent admission and delivery lines, as both neglect heating and cooling 
effects on the gas during its passage into and out of the cylinder. Also that 
in real compressors the pressure of the admission line cannot ever be as high 
as the pressure from which the charge is drawn and the delivery pressure must 
be necessarily higher than that which receives the discharge, in which cases 
the volume of gas admitted, as represented by AB, even if the temperature 
did not change, would not equal the volume taken from the external 
supply, because it would exist in the cylinder at a lower pressure than it 
originally had, and a similar statement would be true for delivered gas. 

Problems. Repeat all the problems of the last section, assuming any numerical 
value for the clearance up to 10 per cent of the displacement. 

5. Single-stage Compressor Isothermal Compression. Capacity, Volu- 
metric Efficiency, Work, Mean Effective Pressure, Horse-power and Horse- 
power per Cubic Foot of Substance, in Terms of Dimensions of Cylinder and 
Clearance. 

Consider first the case where clearance is not zero. Then Fig. 25 is the 
reference diagram. 

Let D = displacement = volume, in cubic feet, displaced by piston in one 
stroke = area of piston in sq.ft. X stroke in ft. = (F&— V&). 
(H. P. Cap.) = high pressure capacity = vol. cu.ft. of gas delivered per 

cycle at temperature equal to that of supply = (V C —Vd); 
(L. P. Cap.) = low pressure capacity = vol. in cu.ft. of gas entering 
cylinder per cycle = (F&— V a ) ; 

u yp , , • ffi • L.P. Cap. V b -V a 
E v = volumetric efficiency = =r = == — ^F J 

" Cl = volume of clearance, cubic ieet=V d 

" c = clearance volume expressed as a fraction of the displacement; 

CI V d 
" =— = — — — whence Cl = cD; 
D Vj)— V a 

W 
" M.E.P.=mean effective pressure, lbs. per square foot = yr-; 

W 

" m.e.p. = mean effective pressure, lbs. per square inch = 



1UD 
N = number of revolutions per minute; 



n = number of cycles per minute; 

N 
z = number of revolutions per cycle = — ; 

I.H.P. = indicated horse-power of compressor; 



S8 ENGINEERING THERMODYNAMICS 

The low-pressure capacity of the single-stage isothermal compressor with 
clearance is, 

(L.P.Cap.) = (7 & -7 a ), but V a = V d X~. 

Whence (L. P. Cap.) = ( V b — Vdj^j for which may be substituted the symbols 
for displacement and clearance volumes, thus 

(L. P. Cap.) = D+cD-cDR p , 

= D(l+c-cR p ) (39) 

For convenience the term, Volumetric Efficiency, E v is introduced. Since 
this is defined as the ratio of the low-pressure capacity to the displacement, 

JBt _ (L.P^C ap .) = 1+e _ c g > (4Q) 

Referring to Eq. (35) it is seen that the value of (L. P. Cap.) can be 
substituted from Eq. (39) and the result is: 

Work per cycle, foot-pounds, in terms of supply pressure, pound spersq uare 
inch, displacement cubic feet, clearance as a fraction of displacement, and ratio 
of delivery to supply pressure is, 

17=144 (sup.pv.)D(l+c-cR p )\og e R P , .... (41) 

or in the terms of the same quantities omitting clearance and introducing 
volumetric efficiency, E v , 

W = 144(sup.pr.) DEAogeRp, (42) 

To obtain the mean effective pressure for the cycle, the work done per cycle 
is divided by displacement, D. 

Mean effective pressure, pounds per square inch, 

W 
(m.e.p.)= niB -, 

whence 

(m.e.p.) = (sup.pr.)(l+c — cR p ) \og e R P (43) 

or 

(m.e.p.) = (sup.pr.)E„ log e R p , . (44) 



WORK OF COMPRESSORS 89 

The indicated horse-power of the isothermal compressor is equal to the 
svork per minute, in ft. -lbs. divided by 33,000. If n cycles are performed 
Der minute, then 

\ LH ' R = 3So = SS (^^)D(l+c-cR p )\o ge R p 

- {sUlp ' W ' )n D(l+c-cR p )\og e R p (45) 



229.2 

(sup.pr.)n 
229.2 



DE v \og e R p , (46) 



Introducing the effective area of the piston, in square inches, a, and the piston 
speed S, feet per minute, then since 

n _ a T _ a S _ aS 
144 f4l 2^~288? 

*- H - p -=^7^*> C*7) 

The same expression for the indicated horse-power may be derived by the 
substitution of the value of (m.e.p.) Eq. (44) in the following general expres- 
sion for indicated horse-power. 

= (m.e.p.)a £ 
1 ' * 33000X2z* 

Example 1. Method of calculating Diagram Fig. 25. 
issumed Data: 

P a =Pb =2116 lbs. per square foot; 
Pd=Pc =18,000 lbs. per square foot; 
c=3 per cent. L.P. Cap. =5 cu.ft. s = l. 

To obtain point D : 

From formula Eq. (39), L.P. Cap.=D(l+c -cR p ) or 5 =D(1 +.03 -.03x8.5), 
)r 

D =6.5 cu.ft. and CI. =.03 X6.5 =.195 or approximately .2 cu.ft. 

:. V d = .2 cu.ft., P d = 18,000 lbs. sq.ft. 

To obtain point A : 

PaVa-PaV a OT V a =^V d =8.5 X .2 = 1.7, 
■L a 

.'. V a = 1.7 cu.ft., P a =2116 1bs. sq.ft. 



90 ENGINEERING THERMODYNAMICS 

Intermediate points D to A are obtained by assuming various pressures and finding the 
corresponding volumes as for V a . 
To obtain point B: 

V b = V a +5=Q.7 cu.ft. P 6 = 2116 lbs. sq.ft. 

To obtain point C: 

F .v.-P,v„ . y.,m = ^f.,^.,, 

.'. V c = .79 cu.ft., Pc = 18,000 lbs. sq.ft. 

Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to 8.5 
atmospheres isothermally in a compressor having 4 per cent clearance. What must 
be the displacement work, per 100 cu.ft. of supplied and delivered air, and horse-power 
of machine? Speed is 150 R.P.M., compressor is double acting and stroke = 1.5 diameters. 
Neglect piston rods. 

D = (L.P.Cap.)-r# and ^ = (1 +.04 -.04x8.5) =.7, 
.'. D = 1000 -r- .7 = 1428 cu.ft. per minute. 

Work per cu.ft. of supplied air = (sup.pr.) 144 log e R p = 144 X 14.7 log e R p =4530 ft.-lbs.; 
Hence the work per 100 cu.ft. =453,000 ft.-lbs. 

Work per cu.ft. of delivered air = 144(sup.pr.)#p log e R p = 144 X 14.7 X8.5 X2.14 =38,550 

ft.-lbs. 

Hence the work per 100 cu.ft. =3,855,000 ft.-lbs. 

4530000 
LH - P '-^300^- 137 ' 2 - 

1428 
£>= =4.76 cu.ft. per stroke. 

-LOU /\ £ 

D = LXA = LddX^-d 2 = 1.18d 3 = 4.76. 
4 



-m*--" 



Hence cylinder diameter = I — — ■- I =1.59 feet = 19.1 inches. 



Prob. 1. How many cubic feet of free air per minute may be compressed isother- 
mally to 100 lbs. per square inch absolute in a compressor having 6 per cent clearance 
if the horse-power supplied is 60? 

Prob. 2. A compressor has a cylinder 18x24 ins., clearance 4 per cent, is double 
acting and runs at 150 R.P.M. If it compresses air from atmosphere to 100 lbs. per 
square inch gage, what will be its high- and low-pressure capacity, its horse-power, and 



WOEK OF COMPEESSOES 91 

how will the horse-power and the capacity compare with these quantities in a 
hypothetical compressor of the same size but having zero clearance? How will 
the horse-power per cubic foot of delivered air compare? 

Prob. 3. A manufacturer gives for a 10jXl2 in. double-acting compressor running 
at 160 R.P.M. a capacity of 177 cu.ft. of free air per minute for pressures of 50-100 lbs. 
per square inch gage. What clearance does this assume for the lowest and highest 
pressure if the compression is isothermal? The horse-power is given as from 23 to 35. 
Check this. 

Prob. 4. Air enters a compressor cylinder at 5 lbs. per square inch absolute and is 
compressed to atmosphere (barometer = 30| ins.). Another compressor receives air 
at atmosphere and compresses it to 3 atmospheres. If each has 6 per cent clearance 
what must be the size of each to compress 1000 cu.ft. of free air per minute, how will 
the total work compare in each machine, and how will the work per cubic foot of high 
and low pressure air compare in each? Assume compression to be isothermal. 

Prob. 5. 1800 cu.ft. of free air per minute is to be compressed isothermally to a pres- 
sure of 100 lbs. per square inch gage. What must be the displacement and horse-power 
of a hypothetical zero clearance compressor, and how will they compare with those of a 
compressor with 6 per cent clearance? 

Prob. 6. Consider a case of a compressor compressing air isothermally from atmos- 
phere to 100 lbs. per square inch gage. Plot curves showing how displacement and 
horse-power will vary with clearance for a 1000 cu.ft. free air per minute capacity 
taking clearances from 1 per cent to 10 per cent. 

Prob. 7. Two compressors of the same displacement, namely 1000 cu.ft. per min- 
ute, compress air isothermally from 50 lbs. per square inch gage to 150 lbs. per square 
inch gage. One has 5 per cent clearance, the other 10 per cent. How will their capaci- 
ties and horse-power compare with each other and with a no clearance compressor? 

Prob. 8. A 9 X 12 in. compressor is compressing air from atmosphere to 50 lbs. gage. 
How much free air will it draw in per stroke, and how much compressed air will it dis- 
charge per stroke for each per cent clearance? 

Prob. 9. The volumetric efficiency of a compressor is .95 as found from the indicator 
card. It is double acting and has a cylinder 18 X 24 ins. What will be its capacity and 
required horse-power for 100 lbs. per square inch gage delivery pressure? 



6. Single-stage Compressor, No Clearance Exponential Compression, 
(Cycle 3) . Work, Capacity and Work per Cubic Foot, in Terms of Pressures 
and Volumes. The cycle of the single-stage exponential compressor without 
clearance is represented by Fig. 26. Referring to work areas on this diagram, 



Net work ABCD = compression work EBCG 
+ delivery work GCDF 
— admission work FABE. 



Algebraically, 



W 



=~lW) >l]w.-P.7.. 



92 

But 

or 

and 



s-l 



ENGINEERING THERMODYNAMICS 

p c v c =p b v b (|?) s ~~ l =p b v b (g) 

s_-l 

PcV c -P b V b = P b V b [(^) s -l]. 



18000 



14400 



$10800 



7200 



3600 
A 



(H.P. Cap 1 1 

c Cold) '{H.P. Cap, 



\ -I 



^ 



'JV 



"1 



2 3 

Volumes in Cubic Feet 



L.P. Cap- 



Fig. 26. — One-Stage Compressor Cycle 3, No Clearance, Exponential. 
Substituting above 

•r-K[(Fr-']+w(£r-i] 



**[©*->]&+>) 



WORK OF COMPRESSORS 93 

iVhence 

* r -.-= i ft*[© i?S -i]' ■•••••• < 48 > 

Eq. (48) gives the work in foot-pounds for the execution of the cycle when 
pressures are in pounds per square foot, and volumes in cubic feet. 
The equivalent expression for pressures in pounds per square inch is 



W 



"^•[fiT- 1 ] (49) 



When there is no clearance, as before, V b represents the entire vol- 
jme of displacement, which is also here equal to the volume admitted 

[L. P. Cap.), p b is the supply pressure (sup.pr.) pounds per square inch 

Pc 

absolute and — is the ratio of delivery to supply pressure, R p . 

Accordingly, the work of an exponential, single-stage compressor with no 
clearance is 

8-1 

Tf=144^-(sup.pr.)(L.P.Cap.)^/ -lj (50) 

The work per cubic feet of low pressure gas, foot-pounds is 

8-1 



w 

144 



jij(Bup.pr.)(B/ -l) (51) 



(L.P.Cap.) s 

Before obtaining the work per cubic foot of high-pressure gas, it is neces- 
sary to describe two conditions that may exist. Since the exponential com- 
pression is not isothermal, it may be concluded that a change in temperature 
will take place during compression. This change is a rise in temperature, 
and its law of variation will be presented in another chapter. 

1. If the compressed air is to be used immediately, before cooling takes 
place, the high-pressure capacity or capacity of delivery will be equal to the 
volume at C, V c and may be represented by (H. P. Cap. hot). 

2. It more commonly occurs that the gas passes to a constant-pressure 
holder or reservoir, in which it stands long enough for it to cool approximately 
to the original temperature before compression, and the volume available 
after this cooling takes place is less than the actual volume discharged from 
the cylinder in the heated condition. Let this volume of discharge when 
reduced to the initial temperature be represented by (H. P. Cap. cold) which 
is represented by Vt, Fig. 26. 



94 ENGINEERING THERMODYNAMICS 

Since B and C in Fig. 26 lie on the exponential compression line, P b V b =P C V C S , 

. w # 

or 

(L. P. Cap.) = (H. P. Cap. hot) (R p y (52) 

Hence, the work in foot-pounds per cubic foot of hot gas delivered from 
compressor is 

W 



(H. P. Cap. hot) 



H4^-(sup.pr.)^^V-lj. ... (53) 



On the other hand, B and K lie on an isothermal and P b V b = PtV^ or 
since P k = P c , 

Vb=Vk % 

whence 

(L. P. Cap.) = (H. P. Cap. cold)£ p (54) 

The work foot-pounds per cubic foot of gas cooled to its original tempera- 
ture is, therefore, 



W 



(H. P. Cap. cold) 8-1 

or 



s-1 



= 144— -(sup.pr.)#Ji2*~r---l , .... (55) 



= 144-*- (del. pr.) (rJ~T - 1 



(H. P. Cap. cold) s-l 

This last equation is useful in determining the work required for the storing 
or supplying of a given amount of cool compressed air or gas, under conditions 
quite comparable with those of common practice. 

Example 1. Method of calculating Diagram Fig. 26. 
Assumed Data: 

p a =p b =2116 lbs. per square foot; P c =Pd = 18,000 lbs. per square foot. 
Cl=Q; V a = Vd=0; L. P. Capacity = 5 cu.ft. ; s = 1.4 (adiabatic value of s). 

To obtain point C: 

p c 7 c i-4 =P b V b 1A or Vc = V b + (~) 1A 

l 
P t /P 6 =8.5; log e 8.5 = .929, and .71 log e 8.5 = .665; (P c /P b ) 1A =4.6, 



WOEK OF COMPRESSORS 95 

hence V e =5*4.6 =1.09 eu. ft. P c = 18,000 lbs. per sq.ft. 

Intermediate points B to C are obtained by assuming various pressures and finding 
the corresponding volumes as for V c . 

Example 2. To compress 5 cu.ft.of air from atmospheric pressure (2116 lbs. per 
square foot) to 8.5 atmospheres (18,000 lbs. per square foot) adiabatically and with no 
clearance requires how many foot-pounds of work? 

P & =2116 lbs. sq.ft., P c = 18,000 lbs. sq.ft., 

V b =5 cu.ft. 

Vh 
V c =jpT 71 =5*4.57 = 1.092 cuit. 

W 
Work of admission is 

P b V b =2116 X5 = 10,585 ft.-lbs. 
Work of compression, using y to represent the adiabatic value of s is, 

^[(g^-l]=»[(8.5)--l]=^x.860=22,350ft.-lbs. 

Work of delivery is 

P C V C = 18,000 X 1.092 = 19,650 ft.-lbs. 

Total work = 19,650+22,350-10,585=31,425 ft.-lbs., 

or by the formula Eq. (50) directly 

TF = 144^(sup. pr.) (L. P. Cap.) (r p ~ -1.) 

= 144+3.46 X2116X5X[(8.5) 29 -1]; 

= 144+3.46 X2116 X5 X.86 =31,450 ft.-lbs. 

Prob. 1. A single-stage zero clearance compressor compresses air adiabatically from 
t to 6 atmospheres. How many cubic feet of free air per minute can be handled if the 
3ompressor is supplied with 25 H.P. net? 

Prob. 2. The same compressor is used for superheated ammonia under the same 
pressure conditions. For the same horse-power will the capacity be greater or less 
md how much? 

Prob. 3. A dry-vacuum pump receives air at 28 ins. of mercury vacuum and delivers 
t against atmospheric pressure. What will be the work per cubic foot of low-pressure 
lir and per cubic foot of high-pressure air hot? Barometer reads 29.9 ins. 



96 ENGINEERING THERMODYNAMICS 

Prob. 4. The manufacturer gives for a 10ixl2 in. double acting compressor run 
ning at 160 R.P.M., a capacity of 177 cu.ft. of free air per minute and a horse-power 
25 to 35 when delivering against pressures from 50 to 100 lbs. Check these figures. 

Prob. 5. A set of drills, hoists, etc., are operated on compressed air. For their opera 
tion 3000 cu.ft. of air at 70 lbs. gage pressure are required per minute. What must b 
the piston displacement and horse-power of a compressor plant to supply this air 
compression is adiabatic and there is assumed to be no clearance? 

Prob. 6. Air is compressed from atmosphere to 60 lbs. per square inch gage by 
compressor having a 12x18 in. cylinder and running at 100 R.P.M. Find its capacity 
and horse-power at sea level and loss in capacity and horse-power if operated at an alti 
tude of 10,000 ft. for zero clearance. 

Prob. 7. 10,000 cu.ft. of free air per minute at a pressure of 15 lbs. above atmo^ 
phere are compressed and delivered by a blowing engine. Find the horse-power requirec 
to do this and find how much free air could be delivered by same horse-power if th 
pressure were tripled. 

Prob. 8. In a gas engine the mixture of air and gas is compressed in the cylinder 
before ignition. If the original pressure is 14 lbs. per square inch absolute, final 
pressure 85 lbs. absolute and compression is adiabatic, what will be the work of 
compression only, per pound of mixture? 

Note: Weight per cubic foot may be taken as .07 and y as 1.38. 

Prob. 9. A vacuum pump is maintaining a 25-in. vacuum and discharging the air 
removed against atmospheric pressure. Compare the work per cubic foot of low pres- 
sure air with that of a compressor compressing from atmosphere to 110 lbs. above atmos- 
phere. 

7. Single-stage Compressor with Clearance, Exponential Compression, 
(Cycle 4). Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
and Volumes. When clearance exists in the cylinder, it is evident that a 
volume equal to the clearance, Va, will not be expelled during the delivery 
of compressed gas, and this volume will expand with fall in pressure as the 
piston returns, causing pressure-volume changes represented by the line DA 
on the diagram, Fig. 27. Until the pressure has fallen to that of supply, the 
admission valve will not open, so that while the total volume in the cylinder 
at end of admission is Vb, the volume V a was already present by reason 
of the clearance, and the volume taken in is (Vb — V a ) which is the low-pressure 
capacity (L. P. Cap.). 

The work area of the diagram is ABCD, which may be expressed as 



Work 3ire2L = JBCL-JADL, 

which areas are of the form evaluated in Section 6. Hence, the above expression 
in algebraic terms is 

s-l s-l 

'Pc\~ 



S— 1 



-i]-^-"4(r) ' -] 
-A**-'' .#;f -] (57) 



WORK OF COMPRESSORS 



97 



This is the general expression for the work of the cycle, in foot-pounds, when 
the pressures are expressed in pounds per square foot, and volumes in cubic 
feet. Using symbolic equivalents 



W = 144 ~i (sup.pr.)(L.P.Cap.) [(£„)¥ -1,1 



(58) 



(H.r.,Cap.Hot) 



1801X) 



1441)0 



c-10800 



7200 



3600 
J 



__j U J — J _.. 1 




-+- -j- 




-4n -U- 




-+U -*Qr 




> r - 




11 




ZJT 




in 




_ 4t 




Xt-~ ::: : : :: 




XX 




X* 








z: X~l~ ~ ' 




1 t v 




X A 




X \ 




\ 




IZ} i--3 




:4;::::::::1::::#:::::::::::::::::: 




:iE: i::::fi:::::::::::::::::: 




i ,5 j U 




n^ r v 




| Ire \ v. 


■— 


--Rf ~ —}--- ::W - — - : : 




=H* _ _rt__ __$_ _ 


1 "3 


-»-T ^ 


Q 


^ 




\ 




. 




^ . 




^ j. 




'T ^v 




\ ^i 




* ^> k 




X r X V * - 




-£-$- __:L -^.-- 




__j -\- j >>« 




\ 


* * s 


--+M ^c 1 


Z = = ^,-___ 


S! *±U iEi rLL^U =44== = = = = = = = = = = 44= = = = = — == = = 

"X A "~ X~~ 


t-wt 






i ii 


a 


1 i J i 1 1 1 Lu i i i 1 1 1 — 1 [I il I 1 i II 1 ■ M 1 1 I 1 


::::::::::::::::::::::t ®:: 



fan' 



Hi G 2 3 

! Volumes in Cubic Feet 

« U.P. Cap.)- 



E;0 



D- 



l^p 

Fig. 27. — One-Stage Compressor Cycle 4, Clearance, Exponential. 

Eq. (58) is identical with Eq. (50), showing that for adiabatic as for 
isothermal compressors, the work done for a given low-pressure capacity is inde- 
pendent of clearance. Due to this fact, the expressions derived for the expo- 
nential compressor without clearance will hold for that with clearance: 



Work, in foot-pounds per cubic foot of low-pressure gas is, 
W 



s-l 



(L.P.Cap.) = 144 ^ (sUp - pr - )W 



1) 



Work, in foot-pounds per cubic foot of hot gas delivered is, 
W 



s-l 



(H. P. C»p. hot) " l«jr i <«W->S> 1 *> 



(»• -,) 



(59) 



(60) 



98 ENGINEERING THERMODYNAMICS 

Work, in foot-pounds per cubic foot of cooled gas to its original temperature is, 

s_-l_ 

(H.P.CaT.coId) - 144 ^I< del ^(^ - 1 )- ' • / • (61) 

The relation of high-pressure capacity either hot or cold to the low-pressure 
capacity is also as given for the case of no clearance, as will be shown. 

In Fig. 27, the high-pressure capacity, hot, is DC=V C — V d . The low- 
11 11 

pressure capacity is AB=V b —V a , but VcP c s = V b P b s and IW =VaP a 8 , or 

i i 

V b = VcRp* and V a = VaRp' . 

f i 
Hence (L. P. Cap.) = (H. P. Cap. hot)R p ' (62) 

If the delivered gas be cooled to its original temperature, then the volume 
after delivery and cooling will be 

(H.P.Cap.cold)~ (L - P - Cap0 t 
K p 

or 

(L.P. Cap.) = (H. P. Cap. co\d)R p .... (63) 

From the work relations given above, it is. seen that in general, the work 
per unit of gas, or the horse-power per unit of gas per minute is independent 
of clearance. 

8. Single-stage Compressor Exponential Compressor. Relation between 
Capacity, Volumetric Efficiency, Work, Mean Effective Pressure, Horse- 
power and H.P. per Cubic Foot of Substance and the Dimensions of Cylinder 
and Clearance. As indicated on Fig. 27, for the single-stage exponential com- 
pressor with clearance, the cylinder displacement D, is (V b — V&). The low- 
pressure capacity per cycle is (L. P. Cap.) = (V b — V a ). The actual volume of 
gas or vapor delivered by the compressor is (H. P. Cap. hot) = (V c — V d ). 
This is, in the case of a gas at a higher temperature than during supply, 
but if cooled to the temperature which existed at B will become a less volume. 
This delivered volume after cooling is symbolized by (H. P. Cap. cold) and is 

,-r ^ ^ s Csup.pr.) (L. P. Cap.) _ . ' 

equal to (L. P. Cap.) X7-— ; or where R P is the ratio of delivery 

(del.pr.) R P 

pressure to supply pressure. 

Volumetric efficiency, E v , already defined as the ratio of low-pressure 

capacity to displacement is 

w V»-V a (L.P. Cap.) 



WORK OF COMPRESSORS 99 

Clearance, c, expressed as a fraction of the displacement is the ratio of 
clearance volume, CI, to displacement, D, and is, 

CI V d 
c 



D V b -V d 

Mean effective pressure, pounds per square foot (M.E.P.), is the mean height 
of the diagram or the work area W, divided by displacement, D. If expressed 
in pounds per square inch the mean effective pressure will be indicated by 

W 
(m.e.p.) = IS5 -. 

Let (I.H.P.) be indicated horse-power of the compressor; 
N the number of revolutions per minute; 
n the number of cycles per minute and 
z the number of revolutions per cycle, whence nXz = N. 

Then, the low-pressure capacity is 

(L. P. Cap.) = (V b -V a ). 

But 

1 



v a =v d x x 

since the re-expansion DA is exponential and similar to compression as to 
value of s, whence j 

1 
(L. P. Cap.) = (V b - Va) = V b - V d R P * ; 

i 
=D+Cl-Vdiy; 

i 
= D+cD-cDR P s ; 

or 



(L. P. Cap.)=2>(l+c-cfi P «) 
From this, by definition, the volumetric efficiency is 



(64) 



E ^ JL.P.Ca.p.) = 1+c _ cR i (6g) 

Referring to Eq.(57), in which may be substituted the value Eq. (64) for 
(Vt— V a ), the work of the single-stage exponential compressor in terms of dis- 



100 ENGINEERING THERMODYNAMICS 

placement, clearance (as a fraction of displacement), and pressures of supply 
and delivery in pounds per square foot is, 

or using pressures, pounds per square inch, and inserting the symbols, this 
may be stated in either of the following forms: 

Tf=144^-(sup.pr.)L>(l+c-c^)r^ 1 V-ll. . . . (67) 
= 144 s -^(sup.pr.)D^[^ £ ^" 1 -l] (68) 

The mean effective pressure in pounds per square foot is this work divided 
by the displacement, in cubic feet, and may be converted to pounds per square 
inch by dividing by 144, whence 

Mean effective pressure, pounds per square inch, 

(m.e.pO=™(sup.prO(l+c-cE^)r^V 1 -ll, . . (69) 



s 



(sup.prO^Ji^V-lj (70) 



The indicated horse-power of the single-stage exponential compressor from 

(67) is, 



TT jp _ Wn __s_ (s\ip.pr.)nDE v \ p ±J_ 
itLr \- 33000 "s-1 229.2 



[Vr 1 - 



(71) 



Where n is the number of cycles per minute, or in terms of piston speed S and 
effective area of piston, square inches, and z the number of revolutions per 
cycle, 

THP _^_ (sup.pr.)aSfl , [ R *^ J (72) 



Since it was found in Section 7, that the work per unit volume of gas is the 
same with clearance as without clearance, the horse-power per cubic foot per 
minute will also be independent of clearance. (See Eqs. (51), (53) and (56) ). 



Horse-power per cubic foot of gas supplied per minute 



I.H.P. s (sup.pr.) 



n(L.P.Cap.) s-1 229.2 



a 



--l] . ... (73) 



WORK OF COMPRESSORS 101 

The horse-power per cubic foot of hot gas delivered per minute is 

I.H.P. _ 8 (sup .pr.) I \ Li* 1 

n(H.P.Cap.hot)~s-l 229.2 Kp [ Kp l \ ' ' ' U4) 

Horse-power per cubic foot of gas delivered and cooled is 

I-H.P. 8 (sup.pr.) [" s _^i 1 , 

rc(H.P. Cap .cold)" s-l 229.2 Kp [ Kp & l \> ' ' ' U5j 

_ s (del.pr.) [ i^i 1 

In the above formulae (del.pr.) and (sup.pr.) indicate delivery pressure and 
supply pressure, in pounds per square inch. 

Example 1. Method of calculating Diagram, Fig. 27. 
Assumed data: 

Pa = Pb =2116 lbs. per square foot. 

P c =P d = 18,000 lbs. per square foot. 

CI. =3.5 per cent. L. P. Capacity =5 cu.ft. s = 1.4. 
To obtain point D: 

L. P. Cap. =D(l+c-cR p 7) or 5 =D (l+. 035 -.035(8.5)) ; 

Hence 

D =-5 -Kl +.035 -.035x4.6) =5.72 cu.ft. and CI = .035x5.72 = .2 cu.ft. 

.*. V d = .2 cui t. ; P d = 18,000 lbs. sq.ft. 



To obtain point A : 



Mr 



= 4.6X.2=.92; 
.*.- F a = .92 cu.ft.; P a =2116 lbs. sq.ft. 

Intermediate points D to A are obtained by assuming various pressures and finding the 
corresponding volumes as for F a . 

To obtain point B: 

V b = V a +L. P. Cap. =.92 +5 =5.92. 

.*. F & = 5.92 cu.ft.; P & =2116 lbs. sq.ft. 



102 ENGINEERING THERMODYNAMICS 

To obtain point C: 

= 5.92 -5-4.6 = 1.29 cu.ft. 
.'. V c = 1.29 cu.ft. ; P c = 18,000 lbs. sq.ft. 

Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to 8.5 
atmospheres absolute so that s = 1.4, in a compressor having 4 per cent clearance. What 
must be the displacement of the compressor, work per 100 cu.ft. of supplied and delivered 
air, hot and cold, and horse-power of machine? Speed is 150 R.P.M., compressor is 
double acting and stroke = 1.5 diameters. 

D =L. P. Cap.-v-^p, and E v = (l+c-cR p 7 ) . 

.*. E v = (l+.04-.04x(8.5)- 71 ) =.86; 
/. D = 1000 -r .86 =1162 cu.ft. per min. 

s-1 

Work per cubic foot of supplied air = 144 (sup.pr)[^ « — 1], 

o JL 

= 144 X3.46 X14.7 X.86 =6300 ft.-lbs. 
.'. Work per 1000 cu.ft. =6,300,000 ft.-lbs. 



Work per cubic foot of delivered air cold is R p times work per cubic foot of supplied air, 
hence work per 100 cu.ft. of delivered cooled air is 5,350,000 ft.-lbs. 

L 
Work per cubic foot of delivered air hot is R p s times work per cubic foot of supplied air, 

hence work per 100 cu.ft. of hot delivered air is 2,800,000 ft.-lbs. 

I.H.P. = ( ^£gS. ,-J; (m.e.p.)=^- 1 (sup.pr.)E{iJ^-l,] 



or 



66,0002 



(m.e.p.) =3.46x14.7 X.86 X.86 =37.7 lbs. per square inch. 

.'. d 3 =5690 or d = 17.85. 
a =250 sq.inches. S =670 ft. per min. .'. I.H.P. =191. 



WORK OF COMPRESSORS 103 

Prob. 1. A dense-air ice machine requires that 4000 cu.ft. of air at 50 lbs. per square 
inch absolute be compressed each minute to 150 lbs. per square inch absolute. The 
compression being such that s = lA, clearance being 6 per cent, find the work required. 
What would be the work if clearance were double? Half? 

Prob. 2. The compressor for an ammonia machine compresses from one atmos- 
phere to 8 atmospheres absolute. With adiabatic compression and 4 per cent clear- 
ance, what will be work per cubic foot of vapor at the low pressure and at the high? 
Assume vapor to be superheated. 

Prob. 3. On a locomotive an air-brake pump compresses air adiabatically from 
atmosphere to 80 lbs. per square inch gage. It is required to compress 50 cu.ft. of free 
air per minute and clearance is 5 per cent. What horse-power must be supplied to it? 

Prob. 4. In a manufacturing process a tank must be maintained with a vacuum of 
29 ins. when barometer reads 30 ins. To do this 100 cu.ft. of carbon dioxide must be 
removed from it per minute and returned under atmospheric pressure. Compression is 
adiabatic and clearance 7 per cent. How much power must be supplied to compressor 
and what should be its displacement? 

Prob. 5. Two compressors each 12x18 in., double acting, with 8 per cent clear- 
ance, and running at 150 R.P.M. compress in the one case air, in the other carbon disul- 
phide. The compression being adiabatic in each case, what (a), is the difference in 
power required, (6), in low-pressure capacities? Take pressures as 2 and 15 atmos- 
pheres of 26 inches mercury. 

Prob. 6. A compressor is supplied with 40 horse-power. If it draws in air from 
atmosphere to what pressure can if 500 cu.ft. per minute be compressed, when s = 1.38 
and clearance 10 per cent? 

Prob. 7. For forcing gas through a main, a pressure of 50 lbs. per square inch gage 
is required. What is the work done per cubic foot of high-pressure gas, if a compressor 
having 6 per cent clearance is used, and s for the gas is 1.36? What should be its dis- 
placement? 

Prob. 8. A gas compressor 20x22 ins. has a volumetric efficiency of 90 per cent, 
supply pressure =4 lbs. per square inch and delivery 110 lbs. per square inch gage. 
What are its L. P., H. P., hot and cold capacities, and its I.H.P. if single acting at 70 
R.P.M. when s = 1.35? 

9. Two-Stage Compressor, no Clearance, Perfect Intercooling, Exponential 
Compression, Best Receiver Pressure, Equality of Stages (Cycle V). Work and 
Capacity in Terms of Pressures and Volumes. The common assumption in con- 
sidering the multi-stage compressor is that in passing from one cylinder to the 
next, the gas is cooled to the temperature it had before entering the com- 
pressor, which has already (Section 2), been defined as " perfect intercooling." 

This condition may be stated in other words by saying that the product 
of pressure and volume must be the same for gas entering each cylinder. If 
then the volume and pressure of gas entering the first stage be determined, 
fixing the volume entering the second stage will determine the pressure of the 
gas entering the second stage, or fixing the pressure of the gas entering the 
second stage will determine the volume that must be taken in. 

Using subscripts referring to Fig. 28, for the no clearance case, 

Pb y b = P dVd (77) 



104 ENGINEERING THERMODYNAMICS 



T he net work of the compressor, area ABC DEF = area ABCH first stage -f- 
area HDEF second stage. Using the general expression, Eq. (48) for these 
work areas with appropriate changes in subscripts 



-l 



W "i=i PtV, [(^) S - 1 ]- • (fi^t stage) 
H — ZTi^"^ (p^) ~~M* ' ( sec o n d stage) 



But from the above, and since P c = P d , 

T ^v4(8) v+ (£r-4 (78) 



which is the general expression for work of a two-stage compressor without 
clearance, perfect intercooling, and may be restated with the usual symbols as 
follows : 



F = 144~- (sup.pr.)(L. P. Cap.)f'(Bjn)V 1 +(i^ 2 )V-2l > 



(79) 



in which (R P i) and R&) are the ratios of delivery to supply pressures for the 
first stage and for the second stage respectively. From Eq. (79), work per 
cubic foot of gas supplied is, 

w s r s_i s ~ i "i 

(L. P Cap O = 144 ^-"I (sup - pr - } [ B > 1 ~ +fr*~ ~ 2 J • • ( 8 °) 
Work per cubic foot of gas discharge and cooled to its original temperature is 

(H. P. Cap. cold) =14 V-T (MP-Pr-)ft [fti ' +**>* s ~ 2 J> 

= 144^- (del.pr.)|^ S ~V ^2^-2! (81) 

The low-pressure capacity stated in terms of high-pressure capacity hot, 
as actually discharged is 

(L. P. Cap.) = (H. P. Cap. hot)R p2 T R P i, (82) 

whence 

Work per cubic foot hot gas discharged 

(H. P. Cap. hot) = l^(s»P-Pr-)ft>*4^ +ft. . -2j . (83) 



WORK OF COMPRESSORS 



105 



Examination of Fig. 28 will show without analysis that there must be some 
best-receiver pressure at which least work will be required. For if the receiver 
pressure approached P b then the compression would approach single stage and 




jootf 9.retibs asd spunojL ut soanssejj 



the compression line approach BCG. The same would be true as the receiver 
pressure approached P g = P e , whereas at any intermediate point C, intercooling 
causes the process to follow BC DE wit h a saving of work over single-stage 
operation represented by the area DCGE. This area being zero when C is at 



106 ENGINEERING THERMODYNAMICS 

either B or G, it must have a maximum value somewhere between, and the 
pressure at which this least-compressor work will be attained is the best-receiver 
pressure. 

By definition the best-receiver pressure is that for which W is a minimum, 
or that corresponding to 

dW 






ip.-°- 






Performing this differentiation upon Eq. (78), equating the result to zero, 
and solving for P c , 

(Best rec.pr.) = (P b Pe)° [(sup. pr.) (del. pr)] § . .... (84) 

Substituting this value in the general expression for work Eq. (78), noting that 

^_(W_/PA* and £ = /PA* 



p» p» 



s-1 

Tr=2^n[g)" 28 -l], .... . (85) 

Eq. (85) is the general expression for two-stage work with perfect inter- 
cooling at best-receiver pressure in terms of pressures and volumes. Sub 
stituting the symbols for the pressures and volumes and noting that as in 
Cycle 1, 

V b =(L. P. Cap.) and 7 e =(H. P. Cap. hot) and using,. (#„) for fj\ 

TF = 288 g ^ r j-(sup.pr.)(L. P. Cap.) (fl/aT-l) ( 8 6) 

This equation gives the same value as Eq. (85), but in terms of different units. 

It should be noted here that the substitution of best-receiver pressure in the 
expressions for the two stages preceding Eq. (78), will show that the work done 
in the two cylinders is equal. 

The work per cubic foot of low-pressure gas, from Eq. (86) is, 

W s , v r Fs-i 

288 



(L.P.Cap.) 



—-(sup.prO^V-l] (87) 



To transform Eq. (85) into a form involving delivery volumes, use the rela- 
tion from the diagram, 



i i 



^m-<mi- 



WORK OF COMPRESSORS 107 

Whence 

V b = V e 



Hft)U) ' 
which for the best-receiver pressure becomes 



£+1 

V b =VeR P 2s . 



Substituting in Eq. (85), 

W = 2-^P b V e (R p ) S ^[Rp^-l\ (88) 

Introducing the symbols, 



-(sup.prO(H. P. Cap. hot)^^^^ 1 -!! . . (89) 



and 

(H.P.clp.hot) =288 ,4 1 ( S "P-P---)^[^ 1 -lj • • • (90) 

The volume of gas discharged at the higher pressure when reduced to its 
original temperature will become such that 

(L. P. Cap.) P e „ 



(H. P. Cap. cold) P b 
or 

(sup.pr.) (L. P. Cap.) = (del.pr.) (H. P. Cap. cold), . . . (91) 

which may be substituted in Eq. (86), 

W = 288^ (del.pr.) (H.P. Cap. cold)lfl/~ir- 1,1. . . . (92) 
from wnich the work per cubic foot of gas delivered and cooled is, 



W 



(H.P. Cap. cold) 



= 288^ (del.pr.) \rJ^T - ll 



(93) 



108 ENGINEERING THERMODYNAMICS 

Example 1. Method of calculating diagram, Fig. 28. 
Assumed data: 

V a =0 cu.ft. P a =2116 lbs. per square foot. 
V f = cu.ft. P c =P d = VPaPe =6172 lbs. sq.ft. 
F 6 = 5 cu.ft. P/=P e =P,= 18,000 lbs. sq.ft. 

5=1.4. 



To obtain point C: 



i_ 

.4 



2.36 cu.ft.: 



or 



To obtain point D : 



To obtain point E: 



but by definition 



\ 7 C =2.36 cu.ft. P c = 6172 lbs. sq.ft 



F (i = nx^=5x|| = 1.71cu.ft. 



F d = 1.71 cu.ft. P d =6172 sq.ft. 

r - v Mr) h . 



hence, 



Ye = 1.71-4-2.14 = .8 cu.ft. P e = 18000 lbs. sq.ft. 

Example 2. To compress 5 cu.ft. of air from one atmosphere (2116 lbs. per square 
foot) to 8.5 atmospheres (18,050 lbs. per square foot) in two stages with best-receiver 
pressure and perfect intercooling requires how much work? 



^=288— -(sup.pr.)(L. P. Cap.)(fl p 2s -1), 
s — 1 

(sup.pr.) =14.7. (L. P. Cap.) =5. i^=8.5. 
/. W =288 X3.463 X14.7 X5 X (8.5 _2s ~ -l) =26,800 ft.-lbs. 



WORK OF COMPRESSORS 109 

Prob. 1. Air at 14 lbs. per square inch absolute is compressed to 150 lbs. per square 
inch absolute by a two-stage compressor. What will be the work per cubic foot of air 
delivered? What will be the work per cubic foot if the air be allowed to cool to the 
original temperature, and how will this compare with the work per cubic foot of sup- 
plied air? Best receiver-pressure and perfect intercooling are assumed for the above 
compressor, s =1.4. 

Prob. 2. A compressor receives air at atmosphere and compresses it to half its vol- 
ume, whereupon the air is discharged to the cooler and its temperature reduced to the 
original point. It then enters a second cylinder and is compressed to 80 lbs. absolute. 
What will be the work per cubic foot of supplied air in each cylinder and how will 
the work of compressing a cubic foot to the delivery pressure compare with the work 
done if compression were single stage, compression being adiabatic. 

Prob. 3. Air is to be compressed from 15 lbs. per square inch absolute to 10 times 
this pressure. What would be the best-receiver pressure for a two-stage compressor? 
How many more cubic feet may be compressed per minute in two stage than one stage 
by the same horse-power? 

Prob. 4. A manufacturer sells a compressor to run at best-receiver pressure 
when (sup.pr.) is 14 lbs. per square inch absolute and (del.pr.) 100 lbs. per square inch 
absolute. What will be the work per cubic foot of supply-pressure air done in each cylin- 
der? Another compressor is so designed that the receiver pressure for same supply 
pressure and delivery pressure is 30 lbs. per square inch absolute, while a third is so 
designed that receiver pressure is 50 lbs. per square inch absolute. How will the 
work done in each cylinder of these machines compare with that of first machine? 

Prob. 5. For an ice machine a compressor works between 50 and 150 lbs. per square 
inch absolute. It is single stage. Would the saving by making compression two stage 
at best-receiver pressure amount to a small or large per cent of the work in case of single 
stage, how much? 

Prob. 6. A compressor has been designed to compress 1000 cu.ft. of carbon dioxide 
per minute from 15 to 150 lbs. per square inch absolute. What horse-power will be 
required at best-receiver pressure? Should delivery pressure change to 200 lbs., what 
power would be required? To 100 lbs. what power? 

Prob. 7. A gas-compressing company operates a compressor which has to draw 
C0 2 gas from a spring and compress it to 150 lbs. per square inch gage. In the morning 
pressure on the spring is 10 lbs. gage, while by evening it has dropped to 5 lbs. absolute. 
If the compressor was designed for the first condition, how will the high-pressure 
capacity cold and horse-power per cubic foot of high-pressure gas at night compare 
with corresponding values in morning? Assume a barometric reading. 

Prob. 8. On a mining operation a compressor is supplying a number of drills and 
hoists with air at 150 lbs. per square inch absolute, the supply pressure being 14 lbs. 
What will be the difference in horse-power per cubic foot of delivered air at compressor 
and per cubic foot received at drills if air is a long time in reaching drills? 

Prob. 9. With a best-receiver pressure of 40 lbs. per square inch absolute and a 
supply pressure of 14 lbs. per square inch absolute, what horse-power will be required 
to compress and deliver 1000 cu.ft. of high-pressure air per minute at the delivery 
pressure for which compressor is designed and what is that delivery pressure? 

10. Two-stage Compressor, with Clearance, Perfect Intercooling Expo- 
nential Compression, Best-receiver Pressure, Equality of Stages, (Cycle 6). 
Work and Capacity in Terms of Pressures and Volumes. The two-stage expo- 



110 



ENGINEERING THERMODYNAMICS 



nential compressor with clearance and perfect inter cooling is represented by 
•the PV diagrams Figs. 29, 30, 31, which are clearly made up of two single-stage 
compression processes, each with clearance. 

.a 





t 

53 

I 

8 



'** ('jj 'pa 








! ' u-j^-dnsy. 




1 CQ(.---^[r:- r 




^ -i h 


/i 




4 4- 






3 i- 






1 I- 






4 r 






h i_ 






4 -i - 






4 I ~ «> 






i t - 






t: f _ 












i- 1 ^ " 






4 A- - 






1 t - 






. i 






-4 I - 






-» £+ - 












7 






r L ' 






4 z2 ■§ 






1 ^H $ 






4 J£/ - w 






• $t o 






1 oSV - S 






J r£3- 3 






-♦-nr -' ^ 








-i 




"I Z^ 'S 












o *- o 






2 : a 






-<1 _ ± 3 






^' i o 






A X >^ 






^' ± ft 






^' 3- r*3 












> ""I ** 




+* _.'* j 


^3 -I • 




° -"* 


S/ 4 .. . J 




ta- W O*'^ -"'" 






■^ a 1- -<rvpOj^V"" 


*? -tl _J :: : 










rf ^ Ol •-'■" t <!X®£.*'* 


- 1 "" 




%£ • vs^Bh^ - 


r r | 




O M -3=^ „*' 


~f qj_ 




. W LU ^ -" »■-' 












w " . i^fl-t* ■"'"'"' 


iz: "2-iS 






4 '- *' h " 




r;, " 








X s * 




I ^ 


'^l^''-" { pr " 


i 


V 




■ 


nil Eii -~ — Z11- -- 


=-r=l===l|L= = T=::-:== 




\xr 


.::„„:iL„^„„:±if ,„ 


L8000 

4400 
0800 


1 2 ^ I " ? 



O 
O 

53 ° 



a? 



O 



sooj aaranbg aad sqi ut sextissaaa 



u 



Applying Eq. (57) to the two stages and supplying proper subscripts, 



referring to Fig. 29, 



W 



3-1 



P»(V b -V a ) 



[©*-'] 



(first stage) 



WORK OF COMPRESSORS HI 

a-1 



}--^P d (V d -V) h 
s 1 



( pM — 1 . . (second stage) 



s-l ~i 



If the condition of perfect intercooling be imposed, it is plain that since 
the weight of gas entering the second stage must equal that entering the first 
stage, and the temperature in each case is the same, 

(V d -V h )P d =(V b -Va)P b , 

and noting also that 

Pc = P d , 

TF= _l_p i(FS _ Fa) [(l_ : ) - +[¥ J - _ 2 j, . . . m 

Eq. (94) is the general expression for the work of two-stage exponential 
compressor with perfect intercooling, P c being the receiver pressure. 

Pc 

As in Section 9, let (R P i) be the pressure ratio — for the first stage and(ff P 2) 

P b 
Pe 

the pressure ratio — for the second stage and using instead of P& its equivalent 

P c 

144 (sup.pr.) lbs. per square inch. 

T^ = 144^ 1 (sup.pr.)(L.P. Cap.) [(^0^ + (^2,^-2,1 . . (95) 

which is identical with (79), showing that for two-stage compressors with perfect 
intercooling (as for single stage, Section 7), the work for a given low-pressure 
capacity is independent of clearance. 

The work per cubic foot of gas supplied is given by Eq. (80); per cubic 
foot of cold gas delivered by Eq. (81) and per cubic foot hot gas delivered by 
Eq. (83). 

The reasoning regarding best-receiver pressure followed out in Section 9, will 

dW 
hold again in this case, and by putting — = in Eq. (94), and solving for P c 

dP c 

it will again be found that best-receiver pressure will be 

(best-rec. pr.) = {P b P e f ......... (96) 

Substitution of this value for P c in Eq. (94), gives the following expression 
forwork of the two-stage exponential compressor with best-receiver pressure, 

8-1 

^ = 2 ,4fW-^)[(g) " -l], (97) 



112 ENGINEERING THERMODYNAMICS 

which may be expressed in terms of supply pressure, pounds per square inch, 
low-pressure capacity, cubic feet, and ratio of compression, 





}* — (-.tj -pa) 


r. rrzr^ H 






1 [t-^-dn S )1 


I 4l 


= H= 5f =* , 




- -, 3- - a 


/v 




1 4- 






3 4- 






ji 4 






4^ i- 






H t- 






H -f - ' 












-j J- >o 






Zt - 






H ■ tt- 






-J-t 






j tt- 






1 t4- 












3 ^ 4- 






1 - Xf 4-- 






d 3; 4-- 






^?> 












?/ "S 






1 ^4 41 - $ 






1 ^ 4- - fc 






i Iy T 
















\ { 


-i / 4- : o 




j 


r w 


^ 
















u .r 1 ® 






•m 4-- S 






^ 1 it- 5 






^ -H 4-- O 






-,£ it- >^ 






-* y M 41- o 






^ i=ilF^T 4- - « 




^^ 


r ^P -- +- N c 










o o Z+Z ^ 


^^^ I p " 




w ^ _L . a \ ,<-- 






"■ S 1 tvera^" 






« ^ «,<p^.--'- 


** /■ 






j< 






j 




j m TT^r>Q t C+--rr' -- '" 


s 










*- ' 4i. ».«"""" , ofV^" 1 * -•'""" 


i 




1 ;: "• wp^--"* 


1 1 «=Ct=:r=-4^~ . V 




: ---*= 


E= = 1 = = " == r<+ ~ - 




L " 


H -- ^ t 




* Vi -h 








- == =s.— i-«-f 'T"iif 


£_ 


y v IJ- jj. 


ZZZZSEEZZiEfZEEE 




_J 


J~l ll_L-^_- o 


L8000 
14400 
10800 


8 2 g p^ "> pV O 

t- CO « „ 



PQ 



q.ootf aauribs jad sqi; m saanssajj 



W = 288 — - (sup.pr.) (L. P. Cap.) 

O J- 



R„* 



■]• 



03 03 

O 03 

03 . 

Ph fin 



0) 

a 
^ 'I 

IS 



H 



a 



Oh 

a 

o 
O 

03 

o3 

I 



(98) 



which is the same as Eq. (86). 

Substitution of the value of best-receiver pressure in. the expression for the 



WORK OF COMPRESSORS 113 

work of the two stages separately will show the equality of work done in the 
respective stages for this case with clearance. 

Work per cubic foot gas supplied to compressor is 
W 



= 288-^sup.pr.) \r p St - ll 



(99) 



(L. P. Cap.) 
Work per cubic foot of high-pressure gas hot is 

( H.p.cI P .hot) ° 288 r^i< 8U p-p r -) g ^[^ i - 1 ]- • • (100) 

The work per cubic foot of air delivered and cooled to its original tem- 
perature is, 

(H.P.cTp- cold) " 288 ;---!^'-^^" 1 ]' • • (101) 

Due to the fact that clearance has no effect upon the work per cubic foot 
of substance, as previously noted, Eqs. (99), (100) and (101) are identical with 
(87), (90) and (93). 

11. Two-Stage Compressor, any Receiver Pressure, Exponential Compres- 
sion. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure and 
Horse-power, in Terms of Dimensions of Cylinders and Clearances. Referring to 
Fig. 29, let Di be the displacement of the first stage cylinder in cubic feet = 
(Vb — Vk), T>2 the displacement of the cylinder of the second stage in cubic 
feet = (Va— V/), ci the clearance of the first stage, stated as a fi action of the 
displacement of that cylinder, so that the clearance of the first stage cubic 
feet = ciDi, and that of the second stage = 02^)2. 

The low-pressure capacity of the first stage (L. P. Cap.) in cubic feet is 
(Vb— V a ) } and, as for the single-stage compressor, is expressed in terms of dis- 
placement, clearance and ratio of compression of the first stage as follows, see 
Eq. (64) : 

(L.P.Cap.i)=Di(l + ci-ci^i 7 )= J Di£' c i (102) 

For the second stage, the low-pressure capacity (L. P. Cap. 2) is 

and is equal to 

(L.?.C£Lp.2)=D 2 {l+C2-C2R P 2 7 ) =D 2 E v2 .... (103) 

Volumetric efficiency of the first stage is given by 

1 
E vl = l+ Cl -ciR P i s (104) 



114 



ENGINEERING THERMODYNAMICS 



Volumetric efficiency for second stage 



E v2 = l+C2 — CoRp2 s 



(105) 





o 

I s 

O p. 


r^ (*J^ "isa) — 






i 

t 


r ^ i<*K*«ttj 




-" = 3= = ==== 5t :° T 




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n 




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1 f 












- 1 ^/ 






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H t - P 






1 -V - fe 






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1 -V - 2 






1 1Z 3 






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: xz : OT -S 






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nnn~T"'M~rTT _ i nr t t 












I" 1 § 1 


| 2 8 ->$&* 



^oo^arenbg aad sq^i tn sa.mssojj 



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a 




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■+3 








a> 




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bti 




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■+3 


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Cu 




h-5 


"r3 


c3 


'43 


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fl 








fl 


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o 

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(7) 


CO 


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Ph 


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W 


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h 





It may be required to find the receiver pressure (incidental to the finding of 
work or horse-power) for a compressor with given cylinder sizes and delivery 
pressure. The condition assumed of perfect intercooling stipulates that 



(L. P. Cap.i)(sup.pr.) = (L. P. Cap.2) (rec.pr.), 



WORK OF COMPRESSORS 115 

whence 

(rec.pr.) = (sup.pr.) ( ( ^;gP^ 

= (sup.pr.)gg (106) 

If the volumetric efficiencies are known or can be sufficiently well approx- 
imated this can be solved directly. If, however, E v \ and E V 2 are not known, 
but the clearances are known, since these are both dependent upon the receiver 
pressure sought, the substitution of the values of these two quantities will give 

Dl r 1+ei _ cl (j^H-)«i 

(rec.pr.) = (sup.pr.)— L Vsup.pr./^ _ _ _ (10y) 



4 + *-(Sg) 7 



an expression which contains the receiver pressure on both sides of the equation. 
This can be rearranged with respect to (rec. pr.), but results in a very complex 
expression which is difficult to solve and not of sufficient value ordinarily to 
warrant the expenditure of much labor in the solution. Therefore, the relations 
are left in the form (107). It may be solved by a series of approximations, 
the first of which is 

(rec.pr.) = (sup.pr.)^ approx (108) 

With this value for the receiver pressure, substitution may be made in the sec- 
ond member of the Eq. (107), giving a result which will be very nearly correct. 
If desirable, a third approximation could be made. 

To find the work of a two-stage exponential compressor in terms of displace- 
ment of cylinders, supply pressure, receiver pressure and delivery pressure, 
pounds per square inch, and volumetric efficiency of the first stage, E v i, from (79) 
or (94) , 

TF = 144^- (sup.pr.)Di^n | RpW+R?^ -2] ... ,(109) 

R P i = 7 — n and R P 2 = -, — - — \. 

(sup.pr.) . (rec.pr.) 



in which 



To solve this the receiver pressure must be found as previously explained and 
the volumetric efficiency must be computed by Eq. (104) or otherwise be known. 
It is impracticable to state work for this general case in terms of displace- 
ment and clearances directly, due to the difficulty of solving for (rec. pr.) in Eq. 



116 ENGINEERING THERMODYNAMICS 

(107). It may, however, be stated purely in terms of supply and delivery pres- 
sures, in pounds per square inch, displacement, in cubic feet, and volumetric 
efficiencies, as follows: 
From Eq. (106), 

D DiE v i 



'2J^v2 



and 



Hence 



p del.pr . ., E> 2 E v2 T) 2 E v2 
Up2 sup.pr. X D 1 £' pl ~ Uv B x EW 



8-1 8-1 



^=144-l I ( S up.p,) Bl ^[(|f : ) * + (ft^) ' -2] . (HO) 

The mean effective pressure of the two-stage compressor referred to the low- 
pressure cylinder is found b}^ dividing the work of the entire cycle Eq. (110), by 
the displacement of the first-stage cylinder, and by 144, to give pounds per 
square inch. 

m.e.p. referred to first-stage cylinder, pounds per square inch is, 

s-l s-l 



w 

U4D L s- 



=^( S up.p,)^[(|£) 1 + (%jg^-a]. . (in) 



It is well to note that this may also be found by multiplying (work done 
per cubic foot of gas supplied) by (volumetric efficiency of the first stage, E v \), 
and dividing the product by 144. 

In terms of the same quantities, an expression for indicated horse-power 
majr be given as follows : 



s-l 

_s_ (sup.pr.) 
s-l 229.2 ? 



^4^r + ( R wr- 2 i (u2) 



where n is the number of cycles completed per minute by the compressor. 
For n may be substituted the number of revolutions per minute, divided by the 
revolutions per cycle, 

N 



The horse-power per cubic foot of gas supplied per minute is 

'I.H.P s (sup.pr.) [ (D x E vl \ ~«~ / D 2 E v2 \ V 1 

n(L. P. CapT) -J^l ~M9J~ [\D 2 eJ ^{^D^J Z \' * {n6) { 



WORK OF COMPRESSORS 
Horse-power per cubic foot of gas delivered and cooled per minute. 



117 



I.H.P. 



s-l 



- 1 229.2 [ \D 2 E v2 ) + [T'DxEJ Z \ 



n(H. P. Cap. cold) s- 
Horse-power per cubic foot of hot gas delivered per minute 



I.H.P. = s sup.pr. / Di^ P i \~r 1 

w(H. P. Cap. hot) s-l 229.2 \D 2 E v2 



2 (114) 



:®a • + «sr -]• • '-> 



For the case where clearance is zero or negligible, these expressions may be 
simplified by putting E v -> and E v \ equal to unity. 



1±ir ' s-l 229.2 n ^ x 



s-l 



D 2 



+ (<;) s;1 - 2 ] 



(116) 



I.H.P. per cubic foot, gas supplied per minute 
I.H.P. s (sup.pr.) 



s-l 



5-1 



n(L.P.Cap.) s-l 229.2 
I.H.P. per cubic foot gas delivered and cooled per minute 

I.H.P. s (del.pr.) 



© • +(*sr- • ■ <-> 



s-l 

Di\~s 
D 2 



n(H. P. Cap. cold) s-l 229.2 
I.H.P. per cubic foot hot gas delivered per minute 

s-l 



+ R 



'-tf-4 



(118) 



I.H.P. 



s sup.pr. ID 



w(H. P. Cap. hot) s-l 229.2 \D 



M(tf +(4r-] <■* 



Example 1. Method of calculating diagram, Figs. 29, 30, 31. 
Assumed data. 

Pa =P& = 2116 lbs. per square foot; 
p c =P d =P h =P k = Ql72 lbs. per square foot. 
p g =p e =p f = 18,000 lbs. per square foot. 
CZ(H. P.) = 7.5 per cent; Cl(L. P.) 7.5 per cent; s = 1.4; L. P. Capacity =5 cu.ft. 



118 ENGINEERING THERMODYNAMICS 

To obtain point K. 

From formula Eq. (64), 

1 
L. P. Cap. = A(l +Ci -cjt p {~ s ) 

5 = Z>i(l + .075-.075X2.14), hence Di = 5.45 cu.ft. 
Ch = V k = 5.45 X .075 = .41 cu.ft. 
.'. V* = A cuit.; Pi =6172 lbs. sq.ft. 
To obtain point A: 

^« = ^(^) 1 ' 4 = .4X2.14 = .856 cu.ft. 

,*. F a = .85 cu.ft.; P a =2116 lbs. sq.ft. 
To obtain point B : 

V b = V a +5 =.85+5 =5.85 cu.ft.; P 6 =2116. lbs. sq.ft. 

To obtain point C: 

V c = V b + (~\ 1A =5.85 +2.14 =2.73, 

.*. V c =2.73 cuit.; P c =6172 lbs. sq.ft. 
To obtain point D : 

Volume at D is the displacement plus clearance of H. P. cylinder. This cannot be 
found until the capacity is known. The capacity is the amount gas which must be 
taken in each stroke and which is also the amount actually delivered by L. P. cylinder 
cooled to original temperature. The amount of cool gas taken in by the second cylinder 
is 

(L. P. Cap. 2 ) = (^)(L. P. Cap. 1 )=||x5 = 1.7 cu.ft. 

But 

(L. P. Cap. 2 ) 1.7 , cc , + 

D2 = (- -^\ = f+ .075-. 075X2.14 = L88cU ' ft ' 

\1 + C2 — C2R P 2 S / 

d 2 = V f = c 2 D 2 = .075x1.88 =.14 cu.ft. 

V d =Cl 2 +D 2 = 1.88 + .14 =2.02 cu.ft. 

P d = 6172 lbs. sq.ft. 

Other points are easily determined by relations too obvious to warrant setting down. 

Example 2. What will be the capacity, volumetric efficiency and horse-power per 
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for the follow- 
ing compressor: Two-stage, double-acting cylinders, 22| and 34^X24 in., running 
at 100 R.P.M. High-pressure clearance 6 per cent, low-pressure 4 per cent. Supply 



WORK OF COMPRESSORS 119 

pressure 14 lbs. per square inch absolute. Delivery pressure 115 lbs. per square inch 
absolute. 

The capacity will be the cylinder displacement times the volumetric efficiency. 

Di =displacement of a 34ix24 in. cylinder, or 12.8 cu.ft. and Z>2 = displacement of 
a 22^+24" cylinder or 5.4 cu.ft. To obtain the volumetric efficiency, make use of 
approximation of formula Eq. (108), 

D 12 8 

(rec.pr.) = (sup.pr.) * = 14 X-=^r =33.2 lbs. sq.in, 

and then by Eq. (107) checking, 

E vl - 1 +ci -Ci(R pl ) ■ from Eq. (104), 

l 

= l+.04-.04x(2.5)~ s =96.8 per cent 

Therefore the capacity will be, 

200 X12.8X. 968 =2480 cu.ft, per minute; 

l 
E V 2 = 1 +c 2 -c 2 {R P i)s from Eq. (105), 

= 1 + .06 - .06 X (3.28)- 714 - 92 percent. 
From Eq. (113), I H.P. per cu.ft, (sup.pr.) air per minute is, 

s-JL 8-1 

s sup.pr. [~ [D l E n \ s .(jjDzEvA » "] 
~ S -1 229.2 1\D 2 eJ + \ Kp D 1 E 1 J, ~ 2 J' 

1.4 14 r/12.8x.968\-286 / 5.4X.92X.286 n 
= A X M92l\~5A^92~) + \ 8 ' 22 12^968/ ~ 2 J = ' 15 ' 

Whence horse-power per 1000 cu.ft. of free air per minute is, =150. 

From Eq. (115) horse-power per cubic foot (del.pr.) air, hot=that of (sup.pr.) 

1 s-l 

— /D W \ — i — 
air X Rp S (^-rr) or 5.85 times that of (sup.pr.) air. 

^1)2-^2 

.'. Horse-power per 1000 cu.ft. of hot (del.pr. air) =150 X5.85 =877. 

Prob. 1. A two-stage double-acting compressor has volumetric efficiencies as shown 
by cards of 98 per cent and 90 per cent for the high- and low-pressure cylinders respect- 
ively. It is running at 80 R.P.M. and compressing from atmosphere to 80 lbs. per square 
inch gage. If the cylinders are 15| X25ixl8 ins., and speed is 120 R.P.M. , what 



120 ENGINEEEING THERMODYNAMICS 

horse-power is being used and how many cubic feet of free and compressed air (hot 
and cold) are being delivered per minute, when s equals 1.41? 

Prob. 2. What horse-power will be needed to drive a two-stage compressor 10| ins. 
and I64 Xl2 ins., double acting, with 5 per cent clearance in each cylinder at 160 R.P.M. 
when the supply pressure is atmosphere, delivery pressure 100 lbs. per square inch 
gage, when s equals 1.35? 

Prob. 3. A thousand cubic feet of free air per minute must be compressed by a two- 
stage compressor to 80 lbs. per square inch gage from a supply pressure of 10 lbs. per 
square inch absolute. The volumetric efficiencies for the high- and low-pressure cylin- 
ders are 85 per cent and 95 per cent respectively, and 'the receiver pressure is 25 lbs. 
per square inch absolute. What will be the displacement of each cylinder and the 
horse-power per cubic foot of (sup.pr.) air? 

Prob. 4. How many cubic feet of free air can be compressed in two-stage compres- 
sor 18ix30ix24 ins. with 5 per cent clearance in high-pressure cylinder and 3 per cent 
in low if (sup.pr.) is atmosphere and (del.pr.) 80 lbs. per square inch gage? How would 
the answer be affected if clearance were taken as zero? Take s = 1.41. • 

Prob. 5. The volumetric efficiency of the low-pressure cylinder is known to be 95 per 
cent, and of the high-pressure cylinder 85 per cent. The cylinder sizes are 15| X25J Xl8 
ins. and speed is 120 R.P.M. What horse-power must be supplied to the machine if 
the mechanical efficiency is 80 per cent and the pressure ratio 10 with a (sup.pr.) of one 
atmosphere? 

Prob. 6. A compressor runs at 120 R.P.M. and is double acting. It is compressing 
air from 14 lbs. per square inch absolute to pressures ranging from 70 lbs. per square 
inch gage to 100 lbs. per square inch gage. The cylinders are 201x32^x24 ins., and 
clearances 8 per cent and 4 per cent. Find the approximate receiver pressure, capacity 
and horse-power for the range o'f discharge pressure, for s = 1.3. 

Prob. 7. The volumetric efficiency of the low-pressure cylinder of a two-stage com- 
pressor is known to be 95 per cent, the receiver pressure as shown by gage is 40 lbs., 
delivery pressure 100 lbs., and supply pressure one atmosphere. What will be the 
horse-power if the machine runs at 120 R.P.M. and the low-pressure cylinder is 18 Xl2 in.? 
s = 1.4. 

Prob. 8. An air compressor appears to require more power to run it than should 
be necessary. It is a double-acting 18x30x24 in. machine running at 100 R.P.M. 
The volumetric efficiencies are 85 per cent and 90 per cent respectively and supply and 
delivery pressures 14 lbs. and 110 lbs. per square inch, both absolute. What would be 
the minimum work per cubic foot of (sup.pr.) air, per cubic foot of (del.pr.) air, hot and 
cold, for adiabatic compression? 

Prob. 9. The efficiency of the driving gear on an electric-driven compressor is 
75 per cent. Power is being supplied at the rate of 150 H.P. How much air should 
be compressed per minute from 4 lbs. per square inch absolute to 100 lbs. per square 
inch gage, if the receiver pressure is 35 lbs. per square inch absolute and the low-pressure 
volumetric efficiency is 90 per cent, s being 1.4? 

12. Two-stage Compressor with Best Receiver Pressure Exponential 
Compression. Capacity,. Volumetric Efficiency, Work, Mean Effective Pres- 
sures and Horse-power in Terms of Dimensions of Cylinders and Clearances. 

For the two-stage exponential compressor with or without clearance, and per- 
fect intercooling, the best-receiver pressure was found to be (Eq. 84), 

(best-rec.pr.) = [(sup.pr.) (del.pr.)]* (120) 



WORK OF COMPRESSORS 121 

This expression Eq. (120) for best-receiver pressure makes it possible to 
evaluate R P \ and R P 2 as follows: 

rj , ,, N best-rec.pr. [(sup.pr.) (del.pr.)]* [del.pr.]* , , . 

ii^i for (best-rec.pr.) = — =- — — — = = #A (121) 

sup.pr. sup.pr. Lsup.pr.J 

and 

„ - n , v (del.pr.) (del.pr.) 

^ 2 for (best-rec. pr.) = - z — \ = r, wVi vir 

p - v ^ (best-rec.pr.) [(sup.pr.) (del.pr. )]■> 

-[(££)]*-*• • (122) 

The use of these values for R p \ and R P 2 in the expressions previously given 
for volumetric efficiency for the general case, Eqs. (104) and (105) results in 

Volumetric efficiency, first stage 

£ P i = (l+ci-ci#A (123) 

and volumetric efficiency, second stage 

E v 2 = (l+c 2 -c 2 R p 2s) (124) 



The work was found to be represented by Eq. (98), which may be stated 
in terms of displacement and volumetric efficiency of the first stage, as follows : 

F = 288^- I (sup.pr.)Z) 1 ^ i r^ £ 2r-ll, .... (125) 

where R p = j — . and where (sup.pr.) is in pounds per square inch. 

If the clearance is known for the first stage this becomes by the use of 
Eq. (104), 

^ = 88 ^(sup.pr.)2)i(l+ci-cii2^)|fip _ 2r--l| J . . . (126) 

which is a direct statement of the work of a two-stage adiabatic compressor 
with perfect intercooling in terms of supply pressure and delivery pressure, 
pounds per square inch, displacement, cubic feet and clearance as a fraction 






122 ENGINEERING THERMODYNAMICS 



of displacement, provided the cylinder sizes and clearances are known to be such 
as to give best-receiver pressures. 

The mean effective pressure reduced to first-stage displacement, in pounds 
per square inch, may be derived from either Eq. (125) or (126) by dividing the 
work by the displacement of the first-stage cylinder, and again dividing by 144. 

w 2s r — i 1 

m - e -P- = 1442)" = JZJ (sup.pr.)E 1 ,! \R P 2* - 1 j 

\. . . . (127) 



= — - -r (sup.pr.) I 1 +Cl — ClRp 2sj\R p 2s — 1 



Since the work done is equally divided between the two cylinders when best- 
receiver pressure is maintained, the mean effective pressure, in pounds per 
square foot, for each cylinder will be, one-half the total work divided by the 
displacement of the cylinder in question, 

w s r — - i 

m.e.p., first stage =7^^=—^ (sup.pr.)^] I Bp & -1 . ..... (128) 

Note that this is one-half as great as the m.e.p. of the compressor reduced 
to first stage, (127), 



m 



W s Di [ s ~ 1 1 

.e.p., second stage = 2g^=— y( su P-P r 0^^i^ 2s -1, . . . (129) 



But 



(sup.pr.)-^^ = (rec.pr.) = (sup.pr.) (del.pr.) *, 

whence, 

m.e.p., second stage =- — (sup.pr.) (deLpr.) lE v2 \r p ~2T—\i. . . . (130) 

It is next necessary to investigate what conditions must be fulfilled to obtain 
the best-receiver pressure, the value of which is stated, Eq. (120) . The condition 
of perfect intercooling provides that the temperature of the gas entering the 
second stage is the same as that entering the first stage, and hence that the 
product (volume entering second stage) X (pressure when entering second stage) 
must be equal to the product (volume entering first stage) X (pressure of supply 
to first stage), or 

(L. P. Cap. 2 )(rec. pr.) = (L. P. Cap.i)(sup. pr.). . . . (131 



WORK OF COMPRESSORS 123 



Combining with Eq. (120) 



(L 



. P. Capg) = [(sup.pr.)(del.pr.)]* ^ F (del.pr.) y = R ^ 
. P. Cap. 2 ) " (sup.pr.) L (sup.pr.) J v ' 



or 



(1) (2) 

h = ( L. P. Cap, i) = DiE vl = ^ 



Rp "(L. P. Cap. 2 ) D 2 E v2 



D_ 



(3) J. 

1+Ci— ClR p 2~s 



J. 
1 + C2 — C2Rv 2s 



(132) 



From this three-part equation proper values may be found to fulfill require- 
ments of best-receiver pressure for: 

1. The ratio of capacities for a given ratio of pressures, or conversely, the 
ratio of pressures when capacities are known; 

2. The ratio of cylinder displacements for known volumetric efficiencies; 

3. The ratio of cylinder displacements when the clearances and ratio of com- 
pression are known, or conversely, with known displacements and clearances 
the ratio of pressures which will cause best-receiver pressure to exist. This 
last case in general is subject to solution most easily by a series of approxi- 
mations. 

There is, however, a special case which is more or less likely to occur in prac- 
tice, and which lends itself to solution, that of equal clearance percentages. If 
c\ = C2 the parenthesis in the numerator of part (3) of Eq. (132) becomes equal 
to the parenthesis in the denominator, and evidently the volumetric efficiency 
of the two cylinders are equal, hence for equal clearance percentages in the two 
tages, 

j£-jy (133) 

A case which leads to the same expression, Eq. (133), is that of zero clearance, 
a condition that is often assumed in machines where the clearance is quite 
small. 

The work per cycle, Eq. (126), when multiplied by the number of cycles 
performed per minute, n, and divided by 33,000, gives 

IJHJP.=^^^^fiDi(l+.ft-CiiW(ft^--l), • (134) 

from which are obtained the following: 
I.H.P. per cubic foot supplied per minute 

I.H.P. . (sup.pr.) ( ^_ 1)? _ ; (135) 



n(L.P. Cap.) s-1 114.6 
I.H.P. per cubic foot delivered and cooled per minute 



Ll-P. s (del.pr.) f^-1 

n(H. P. Cap. cold) "5=1 "Hi^^ 2 * "^ ' ' ' (136 > 



124 [ENGINEEEING THERMODYNAMICS 

and I.H.P. per cubic foot delivered hot per minute 

I.H.P. s (sup.pr.) „ £±i £=i 



^^»W-i-(^-l).. (137) 



n(H. P. Cap. hot) s 

These expressions, Eqs. (165), (166) and (167) are all independent of clear- 
ance. 

Example. What will be the capacity, volumetric efficiency and horse-power per 
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for the 
following compressor for s = 1.4? Two-stage, double-acting, cylinders 22| X34| X24 ins., 
running at 100 R.P.M. Low-pressure clearance 5 per cent, high-pressure clearance 
such as to give best-receiver pressure. Supply pressure 15 lbs. per square inch abso- 
lute, delivery pressure 105 lbs. per square inch absolute. 

Capacity will be cylinder displacement times low pressure volumetric efficiency, or, 
200AX#t,i. 

Dj = 17.5 cu.ft. 

i_ 

E Vl from Eq. (123) = (1 +Ci -c,R P 2s) 

= l+.05-.05x7 357 =95 per cent. 

Therefore low pressure capacity =200x12.8 X. 95 =2430 cu.ft. per minute. 
Horse-power per cubic foot of (sup.pr.) air per minute is from Eq. (135) 

_s_sup : pr. 1=1 
s-1 114.6 {Rp2s ~ 1)j 

= --X — (7^-1) =.16. 
.4 105 K ' 

Therefore, horse-power per 1000 cu.ft. of sup.pr. air = 160. 

Horse-power per cubic foot of (del.pr.) air, hot, is from Eq. (137) 

1+8 

R p 2 S times power per cu.ft. of (sup.pr.) air, 

hence, 

160 X5.3 =850 =horse-power per 1000 cu.ft. of (del.pr.) air, hot, per minute. 

Problem Note. In the following problems, cylinders are assumed to be proportioned 
with reference to pressures so as to give best-receiver pressure. Where data conflict, 
the conflict must be found and eliminated. 

Prob. 1. Air is compressed adiabatically from 14 lbs. per square inch absolute to 
80 lbs. per square inch gage, in a 20^x32^x24 in. compressor, running at 100 R.P.M. , 
the low-pressure cylinder has 3 per cent clearance. What will be horse-power re- 
quired, to run compres or and what will be the capacity in cubic feet of low pressure 
and in cubic feet of (del.pr.) air? 

Prob. 2. What must be the cylinder displacement of a two-stage compressor with 5 
per cent clearance in each cylinder to compress 500 cu.ft. of free air per minute from 
14 lbs. per square inch absolute to 85 lbs. per square inch gage, so that s equals 1.4? 
What will be the horse-power per cubic foot of (del.pr.) air hot and cold? 

Prob. 3. A two-stage compressor is compressing gas with a value of s = 1.25 
from 10 lbs. per square inch gage to 100 lbs. per square inch gage. The cylinders are 
18jx30ix24 ins., and speed is 100 R.P.M. If the clearance is 5 per cent in the low- 



WORK OF COMPRESSORS 125 

pressure cylinder and 10 per cent in the high, what will be the cubic foot of (sup.pr.) gas 
handled per minute and what will be the horse-power at best receiver pressure? 

Prob. 4. A manufacturer states that his 201x321x24 in. double-acting compres- 
sor when running at 100 R.P.M. at sea level will have a capacity of 2390 cu.ft. of free air 
per minute, pressure range being from atmosphere to 80 lbs. per square inch gage. At 
best-receiver pressure what clearance must the compressor have, compression being 
adiabatic? 

Prob. 5. The cylinder sizes of a two-stage compressor are given as 10| Xl6jxl2 
ins., and clearance in each is 5 per cent. What will be the best-receiver pressures when 
operating between atmosphere and following discharge pressures, 60, 70, 80, 90, 100 
and 110 lbs. per square inch gage, for s equal 1.4? 

Prob. 6. 1500 cu.ft. of air at 150 lbs. per square inch gage pressure are needed per 
minute for drills, hoists, etc. The air is supplied from 3 compressors of the same size 
and speed, 120 R.P.M. Each has 4 per cent clearance in each cylinder. What will 
be sizes of cylinders and the horse-power of the plant for best-receiver pressure, when 
§ = 1.41? 

Prob. 7. The cards taken from a compressor show volumetric efficiencies of 95 per 
cent and 80 per cent in low- and high-pressure cylinder respectively. What will be 
(del.pr.) for best-receiver pressure if compressor is 15ix25jXl8 ins., and (sup.pr.) 15 
lbs. per square inch absolute to 10 lbs. absolute, and what will be the work in each case, 
£ being 1.35? 

Prob. 8. A manufacturer gives a range of working pressure of his 10jXl6jXl2 in. 
compressor from 80-100 lbs. per square inch page. If clearances are, low 4 per cent, 
high 8 per cent, and (sup.pr.) is atmosphere, find by trial which end of the range comes 
nearest to giving best-receiver pressure? If clearances were equal which would give 
best-receiver pressure? 

Prob. 9 A 16 J X25 J X 16 in. compressor is rated at 1205 cu.ft. free air per minute at 
135 R.P.M. at sea level. What would be> the clearance if compressor were compressing 
air from atmosphere to 100 lbs. gage at sea level? With same clearance what would be 
the size of a low-pressure cylinder to give the same capacity at altitude of 10,000 ft. with 
the same clearance and the same (del.pr.) , best-receiver pressure always being maintained? 

13. Three-Stage Compressor, no Clearance, Perfect Intercooling Expo- 
nential Compression (Cycle 7), Best Two Receiver Pressures, Equality of 
Stages. Work and Capacity, in Terms of Pressures and Volumes. The three- 
stage exponential compressor cycle with no clearance, perfect intercooling Cycle 
7, is shown in Fig. 32. The net work area, ABCDEFGHJKA, is made up of 
three areas which may be computed individually by the formulae for single stage 
Eq. (48), provided the requisite pressures and volumes are known, as follows: 

£— 1 

W = ^ P > V *[(^) S - 1 ] (first stage) 

s— 1 

+^Ti^^|"(f1) ' _1 ] ( second sta § e ) i * ' (138) 

s—1 



126 



ENGINEERING THERMODYNAMICS 



But the condition of perfect intercooling provides that for no clearance, 

P b V b = P d Y d = P f Vf . (139) 




and it may be noted that P d = P c , and P r =P e . Accordingly, 

s-1 s-1 r s-1 



W=~P b V b 
s—l 



+<r+g 



• (140) 



Pressures in this expression are in pounds per square foot. 



WORK OF COMPRESSORS 127 

Changing the equation to read in terms of supply pressure pounds per square 
inch, low-pressure capacity cubic feet, and ratios of pressures, first stage 
(R P i), second stage (R p2 ) and third stage (R P z), it becomes 

Work done by three-stage compressor, perfect intercooling 

FT=144^j(sup.pr.)(L. P. Cap.) ^(R p i)^ + (rJ~s 1 +(rJ z t-3\ ) (141 

From this the following expressions are derived: 
Work per cubic foot supplied 

( L.P.Cap.) = 144 ^Ti( su P-P r -) [(flpi) s +(R P 2) '• +(R P s) s~-S ]. . . (142) 

Work per cubic foot gas delivered and cooled 

o s — i s— i s in 

(lDTCa^oTd) = 14 V^ ( 143 ) 

Work per cubic foot gas, as delivered hot 

(H. P. Cap. hot) = 144 ^l( s ^P-P r >)(^i)(^2)(^ 3 )^ [ (R Pl ) s 

+ (Rp2)~+(R P a) L T-s\ (144) 

4 

Best Two Receiver Pressures. Referring to Fig. 32, P c is the pressure in the 
first receiver (1 rec.pr.) and P e is the pressure in the second receiver, (2 rec.pr.). 
It is evident that if either receiver pressure be fixed and the other is varied^ 
the work necessary to compress a given initial volume of gas will be varied^ 
and will have a minimum value for some particular value of the varying receiver 
pressure. By a variation of both receiver pressures a minumim may be found 
for the work when both receiver pressures have some specific relation to supply 
and delivery pressures. For instance, assume that P c is fixed. Then a change 
in P e can change only the work of the second and third stages, and the three- 
stage compressor may be regarded as consisting of 

One single-stage compressor, compressing form P b to P Ca 
One two-stage compressor, compressing from P c to P g . 

_ In this two-stage compressor, best-receiver pressure is to exist, accord" 
mg to Eq. (84), 

Pe = (best 2 rec .pr.) = {PcP g )\ . ...... (145) 



128 ENGINEERING THERMODYNAMICS 



Id show 



Similar reasoning, assuming P e fixed and making P c variable, woul< 
that 

P c =(best 1 rec.pr.) = (P e P & )* (146) 

Eliminate P c from Eq. (145) and the expression becomes, 

P e =(best2rec.pr.) = (P & P, 2 ) = [(sup.pr.)(del.pr.) 2 ]' (147) 

Similarly, from Eq. (146) 

P c = (best 1 rec.pr.) = {P, 2 P g ) = [(sup.pr.) 2 (del.pr.)] * (148) 

From these expressions may be obtained, 



P c = i^ = P, = /PA* 
P b P c P e \Pj 
or 

Rpi = R P 2 = Rp3 = Rp* 



Substitution in Eq. (140) gives, 

Work, three-stage, best-receiver pressure no clearance 

s-l 



(149) 



Arranging this equation to read in terms of supply pressure, pounds per square 
inch, low-pressure capacity, cubic feet, and ratio of pressures 

Work, three-stage best-receiver pressure 



s — * 



Tf = 432-^ 1 (sup.pr.)(L.P. Cap.)Ci2p"^"-l), . . . (151) 

The work of the compressor is equally divided between the three stages 
when best-receiver pressures are maintained, which may be proven by substitu- 
tion of Eq. (149) in the three parts of Eq. (138), and 

Work of any one stage of three-stage compressor with best-receiver pressure. 

If 1 = If 2 = If3 = 144^-(sup.pr.)(L. P. Cap.XP/sT-l). . (152) 



WORK OF COMPRESSORS 129 

From Eq.(151), may be derived the expressions for work per unit of capacity. 
Work per cubic foot low-pressure gas is, 

. (O?^)^ 32 ^!^-^^- 1 ] (153) 

Since 

(L. P. Cap.) = (H. P. Cap. co\d)R P . 

Work per cubic foot cooled gas delivered is, 

( H. P. Cap. cold r 432— l(s up.pr.)^(fi p -37 -1). . . . (154) 
Again, from Fig. 32, 



1 1 



,.( ? f-w.(g),o,n-r.(fi)(g 

which is to say that, when best-receiver pressures are maintained, 



or 



(L. P. Cap.) = (H. P. Cap. hot)R p TRjr S) 



2s + l 

= (H. P. Cap. hot)^~3— , .... (155) 



hence 

Work per cubic foot hot gas delivered 



W S ^± x i^. 1 

(H.P. Cap.hotr 432 g ^I (sup -P r - )fi * "'(^"-D- • • (156) 



Example 1. Method of calculating Diagram, Fig. 32. 
Assumed data. 

P a =Pb =2116 lbs. per sq.ft. 

P c =Pd =best first-receiver pressure =PjP g * =4330. 

P e =P/=best second-receiver pressure =P b %P g * =8830. 

Pa =P h = 18,000 lbs. per sq.ft. 

F a = T^=0cuit. 7*=5cu.ft. s = 1.4. 



130 ENGINEERING THERMODYNAMICS 

To obtain point C: 

V c = V h + (?f\ 1A =5 + 1.67 =3 cuit. 

/. V C =S cu.ft. P c =4330 lbs. sq.ft. 

Intermediate points B to C may be found by assuming variousjpressures and finding 
the corresponding volumes as for V c . 
To obtain point D: 

V d = V b X^ =5X^=2.44 cu.ft. 

JTd *±oo\J 

/. V d =2.44 cu.ft., P d =4330 lbs. sq.ft. 

To obtain point E: 



T/ T/ /PA 1.4 P e Ep d 

V-V' + (p t ) > but -=-, 



by assumption of best-receiver pressure. 

Hence V e = 2.44 + 1.67 =1.46 cu.ft., an P e =8830 lbs. sq.ft. 

Intermediate points D to E may be found by assuming various pressures and finding 
corresponding volumes as for V e , and succeeding points are found by similar methods 
to these already used. 



7, =.72, P, = 18,000, 



-ee 



Example 2. What will be the horse-power required to compress 100 cu.ft. of free 
air per minute from 15 lbs. per square inch absolute to 90 lbs. per square inch gage in a 
no-clearance, three-stage compressor if compression be adiabatic? What will be the 
work per cubic foot of (del.pr.) air hot or cold? 

From Eq. (153) work per cubic foot of (sup.pr.) air is, 

432 — -(sup.pr.) (fl* 3 S -1), 
s — 1 

=432x^Xl5x(7 0952 -l) =4500 ft.-lbs., 

or 

4^00 vlOO 
H.P. for 100 cu.ft. per minute = 09 Z^ =13.6. 

00,1)00 

From Eq. (154) work per cubic foot of (del.pr.) air cold is R p times that per cubic 
foot of (sup.pr.) air, or in this case is 31,500 ft.-lbs. 

2s+l 

! From Eq. (156) work per cubic foot of (del.pr.) air hot is R p 3s times that per cubic 
foot of (sup.pr.) air, or in this case 5.8x45,000 =46,200 ft.-lbs. 



WORK OF COMPRESSORS 131 

Prob. 1. What work will be required to supply 2000 cu.ft. of air at 200 lbs. per inch 
gage pressure if compressing is done adiabatically by three-stage compressors, taking 
air at atmosphere, neglecting the clearances? 

Prob. 2. A motor is available for running a compressor for compressing gas, for 
which s equals 1.3. If 60 per cent of the input of the motor can be expended on the 
air, to what delivery pressure can a cubic foot of air at atmospheric pressure be com- 
pressed in a zero clearance three-stage machine? How many cubic feet per minute 
could be compressed to a pressure of 100 lbs. gage per H.P. input to motor? 

Prob. 3. Two compressors are of the same size and speed. One is compressing 
air so that exponent is 1.4, the other a gas so that exponent is 1.1. Each is three stage. 
Which will require the greater power to drive, and the greater power per cubic foot 
of (sup.pr.) gas, and per cubic foot of (del.pr.) gas, hot, and how much more, neglect- 
ing clearance? 

Prob. 4. How will the work per cubic foot of (sup.pr.) air and per cubic foot of (del. 
pr.) air differ for a three-stage compressor compressing from atmosphere to 150 lbs. per 
square inch gage from a single- and a two-stage, neglecting clearance? 

Prob. 5. A table in " Power " gives the steam used per hour in compressing air to 
various pressures single stage. A value for air compressed to 100 lbs. is 9.9 lbs. steam 
per hour per 100 cu.ft. of free air. Using the same ratio of work to steam, find the value 
for the steam if compression had been three-stage, zero clearances to be assumed. 

Prob. 6. A 5 in. drill requires 200 cu.ft. ef free air per minute at 100 lbs. per square 
inch gage pressure. What work will be required to compress air for 20 such drills if 
three-stage compressors are used, compared to single-stage for no clearance? 

Prob. 7. What would be the steam horse-power of a compressor delivering 150 
cu.ft. of air per minute at 500 lbs. per square inch pressure if compression is three-stage, 
adiabatic, clearance zero, and mechanical efficiency of compressor 80 per cent? 

14. Three-stage Compressor with Clearance, Perfect Intercooling Expo- 
nential Compression (Cycle 8), Best-receiver Pressures, Equality of Stages. 
Work and Capacity in Terms of Pressures and Volumes. The pressure- 
volume diagrams of the three-stage compression is shown in Figs. 33, 34 and 
35, on which the clearance volume and displacements, low-pressure capacity 
and high-pressure or delivery capacity for hot gas are indicated. 

If perfect intercooling exists, as is here assumed, 

(V*-Va)P> = (V a -Vi)P a =(V f -Vj)P f ) 
and also k . . . . (157) 

(L. P. Cap.)P & =(H.P. Cap. cold)iV J 

Apply Eq. (57) to the three stages and the entire work done is, 
W = ^P b (V b -V a )[(^y-lT-l] (first stage) 



s-1 

+■ jrj p *( 7 ' ~ F <) [ (jrj S ^~ 1 ] (second stage) 
+ s 4l P /( 7 /-^) [(Jj 1 ^ 1 - 1 ] ( third ^age) 



(158) 



132 



ENGINEERING THERMODYNAMICS 



By use of the above conditions of perfect intercooling Eq. (157) this 
expression becomes, 



W 



=~iPt>(V»-Va)\Rpi L ^ '+Rp2 L r +R P <hr -3] 



(159) 































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WORK OF COMPRESSORS 



133 



In terms of supply pressure, pounds per square inch, low-pressure capacity, 
cubic foot and ratios of pressures as above, the work of a three-stage com- 
pressor with perfect intercooling and with clearance is 

ft =144-^j (sup.pr.)(L.P. Cap.) F (ftiJ^ + C^) W^V-sl, (160) 












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which is identical with Eq. (141), showing that clearance has no effect upon the 
ir™rh for a given capacity. 



134 



ENGINEERING THERMODYNAMICS 



It readily follows that the work per unit of gas is independent of clearance, 
and hence Eqs. (142), (143) and (144), will give a correct value for the work 






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per cubic foot of gas supplied, per cubic feet delivered and cooled, and per 
cubic foot as delivered hot, respectively. 

Since in two-stage compressors the reasoning leading to the determination 
of best-receiver pressure applies equally well with and without clearance, and 
since the value of best-receiver pressures for three-stage are found by con- 



WORK OF COMPRESSORS 135 

sidering the three-stage a combination of one- and two stage-compressors, the 
same expressions for best-receiver pressures will hold with clearance as without; 
see Eqs. (147) and (148). 

P e =(best 2 rec.pr.) = [(sup.pr.)(del.pr.) 2 ]*. 
P c = (best 1 rec.pr.) = [(sup.pr.) 2 (del.pr.)] 4 . 

The use of these expressions for best-receiver pressures leads to the same 
result as for no clearance Eq. (150), except for the volumes, 

Work, three-stage] best-receiver's pressure with clearance 

8—1 

PF=3 ^T Ps(y6 " Fa) [0 ^ _1 ] •••'•■ ( 161 ) 

which is stated below in terms of supply pressure, pounds per square inch low- 
pressure capacity, cubic foot, and ratio of compression R p , 

Work, three-stage best-receiver pressure. 

TF = 432^ T (sup.pr.)(L.P.Cap.)(i2^ 1 -l) ... (162) 
s J. 

which is identical with Eq. (151). 

Frorn this may be obtained expressions for the work per cubic foot of low- 
pressure gas supplied to compressor per cubic foot of gas delivered and cooled, 
and per cubic foot of gas as delivered hot from the compressor, when the re- 
ceiver-pressures are best, and these will be respectively identical with Eqs. 
(153), (154), and (156), in the foregoing section. 

15. Three-stage Compressor, any Receiver-pressure Exponential Com- 
pression. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure, 
and Horse-power in Terms of Dimensions of Cylinders and Clearances. 

Di = displacement of the first-stage cylinder, in cubic feet=(F 6 — V m ); 
Z>2 = displacement of the second-stage cylinder, in cubic feet = (F d — F*); 
D3 = displacement of the third-stage cylinder, in cubic feet = (V f — V h ). 

ci, C2, C3 are the clearances of the first, second and third stages respectively, 
stated as fractions of the displacement, so that, 

Clearance volume, 1st stage, in cubic f eet = V m = C1D1 ; 
Clearance volume, 2d stage, in cubic f eet = V* = C2-D2 ; 
Clearance volume, 3d stage, in cubic feet = F/j = C3£>3. 






136 ENGINEERING THERMODYNAMICS 

The low-pressure capacity of the first stage, and hence for the compressor 
is (Vi—V a )j and in terms of clearance, a, and displacement D\ of the first 
stage is, according to Eq. (64), 

(L. P. Cap.i)=£>i(l+ci-cAi *)=DiE vl (163) 

For the second stage, the low-pressure capacity is (V d — Vi) and is equal to 

(L. P. CsLp.2)=D 2 (l+C2-C2Rp2*)=D 2 E V 2, .... (164) 
and for the third stage (V f — Vj) or, 

(L. P. Ca P . 3 )=£)3(l + C3-C3^3 s )=^3^3. .... (165) 

The volumetric efficiency of 1st stage is 

^i = (l+ci-cAi s ) (166) 

Volumetric efficiency of second stage is 

E v2 = (l+c2-c 2 Rp2 s ). ....... (167) 

Volumetric efficiency of third stage is 

^3=(1 + C3-C3^3 S ), (168) 

The work of the three-stage compressor with the assistance of Eq. (163) may 
be stated in terms of supply pressure, pounds per square inch, displacement 
of first-stage cylinder, in cubic feet and volumetric efficiency of first stage, and 
also ratios of compression existing in the first, second, and third stages, 

W = 144^-(sup.pr.)£>i^i \(R pl ) '-IT + (R p2 ) ^+ (R p s) V-3 J . (169) 

To make use of this formula for the work of the compressor the two 
receiver pressures must be known, and it is, therefore, important to derive 
relation between receiver pressures, displacements and clearances or volumetric 
efficiencies. 

The assumption of perfect intercooling which has already been made use 
of in obtaining Eq. (169), regardless of the receiver-pressure, requires that— 
see Eq. (157): 

(L.P. Cap.i) (sup.pr.) = (L.P. Cap. 2) (1 rec.pr.) 

= (L.P. Cap. 3 )(2 rec.pr.). . (170) 



WORK OF COMPRESSORS 



137 



Using values of capacities in Eqs. (163), (164), and (165) and solving for 
first- and second-receiver pressures. 



and 



Then 



v , v(L. P. Cap.i) ( J>iE vl 

(1 rec.pr.) = (sup.pr.) (L p g^ = (sup.pr.)-^, 



JL. P. Cap.i) , .DiE* 

(2 rec.pr.) = (sup.pr.)^- p- g^ = (sup.p 



DzEvz 



R P i = 



Rd2 = 



(1 rec.pr.) = DiE v i 
(sup.pr.) 7^2^,2' 
(2 rec.pr.) D 2 E v2 



(1 rec.pr.) D 3 ^ 3 ' 
By definition, (del.pr.) = R P (sup.pr.), 



(171) 

(172) 

(173) 
(174) 



(del.pr. _ _ (sup.pr.) DA 

tp3 (2 rec.pr.) ""(2 rec.pr.) ""Di^.i' 



(175) 



The work of the three-stage compressor may then be stated in terms of 
supply pressure, pounds per square inch, displacements, cubic feet, volumetric 
efficiencies, and overall ratio of compression, R p , as follows : 



lf = 144 sZ ^(sup.pr.)i)i^i 



DiE vl 
D 2 E v2 



s-l 



s-l 



(D 2 E v2 

sEv3 



+ U>^ 



s-l 



+ 



4£) s -4 • < i76 > 



7n Terms of Pressures, Displacements, and Clearances, an expression can be 
written by substitution of values of E v \, E v2 and E v s from Eqs. (166), (167) and 
(168), but it becomes a long expression, further complicated by the fact that 
R P i, Rp2 and R P 3 remain in it. This may be solved by the approximation 
based first upon the assumption that all volumetric efficiencies are equal to each 
other or to unity when 



Rp\ = 

Rp2 = 



5i 
D 2 

D2 , (If volumetric efficiencies are each equal to each other 
D s * or to unity) (177) 



R *~D 1} 



138 ENGINEERING THERMODYNAMICS 

This process amounts to the same thing as evaluating E v i, E v2 , and E v % from 
Eqs. (166), (167) and (168), making use of the approximation Eq. (177) and 
substituting the values found in Eq. (176). 

Since the above can be done with any expression which is in terms of volu- 
metric efficiencies, the following formulae will be derived from Eq. (176), as it 
stands. 

The mean effective pressure of the three-stage compressor reduced to the first- 
stage cylinder is found by dividing the work of the entire cycle, Eq. (176) by 
displacement of the first stage, and by 144 to reduce to pounds per square inch. 

(m.e.p.) reduced to first stage cylinder, 

W s , _ \(D x E vl \^ , /D 2 E v2 \ s — f 



144Di s-l v ^ J VL \\D 2 E v2 . ' \DsE t 



[(DiEvi 
l\D 2 E V 2 



(sup.pr.)^ \l^rP — + 



+Kf:F-4 ■ < ire » 



Note here that this may also be obtained by multiplying (work per cubic foot 
supplied) by (volumetric efficiency of first stage) and dividing the product by 144. 
The indicated horse-power of a compressor performing n cycles per minute 
will be equal to the work per cycle multiplied by n and divided by 33,000, or, 
for the three-stage compressor with general receiver pressures, 



For n may be substituted the number of revolutions per minute, N, divided 
by the revolutions required to complete one cycle 

• N 
n = —. 

z 



The horse-power per cubic foot of gas suppliedjper minute is 

I.H.P. _^_ (sup.pr.) T (BjMA •-=* , (D2E»2\ '-=* 

n(L.P.Cap.) s-1 229.2 l\D 2 E v2 ) " ^~\D 3 eJ ' 

+ (*ot)^- 8 ]- • < 180 > 

Horse-power per cubic foot gas delivered and cooled per minute is 

LH.P. 8 (del.pr.) V/ D^ Yir ( D 2 E v2 " 

w(H.P.Cap.cold) s-1 229.2 [\D 2 E v2 J ^[DsE^ 

+ 



«fer-4 • ^ 



WORK OF COMPRESSORS 
Horse power per cubic foot hot gas delivered per minute is 



139 



LH-P. = s (sup.pr.) /D^aV 1 

n(H. P. Cap. hot) 6—1 229.2 \D S E V J UpS 



(DiEjA 

\d 2 eJ 



s-l 



+(§i:f+(*.glr-4 



(182) 



The last equation is obtained by means of the relation 
(L. P. Cap.) = (H. P. Cap. hot) X Q^EL V 



rec.pr 



sup.pr. 



= (H. P. Cap. hot) X ( R DsEv ^ s (DlEvl 



DiEj \D S E 



W 



(H.P.Cap.hot)Xi^(^f^ 

\L>2E V 2 



8—1 

s 



If clearance is zero or negligible, these expressions may be rewritten, 
Ev, Evo and Evs each equal to unity. 



. (183) 
putting 



LH.P.= 



s (sup.pr.) 



nD 



%f 4 W<*>if->\ <m 



s-1 229.2 
H.P. per cubic foot of gas supplied per minute is 

_JJLP^ = (sup^r/^\ £ ^ 1 , /D 2 \T / D 3 \T "I 
n(L.P. Cap.) 229.2 \\d 2 ) + \d^) + [ R »DI) ~ 3 J 

H.P. per cubic foot delivered and cooled per minute is 



(185) 



I.H.P. 



_ (del. pr.) 



dI) + \d1 



n(H. P. Cap. cold) 229.2 
H.P. per cubic foot hot gas delivered per minute is 



+ R 



s-l 



(186) 



!-l 



I-H.P. _ 8 (sup.prj (DA — Ar/Z)AV m 

n(H.P. Cap.hot) s-l 229.2 \D 3 J *l\p 2 ) + \D 



+ 



s-l 

(5t) s - 3 J- • • < 187 ) 



140 ENGINEERING THERMODYNAMICS 

\ 

Example 1. Method of calculating Diagram, Fig. 35. 
Assumed data: 

P a =Pb =2116 lbs. per square foot. 

Pc=Pd =Pi =P m =4330 lbs. per square foot. 

p e =P f = p j =p ]c =8830 lbs. per square foot. 

P g =Pfi = 18,000 lbs. per square foot. 
ci=7.5 per cent for all cylinders; 8=1.4. 
L.P. capacity 5 cu.ft. 

To obtain point M: 

l 

From formula Eq. (163) L. P. Capi. =Di(l +d -ciR p i *) 

or 

5=D 1 (l+.075-.075xl.67) or A =5.3 cu.ft. and clearance volume 

F ra =5.3 X.075 = .387 cu.ft. 
Therefore, 

7 W = .39 cuit.; P w =4330 lbs. sq.ft.; 
To obtain Point A : 



V a = V m x(jpr) 1A - =.39X1.67 = .67 cuit. 



Additional points M to A may be found by assuming pressures and finding corre- 
sponding volumes as for V a - 

To obtain point B : 

V b = V A + (L. P. Capi.) =.67+5 =5.67 cu.ft. 



Therefore, 

To obtain point C: 

Therefore, 



F & =5.67 cu.ft.; P 6 =2116 lbs. sq. ft.; 



-( y-j A =5.67 -s-1.67 =3.45 cu.ft. 



F c = 3.45 cu.ft.; P c =4330 lbs. sq.ft. 



Intermediate points B to C may be found by assuming various pressures and finding 
corresponding volumes as for Vt>. 



WORK OF COMPRESSORS 141 



To obtain point D : 



Volume at D is the displacement plus clearance of the intermediate cylinder. This 
cannot be found until the capacity is known. Applying the same sort of relations as 
were used in calculating the diagram for the two-stage case with clearance, 

i_ 

D 2 (l +C 2 -C 2 R P 2) s =2.44 or D 2 =2.57, 

and clearance volume. 

V k =.075X2.57 = .192 cuit., 

hence, 

V d =2.57 +.19 =2.76 cu.ft. 
Therefore, 

V d =2.76 cu.ft; P d =4330 lbs. sq.ft.; 

The rest of the points are determined by methods that require no further explana- 
tion and as pressures were fixed only volumes are to be found. These have the following 
values, which should be checked: 

V e = lM; Fr = 1.32; 7, = .79; 7» = .09; Vj = .15; 7, = .32; V X = M) 7^ = 1.23; V 2 = .U. 

Example 2. A three-stage compressor is compressing air from atmosphere to 140 
lbs. per square inch absolute. The low-pressure cylinder is 32 x24 ins. and is known to 
have a clearance of 5 per cent. From gages on the machine it is noted that the first- 
receiver pressure is 15 lbs. per square inch gage and the second-receiver pressure is 
55 lbs. per square inch gage. What horse-power is being developed if the speed is 
9100 R.P.M. and s =1.4? From the formula Eq. (169), 

W = 144^-(sup.pr.)A^i \R P rT-+Rprr+Rp$ L 7--Z . 

From gage readings 

R pl =—=2. R P 2=—=2.33; R P z = ——=2. 

l 
E H = {l+c l -c l R plS ) fromEq. (166), 



or, 



Hence, 



# cl = (l+.05-.05xl.65) =67.5 per cent. 



W = 144x^ Xl5xll.2x.675(1.22 + 1.28+1.22-3); 

= 59,200 ft.lbs. per stroke or 200 X59,200 ft. = lbs. per minute; 
=358 I.H.P. 



142 ENGINEERING THERMODYNAMICS 

Example 3. Another compressor has cylinders 12x20x32x24 in. and it is known 
that the volumetric efficiencies of the high, intermediate and low-pressure cylinders are 
respectively 70 per cent, 85 per cent and 98 per cent. The (del.pr.) is 150 lbs. per square 
inch absolute. What is the horse-power in this case if the speed is 100 R.P.M.? 

From the formula Eq. (176), 

^ = 144— (sup.pr.)!)^— J s + (— J . + (B,^r) . -3J 

= 144x^l5xl,2x,8[( S fJ|)-- + (ggi)-- 

-WiSS)' 286 -] 

= (1.309+1.495+1 -3) =66,400 ft.-lbs. per stroke, 

TnD 200X66,400 A ^ 
Whence I.H.P. = ^^ =402. 

Prob. 1. What will be the horse-power required to drive a 12 X22 X34 X30 in. three- 
stage compressor with volumetric efficiencies of 75, 85, and 95 per cent in the high, 
intermediate and low-pressure cylinders, at 100 R.P.M. when compressing natural gas 
from 25 lbs. per square inch gage to 300 lbs. per square inch gage, adiabatically? 

Prob. 2. A three-stage compressor for supplying air for a compressed-air locomo- 
tive receives air at atmosphere and delivers it at 800 lbs. per square inch gage. Should 
the receiver pressures be 50 lbs. and 220 lbs. respectively in the first and second and the 
volumetric efficiency of the first stage 90 per cent, what would be its displacement 
and horse-power when compressing 125 cu.ft. of free air per minute, adiabatically? 
What are the cylinder displacements? 

Prob. 3. Find the work done on a gas, the value for s of which is 1.3, in compressing 
it from atmosphere to 7 atmospheres absolute, adiabatically in three stages, the com- 
pressor having a low-pressure cylinder displacement of 60 cu.ft. per minute and a volu- 
metric efficiency of 95 per cent, first receiver pressure being 2 atmospheres absolute, 
and second-receiver pressure 4 atmospheres absolute. If air were being compressed 
instead of the above gas, how would the work vary? 

Prob. 4. The cylinders of a compressor are 8x12x18x24 ins. and clearance such 
as to give volumetric efficiencies of 80, 90 and 98 per cent in the different cylinders in the 
order given. Compressor is double acting, running at 120 R.P.M. and compressing air 
adiabatically from 14 lbs. per square inch absolute to 150 lbs. per square inch gage. 
What is the capacity in cubic feet per minute, work done per cubic feet of (sup.pr.) air, 
(del.pr.) air hot and cold and the horse-power of the compressor? What would be 
the effect on these quantities if the clearances were neglected? 

Prob. 5. If the cylinders of a compressor are 10x14x20x18 ins., and clearances 
are 8, 5 and 3 per cent, what work is being done in adiabatically compressing air from 
10 lbs. per square inch absolute to 100 bs. per square inch gage? 

Note : Solve by approximate method. 

Prob. 6. For special reasons it is planned to keep the first-receiver pressure of a 
three-stage compressor at 30 lbs. per square inch absolute, the second-receiver pressure at 
60 lbs. per square inch absolute, and the line pressure at 120 lbs. per square inch absolute 






WOEK OF COMPRESSOES 143 

The (sup.pr.) is 14 lbs. per square inch absolute. If the clearances are 4 per cent in 
the low and 8 per cent in the intermediate and high-pressure cylinders, what must be 
the cylinder sizes to handle 500 cu.ft. of free air per minute at 120 R.P.M. and what 
power must be supplied to the compressor on a basis of 80 per cent mechanical effi- 
ciency, for a value of s equal to 1.39? 

Prob. 7. Should the above pressures (Prob. 6) be gage pressures instead of absolute, 
how would the quantities to be found be affected? 

Prob. 8. The receiver pressures on a C0 2 gas compressor are 50 lbs. per square inch 
absolute, and 200 lbs per square inch absolute, the (del.pr.) being 1000 lbs. per square 
inch absolute. The mach ne has a low-pressure cylinder 8x10 ins. with 3 per cent 
clearance. What horse-power will be required to run it at 100 R.P.M and what would 
be the resultant horse-power and capacity if each pressure were halved? (Sup.pr.) = 14.7 
lbs. per square inch. 

16. Three-stage Compressor with Best-receiver Pressures Exponential 
Compression. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure 
and Horse-power in Terms of Dimensions of Cylinders and Clearances. It 

was found that for the three-stage adiabatic compressor with perfect inter- 
cooling, the work was a minimum if the first and second receivers had pressures 
defined as follows, see Eqs. (147) and (148) : 

(best 1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]* (188) 

(best 2 rec.pr.) = [(sup.pr.) (del.pr.) 2 ]* (189) 

(best iree P rj = /deLpr : y = _ _ 

(sup.pr.) \sup.pr./ 

(best 2 rec.p r, ) / deLprA* _ _ 

(best 1 rec.pr.) \sup.pr./ 

Bp3== (^L_-(M^Y =Bpi (192) 

(best 2 rec.pr.) \ sup.pr./ 



The use of these values in connection with expressions previously given 
for volumetric efficiency, Eqs. (166), (167) and (168), gives, 



Volumetric efficiency of first stage =E v i = (l+ C\ — CiR P Ss ) .... (193) 



Volumetric efficiency of second stage = E v2 = (l+c 2 — c 2 R/ s ) .... (149) 



Volumetric efficiency of third stage =E v s= (l+c 3 — c s R p 3s ) .... (195) 



144 ENGINEERING THERMODYNAMICS 

The work of the three-stage compressor with best-receiver pressures, Eq. 
(162), when expressed in terms of displacement and volumetric efficiency becomes 

W = 4S2^-(sup.pr.)D 1 E vl (Ris^-l) (196) 

O J- 

where 

R = (del.pr.) 
v (sup.pr.) 

If clearance is known, the value of E v \ may be ascertained by Eq. (193) 
and inserted in Eq. (196). Since this may be so readily done the substitution 
will not here be made. 

The mean effective pressure of the compressor referred to the first stage is 
obtained by dividing the work Eq. (196) by 144 D\\ 

(m.e.p.) referred to first-stage cylinder 

W s s- 1 
I 4 4 - 5 - i =3- 1 (sup.pr.)^ 1 (^3J--l) (197) 

The mean effective pressures of the respective stages, due to the equality 
of work done in the three stages will be as follows : 

For first stage 

(m.e.p.) =-^ T (sup.pr.)^i(i2p"3T-l) (198) 

For second stage 

(m.e.p0=^(siip.pr.)g-^i(i2p £ ^"-l) (199) 

For third stage 

s Di s- i 
(m.e.p.) =—— (sup.pr.) w E vX (Rv ^ -1) (200) 

But 

7) IF 1 

(sup.pr.)^£^ = (1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]*, 

and also 

(sup.pr.) j^g^ = (2 rec.pr.) = [(sup.pr.) (del.pr.) 2 ]*. 



WORK OF COMPRESSORS 



145 



Hence 

For second stage 

For third stage 



s-l 



(m.e.p.) = -f rr [(sup.pr.) 2 (del.pr.)]*^2(fip'3r - 1). . (201) 

O J- 



S-l 



(m.e.p.) =- zrr [(sup.pr.)((lel.pr.) 2 ]*^3(^ *° -1). 



(202) 



Conditions to Give Best-receiver Pressures. All the foregoing discussion 
of best-receiver pressures for the three-stage compressor can apply only to cases 
in which all the conditions are fulfilled necessary to the existence of best-receiver 
pressures. These conditions are expressed by equations (173), (174), (175), 
(190), (191), and (192), which may be combined as follows: 



RJ 



(1) (2) (3) (4) 

(L . P. Cap.i; (L. P. Cap. 2 ) = DiE vl = LhEa 
(L. P. Cap. 2 ) (L. P. Cap's) D 2 E v2 D 3 E v3 

(5) (6) 

Di(1+ci-clRJE) D 2 (l+c 2 -c 2 R p zS) 



D 2 ( 1 + c 2 — c 2 R p - 6s j D 3 1 1 -f c 3 — czRj,-^ J 



(203) 



Parts (1) and (2) of this equation state the requirements in terms of 
capacities; (3) and (4) in terms of displacements and volumetric efficiencies; 
(5) and (6) in terms of displacements and clearances. In order, then, that best- 
receiver pressure may be obtained, there must be a certain relation between 
the given ratio of compression and dimensions of cylinders and clearances. 
Since, after the compressor is once built these dimensions are fixed, a given 
multi-stage compressor can be made to give best-receiver pressures only when 
compressing through a given range, i.e., when R v has a definite value. If R v 
has any other value the receiver pressures are not best, and the methods of the 
previous Section (15) must be applied. 

When clearance percentages are equal in all three cylinders, ci=c 2 = C3, and 
the volumetric efficiencies are all equal then, when best-receiver pressures exist, 
Eq. (203) becomes, 

5^=^- = -^- = for equal clearance per cent. . . (204) 

Evidently this same expression holds if clearances are all zero or negligible. 
What constitutes negligible clearance is a question requiring careful thought 
and is dependent upon the ratio of compression and the percentage of error 
allowable. 



146 ENGINEERING THERMODYNAMICS 

Indicated horse-power of the compressor is found by multiplying the work 
per cycle, Eq. (196) by the number of cycles per minute, n, and dividing the 
product by 33,000. 

LH.P. = ^ (s 'Z'Y' ) nD 1 E vl (Rjw - 1) .... (205) 
s—l 70.4 



From this are obtained the following: 
H.P. per cubic foot supplied per minute 






n(L. P. CapO = i=I ^6A- (R ° ~ 1] (206) 

H.P. per cubic foot delivered and cooled per minute 

I.H.P. . (del.pr.)^-^ ^ 



w(H.P.Cap.cold) s-1 76.4 

H.P. per cubic foot delivered hot per minute 

LH.P. 8 (sup.pr.) gi±l i^ 1 

n(H.P.Cap.hot) = ^l-76^^ 3 * (ft * _1) ' ' " (208) 

(See Eq. (156)). 

It is useful to note that these expressions are all independent of clearance, 
which is to be expected, since the multi-stage compressor may be regarded as a 
series of single-stage compressors, and in single stage such an independence 
was found for work and horse-power per unit of capacity. 

Example. If the following three-stage compressor be run at best-receiver pressures 
what will be the horse-power and the best-receiver pressures? Compressor has low- 
pressure cylinder 32x24 ins. with 5 per cent clearance, is compressing air from atmos- 
phere to 140 lbs. per square inch absolute, so that s equals 1.4 and it runs at 100 R.P.M. 

From the formula Eq."(196) 

4Q9<? /s—l \ 

W = s ^- I (sup.pr.) D 1 E vl [Riss- - 1) 

From the formula Eq. (188) 

(best 1 rec.pr.)=[(sup.pr.) 2 (del.pr.)]$ 
= (15) 2 X140]J=31.6. 



WORK OF COMPRESSORS 147 

From Eq. (189) 

(best 2 rec.pr.) = [(sup.pr.) (del.pr.) 2 ]* 
= [15x(140) 2 ]*=66.5. 
From Eq. (193) 

E V1 = ( l+cx-dtf/ 8 



1 + 



.05-.05x(^) 3fi =96.5; 



hence, 



W =432 X^ X 15 X 1 1.2 X96.5 X (9.35' 95 - 1) = 59,000 ft.-lbs., 
.4 ■ 

or, 

TTTT3 59,000x200 Q - Q 
IKF '~ 33,000 ~ 358 

Prob. 1. There is available for running a compressor 175 H.P. How many cu.ft. 
of free air per minute can be compressed from atmosphere to 150 lbs. per square inch 
gage by a three-stage adiabatic compressor with best-receiver pressures? 

Prob. 2. The low-pressure cylinder of a three-stage compressor has a capacity of 
4i cu.ft. per stroke. If the stroke of all three cylinders is 18 ins., what must be the 
diameters of the intermediate and high to insure best-receiver pressures, if clearance 
be neglected, and (sup.pr.) be 1 lb. per square inch absolute and (del.pr.) 15 lbs. per 
square inch absolute, s being 1.4. 

Prob. 3. The above compressor is used as a dry-vacuum pump for use with a sur- 
face condenser. If 800 cu.ft. of (sup.pr.) gas must be handled per minute what horse- 
power will be needed to run it? What will be the horse-power per cubic foot of atmos- 
pheric air? 

Prob. 4. Will a 15 X22 X34 X24 in. compressor with clearances of 3, 5 and 8 per cent 
in low, intermediate and high-pressure cylinders respectively be working at best-receiver 
pressures when (sup.pr.) is 15 lbs. per square inch absolute and (del.pr!) 150 lbs. per 
square inch absolute? If not, find by trial, the approximate (del.pr.), for which this 
machine is best, with s equal to 1.4? 

Prob. 5. For the best (del.pr.) as found above find the horse-power to run the 
machine at 100 R.P.M. and also the horse-power per cu.ft. of (del. pr.) air cold? 

Prob. 6. Should this compressor be used for compressing ammon'a would the 
best (del. pr.) change, and if so what would be its value? Also what power would Le 
needed for this case? 

Prob. 7. Compare the work necessary to compress adiabatically in three stages from 
20 lbs. per square inch absolute to 200 lbs. per square inch absolute, the following gases: 

Air; Oxygen; Gas-engine mixtures, for which s=1.36. 

Prob. 8. For 5 per cent clearance in all the cylinders what must be the cylinder ratio 
for best-receiver pressure and a pressure ratio of 10? 



148 ENGINEERING THERMODYNAMICS 

Prob. 9. A compressor, the low-pressure cylinder of which is 30x20 ins. with 5 per 
cent clearance is compressing air adiabatically from atmosphere to 150 lbs. per square 
inch gage, at best-receiver pressure. Due to a sudden demand for air the (del. pr.) 
drops to 100 lbs. per square inch gage. Assuming that the (1 rec. pr.) dropped to 5 lbs. 
per square inch gage and (2 rec.pr.) dropped to 40 lbs. per square inch gage, how much 
would the speed rise if the power supplied to machine was not changed? 

17. Comparative Economy or Efficiency of Compressors. As the prime 

duty of compressors of all sorts is to move gas or vapor from a region of low 
to a region of high pressure, and as this process always requires the expenditure 
of work, the compressor process which is most economical is the one that 
accomplishes the desired transference with the least work. In this sense, then, 
economy of compression means something different than efficiency, as ordi- 
narily considered. Ordinarily, efficiency is the ratio of the energy at one point 
in a train of transmission or transformation, to the energy at another point, 
whereas with compressors, economy of compression is understood to mean the 
ratio of the work required to compress and deliver a unit of gas, moving it 
from a low- to a high-pressure place, to the work that would have been required 
by some other process or hypothesis, referred to as a standard. This economy 
of compression must not be confused with efficiencyof compressors as machines, 
as it is merely a comparison of the work in the compressor cylinder for an actual 
case or hypothesis to that for some other hypothesis taken as a standard. The 
standard of comparison may be any one of several possible, and unfortunately 
there is no accepted practice with regard to this standard. It will, therefore, 
be necessary to specify the standard of reference whenever economy of compres- 
sion is under consideration. The following standards have been used with 
some propriety and each is as useful, as it supplies the sort of information really 
desired. 

First Standard. The work per cubic foot of supply gas necessary to com- 
press isothermally (Cycle 1), from the supply pressure to the -delivery pressure 
of the existing compressor and to deliver at the high pressure is less than that of 
any commercial process of compression, and may be taken as a standard for 
comparison. Since, however, actual compressors never depart greatly from 
the adiabatic law, their economy compared with the isothermal standard will 
always be low, making their performance seem poor, whereas they may be as 
nearly perfect as is possible, so that it may appear that some other standard 
would be a better indication of their excellence. 

Second Standard. The work per cubic foot of gas supplied when compressed 
adiabatically in a single stage (Cycle 3), if taken as a standard, will indicate a 
high economy, near unity for single-stage compressors, and an economy above 
unity for most multi-stage compressors. For the purpose of comparison it 
will be equally as good as the first standard,and the excess of the economy over 
unity will be a measure of the saving over single-stage adiabatic compression. 
Since, however, single-stage adiabatic compression is not the most economical 
obtainable in practice for many cases, this standard may give an incorrect 
idea of the perfection of the compressor. 



WORK OF COMPRESSORS 149 

Third Standard. Due to the facts noted above, it may be a better indica- 
tion of the degree of perfection of the compressor to compare the work per 
cubic foot of gas supplied with that computed for the standard adiabatic cycle 
most nearly approaching that of the compressor. This standard is, however, 
open to the objection that a multi-stage compressor is not referred to the same 
cycle as a single-stage compressor, and a multi-stage compressor with other 
than best-receiver pressure is not referred to the same cycle as another operating 
with best-receiver pressure. This is, therefore, not a desirable standard for 
comparing compressors of different types with one another, although it doe s 
show to what extent the compressor approaches the hypothetical best condi- 
tion for its own type and size. 

Other standards might be chosen for special reasons, each having a value 
in proportion as it supplies the information that is sought. 

It is seen from the discussion of the second standard that its only advantage 
over the first is in that it affords a measure of the saving or loss as compared 
with the single-stage adiabatic compressor cycle. 

If the first standard, that of the isothermal compressor cycle, be adopted 
for the purpose of comparison, it at once gives a measure of comparison with 
the isothermal, which is more and more nearly approached as the number 
of stages is increased, though never quite reached, or as the gas is more effect- 
ively cooled during compression. It may be regarded as the limiting case of 
multi-stage compression with perfect intercooling, or the limiting case of con- 
tinuous cooling. 

In order to ascertain how nearly the actual compressor approaches the 
adiabatic cycle most nearly representing its working conditions, the economy of 
of the various reference cycles heretofore discussed may be tabulated or charted, 
and the economy of the cycle as compared with that of the actual performance 
of the compressor will give the required information. The process of com- 
putation by which this information is obtained will depend upon the nature 
of information sought. The economy of actual compressor compared with the 
isothermal may be stated in any of the following ways: 



Computed work per cubic foot supplied, isothermal . 

Indicated work per cu.ft. actual gas supplied to compressor 

I.H.P. per cubic foot per minute supplied, isothermal 
I.H.P. per cubic foot per minute actual supplied 



Single stage 

(m.e.p.) isothermal, pounds per square inch, no clearance , N 

f c ) 

(m.e.p.) actuals true volumetric efficiency 

Multi-stage 

(m.e.p.) isothermal, no clearance 



(209) 



(m.e.p.) reduced to first staged first stage vol. eff. 



(d) 



150 ENGINEEEING THERMODYNAMICS 

In this connection it is useful to note that for the case of the no-clearance 
cycles, the work per cubic foot of supply is equal to the mean effective pressure 
(M.E.P.) in pounds per square foot, and when divided by 144 gives (m.e.p.) 
in pounds per square inch. Also, that in cases with clearance, or even actual 
compressors with negligible clearance, but in which, due to leakage and other 
causes, the true volumetric efficiency is not equal to unity, 

Work per cubic foot gas supplied X#t>= 144 (m.e.p.). . . (210) 

The information that is ordinarily available to determine the economy 
of the compressor will be in the form of indicator cards from which the (m.e.p.) 
for the individual cylinders may be obtained with ordinary accuracy. The 
volumetric efficiency may be approximated from the indicator cards also, but 
with certain errors due to leakage and heating, that will be discussed 
later. If by this or other more accurate means the true volumetric 
efficiency is found, the information required for the use of Eq. (209) (c) 
or (d) is available. Evaluation of the numerator may be had by Eq. (31), 
which is repeated below, or by reference to the curve sheets found at the end 
of this chapter. (Fig. 50.) 

» Mean effective pressure, in pounds per square inch for the isothermal com- 
pressor without clearance is given by 

(m.e.p.) isothermal = (sup.pr.) log e R p (211) 

The curve sheet mentioned above also gives the economy of adiabatic 
cycles of single stage, also two and three stages with best-receiver pressures. 
The value of s will depend upon the substance compressed and its condition. 
The curve sheet is arranged to give the choice of the proper value of s applying 
to the specific problem. 

If it is required to find the economy of an actual compressor referred to 
the third standard, i.e., that hypothetical adiabatic cycle which most nearly 
approaches the actual, then 



Economy by third standard is 

Econ. actual referred to isothermal 



Econ. hypothetical referred to isothermal' 



(212) 



It is important to notice that for a vapor an isothermal process is not one 
following the law PXV = constant. What has, in this section, been called an 
isothermal is correctly so called only so long as the substance is a gas. Since, 
however, the pressure-volume analysis is not adequate for the treatment of 
vapors, and as they will be discussed under the subject of Heat and Work, 
Chapter VI, it is best to regard this section as referring only to the treatment 
of gases, or superheated vapors which act very nearly as gases. However, 
it must be understood that whenever the curve follows the law PXV = constant, 
the isothermal equations for work apply, even if the substance be a vapor 
and the process is not isothermal. 



WORK OF COMPRESSORS 151 

18. Conditions of Maximum Work of Compressors. Certain types of com- 
pressors are intended to operate with a delivery pressure approximately con- 
stant, but may have a varying supply pressure. Such a case is found in pumps 
or compressors intended to create or maintain a vacuum and in pumping 
natural gas from wells to pipe lines. The former deliver to atmosphere, thus 
having a substantially constant delivery pressure. The supply pressure, 
however, is variable, depending upon the vacuum maintained. In order that 
such a compressor may have supplied to it a sufficient amount of power to 
keep it running under all conditions, it is desirable to learn in what way this 
power required will vary, and if it reaches a maximum what is its value, and 
under what conditions. 

Examine first the expression for work of a single-stage adiabatic compressor 
with clearance. The work per cycle will vary directly as the mean effective 
pressure. Eq. (69.) 

(m ..p J =^ 1 «„p.p„[ 1+ «- t (|^)-][(ggf-]. m 



This will have a maximum value when 

d(m.e.p.) 

or when 

s-l 



c?(sup.pr.) 



= 0, 



/deLpr.X . _ s [ 1+C _ C ^U^FLY] = . . . . (214) 
\sup.pr./ l+c L s \sup.pr./ J 

Solving this for the value of supply pressure will give that supply pressure 
at which the work will be a maximum, in terms of a given delivery pressure, 
clearance and the exponent s. 

The assumption most commonly used is that clearance is zero. If this is 
true r or the assumption permissible, the above equation becomes simplified, 

(sup.pr.) ' ' • ^ ' 

The value of s for air, for instance, is 1.406, and hence the ratio of compression 
for maximum work for the hypothetical air compressor is 

(1.406) 3 ' 46 = 3.26 (216) 

It may be noted that when s = 1 in the above expression, the value of the 
ratio of compression become indeterminate. To find the supply pressure for 
maximum work in this case, take the expression for mean effective pressure 
for the isothermal compressor (s = l), Eq. (43), 

(m.e.p.) = (sup.pr.)fl+c-c(^H,)llog e (^^-). . . . (217) 

- . L \sup.pr,/J - \sup.pr./ v ' 



152 ENGINEERING THERMODYNAMICS 

Differentiate with respect to (sup.pr.) and place the differential coefficient equal 
to zero. This process results in the expression 



\sup. 



.pr./ 1 + c \sup.pr. 
When c = 0, this becomes, 



del.pr.\ + _c_/del.pr. ) _ 1 = Q (21g) 



/del.pr. 

log e ^~— 

\sup.pr. 



I '^^M =2.72. . . 
sup.pr. 



The expressions Eqs. (215) and (219) given are easily solved, but Eqs. (214) 
and (218) are not, and to facilitate computations requiring their solution 
the results of the computation are given graphically on the chart, Fig. 48, at 
the end of this chapter. 

The mean effective pressure for a compressor operating under maximum 
work conditions may be found by substituting the proper ratio of compression, 
found as above, in Eq. (213) or (217). In Fig. 48 are found also the results 
of this computation in the form of curves. Note in these curves that the mean 
effective pressure is expressed as a decimal fraction of the delivery pressure. 

The discussion so far applies to only single-stage compressors. The problem 
of maximum work for multi-stage compressors is somewhat different, and its 
solution is not so frequently required. Moreover, if the assumption of perfect 
intercooling is made, the results are not of great value, as a still greater amount 
of power might be required, due to the failure for a period of time of the supply 
of cooling water. Consider this case first. 

If intercooling be discontinued in a multi-stage compressor, the volume 
entering the second stage will equal that delivered from the first, and similarly 
for the third and second stages. The entire work done in all stages will be the 
same as if it had all been done in a single stage. It might be questioned as 
to whether this would hold, when the ratio of compression is much less than 
designed. The first stage will compress until the volume has become as small 
as the low-pressure capacity of the second stage. If the delivery pressure is 
reached before this volume is reached, there is no work left to be done in the 
second or any subsequent stages, and, due to the pressure of the gas, the valves, 
if automatic, will be lifted in the second and higher stages, and the gas will 
be blown through, with only friction work. It appears then that under the 
condition of no intercooling the multi-stage compressor acts the same as a 
single stage, and the conditions of maximum work will be the same. 

If intercooling is maintained perfect there will still be a range of pressures 
on which all the work of compression is done in the first stage, merely blowing 
the discharge through receivers, valves and cylinders in the upper stages. If 
this range is such that this continues beyond a ratio of pressures, which gives a 
maximum (m.e.p.) for the single stage, then the maximum will have been reached 
while the compressor is operating single stage, and the single-stage formulae 
and curves may be applied to this case also. 



WORK OF COMPRESSORS 



153 



That this condition frequently exists with multi-stage compressors of 
ordinary design is shown by the fact that the ratio of compression in each stage 
is seldom less than 3, and more frequently 3.5, 4 or even more. The ratio 
of compression giving maximum work for single stage, has values from 2.5 
to 3.26, dependent on clearance and the value of 6' for the gas compressed, and 
is, therefore, less for the majority of cases. 

If a curve be drawn, Fig. 36, with ratio of compression as abscissas and 
(m.e.p.-i-del.pr.) as ordinates, so long as the action is single stage, a smooth curve 
will result, but when the ratio of compression is reached above which the second 
cylinder begins to act, the curve changes direction suddenly, falling as the ratio 






.42 



.40 



.38 









*$£> 


•ov Single 


Sth ff e 




































^^ M 


ix for Tw 


o Stage C( 
Cyli 


impressor 
lder liati( 


) 2.5 : 1 





























































Values of R p . 



Fig. 36. — Curve of Relation between Mean Effective Pressure per Pound of Delivery Pressure 
in Terms of Pressure Ratio for Air, showing Maximum Value. 



of compression increases. Hence, if the ratio of cylinders is such that the single- 
stage maximum is not reached before the second stage begins to operate, the highest 
point of the curve, or maximum work for a given delivery pressure will occur 
when the ratio of supply and delivery pressures is such as to make first-receiver 
pressure equal to delivery pressure. 

19. Actual Compressor Characteristics. Air or gas compressors are very 
commonly made double acting, so that for a single cylinder, two cycles will 
be performed during one revolution, one in each end of the cylinder. If a rod 
extends through one of these spaces and not through the other, the displace- 
ment of that end of the cylinder will be less than the other by a volume equal 
to the area of rod multiplied by the stroke. To avoid mechanical shock 
at the end of either stroke, it is necessary to leave some space between the 



154 ENGINEERING THERMODYNAMICS 

piston and cylinder head. Passages must also be provided, communicating 
with inlet and discharge valves. The total volume remaining in this space j 
and in the passages when the piston is at the nearer end of its stroke constitutes 
the clearance. The amount of this clearance volume varies from .5 or .6 of 
one per cent in some very large compressors to as much as 4 or 5 per cent of 
the volume of displacement in good small cylinders. 

In order to study the performance of an actual compressor and to compare 
it with the hypothetical cycle, it is necessary to obtain an indicator card, and 
knowing the clearance and barometric pressure to convert the indicator card 
into a pressure volume diagram, by methods explained in Chapter I. Fig. 37, 
is such a diagram for a single-stage compressor. In the pipe leading to the 



Press. 



H.P. Cap Hot (Apparent) 
.(Del. Pr.) 




fc;| C (Sup. Pr.) 

-Vol. 
Fig. 37. — Compressor Indicator Card Illustrating Departure from Reference Cycle. 

intake valve the pressure is determined and a horizontal line AB is drawn 
on the diagram at a height to represent the supply pressure. Similarly, 
discharge pressure is determined and drawn on the diagram, KE. Consider 
the four phases of this diagram in succession. 

1. Intake Line. At a point somewhat below A the intake valve opens, 
say at the point F. This remains open till a point H is reached at or near the 
end of the stroke. The line connecting these two points indicates variations 
of pressures and volumes throughout the supply stroke. In general this line 
will lie below the supply-pressure line AB due to first, the pressure necessary 
to lift the inlet valve from its seat against its spring and inertia, if automatic, 
and support it, and second, friction in the passages leading to the cylinder from 
the point where the supply pressure was measured. While the former is nearly 
constant the latter varies, depending upon the velocity of gases in the passages. 
The piston attains its highest velocity near the middle of the stroke, thus 



WORK OF COMPRESSORS 155 

causing the intake line to drop below the supply pressure more at this part 
of stroke. These considerations do not, however, account completely for the 
form of the intake line. Frequently the first portion of the line lies lower than 
the last portion, even at points where the piston velocities are equal. This 
is more prominent on a compressor having a long supply pipe, and is due to 
the forces required to accelerate the air in the supply pipe while piston velocity 
is increasing, and to retard it while piston velocity is decreasing. In com-* 
pressors where the inlet valve is mechanically operated and the supply pipe 
long, it is possible to obtain a pressure at the end of the intake, line H, even 
in excess of the supply pressure. The effect of this upon volumetric efficiency 
will be noted later. 

The apparent fluctuations in pressure during the first part of the intake 
line may be attributed, first, to inertia vibrations of the indicator arm, in which 
case the fluctations may not indicate real variations of pressure; second, the 
indicator card may show true variations of pressure due to inertia of the gases 
in the supply pipe, since a moment before the valve opened at F the gases were 
stationary in the supply pipe. When F is reached the piston is already in motion 
and a very considerable velocity is demanded in the supply pipe to supply 
the demand. This sudden acceleration can be caused only by a difference in 
pressure, which is seen to exist below and to the right of F on the diagram. 
The suddenness of this acceleration may start a surging action which will cause 
rise and fall of pressure to a decreasing extent immediately after. A third cause 
is possible, that is, a vibration of the inlet valve due to its sudden opening 
when it is of the common form, mechanical valves change the conditions. It 
is closed by weight or a spring and opened by the pressure difference. Between 
these forces the valve disk may vibrate, so affecting the pressure. 

2. Compression Line. From the time the inlet valve closes at the point 
H until the discharge valve opens at the point G, the gases within the cylinder 
are being compressed. The compression is very nearly adiabatic in ordinary 
practice, but due to the exchange of heat between the cylinder walls, at first 
from walls and later from gas to walls, which are cooled by water jacket to 
i prevent the metal from overheating, there is a slight departure from the adia- 
t batic law almost too small to measure. 

p A second factor which influences the form of this curve to a greater extent 

is leakage. This may occur around the piston, permitting gas to escape from 

$ one end of the cylinder to the other. During the compression process there is 

3 first an excess of pressure in the other end of the cylinder due to reexpansion, 

E tending to increase pressures on the first part of compression. Later, the 

B pressure rises and the pressure on the other side of the piston falls to supply 

<; pressure. During this period leakage past piston tends to decrease successive 

t pressures or lowers the compression line. Leakage also occurs through either 

discharge or inlet valves. The former will raise the compression line, while 

excessive leakage of the inlet valve will lower it. 

It is then evident that unless the nature of the leakage is known, it is 
I impossible to predict the way in which it will change this line. It is, however, 



156 ENGINEERING THERMODYNAMICS 

more frequently the case that the piston and inlet leakage are large as compared 
with the discharge valves, in which case the actual compression line has a 
tendency to fall lower than the adiabatic as the volumes are decreased. Com- 
pression lines lowered by leakage are often mistaken for proofs of effective 
cooling, and cases have been known where isothermal compression of air was 
claimed on what proved to be evidence only of bad leakage. 

3. Delivery Line. After the delivery pressure of the compressor has been 
exceeded sufficiently, the discharge valve is opened and the gas is delivered to 
the discharge pipe or receiver till the end of the stroke is reached and at the 
point J the valve closes. The same group of factors influence the form of this 
line as act upon the intake line; spring resistance of discharge valve: friction 
in discharge passages varying with piston velocity; inertia of gases in delivery 
pipe ; sudden acceleration of gases in delivery pipe when discharge valve opens, 
and inertia of indicator arm, but in addition a strong tendency for the valve 
to chatter or jump open and shut alternately. Leakage also occurs through 
tntake valve and past piston during this process, with the result that less gas 
passes through the discharge valve than is shown on the indicator card. 

4. Reexpansion Line. From the time the discharge valve closes, at J, 
till the intake valve opens, at F, the gas which remained in the clearance space 
after delivery expands, due to the advancing of the piston, till the pressure 
has fallen to such an amount that the intake valve will open. The same factors 
influence this line as the compression line. Heat is exchanged with the jacketed 
cylinder walls, at first cooling and later heating the gas as the pressure falls. 
This, for any given volume, changes the pressure. Leakage occurs inward 
through the discharge valve and outward through intake valve and past piston. 
If these last two are in excess, the pressures will fall more rapidly than if the 
expansion were that of a constant quantity of gas. 

Work due to gas friction and inertia, it should be noted, is fully represented 
on the indicator card, and may be regarded as being equal to that extra area 
below the supply-pressure line and above the delivery-pressure line. In the 
combined card of a two-stage compressor there would be an overlapping of 
the diagrams due to this frictional loss. 

Low-pressure Capacity. Referring to the adiabatic compression and expan- 
sion lines, CD and KL, Fig. 37, it is seen that the low-pressure capacity of the 
hypothetical cycle is the volume, LC. 

The apparent loiv-pressure capacity of the actual compressor, measured at 
the supply pressure is AB. This is not, however, the true volume of gas at 
supply pressure and temperature that is taken in, compressed, and finally 
delivered per cycle. First, the valves, passages and walls are not at the same 
temperature as the entering gas, due to the heat left from the compression of 
the previous charge. This causes the temperature of the gas within the cylin- 
der to be something higher than the supply gas outside. This causes it to 
be less dense, and hence an equivalent weight of gas at supply temperature 
and pressure would occupy a volume somewhat less than AB. Second, the gas 
which occupies the volume AB has not all entered the cylinder through the intake 



WORK OF COMPRESSORS 157 

valve. After reexpansion is completed the intake valve opens and gas enters 
the end of the cylinder under consideration. At the same time compression 
is taking place in the other end, and later delivery'. During these processes 
whatever gas leaks past the piston tends to fill the end of the cylinder in which 
intake is going on. Leakage past the discharge valve also tends to fill the cylin- 
der with leakage gas. Both of these tendencies decrease the quantity of gas 
entering through this intake valve, and its true amount when reduced to external 
supply pressure and temperature is, therefore, less than the volume AB. 

The true low-pressure capacity of the compressor is the true volume of gas 
under external supply conditions that enters the cylinder for each cycle. This 
cannot be determined from the indicator card except by making certain assump- 
tions which involve some error at best. It can, however, be ascertained by 
means of additional apparatus, such as meters or calibrated nozzles or receivers, 
by means of which the true amount of gas compressed per unit of time is made 
known. This reduced to the volume per cycle under supply pressure and tem- 
perature will give the true low-pressure capacity. 

Volumetric efficiency is defined as being the ratio of low-pressure capacity 
to displacement. On the diagram, Fig. 37, the displacement is represented 
to the volume scale by the horizontal distance between verticals through the 
extreme ends of the diagram, K and H. Since there are three ways in which 
the low-pressure capacity may be approximated or determined, there is a 
corresponding number of expressions for volumetric efficiency. 

1. The volumetric efficiency of the hypothetical cycle is 

^(hypothetical) = ^^ L - P ; ^ (220) 

v J ^ ' (displacement) 

and this is evaluated and used in computations in the foregoing sections of 
this chapter. 

2. The apparent volumetric efficiency is 

fi.(apparent) = <*»%«? ^ R ^ , (221) 

(displacement) 

and would be very nearly equal to the true volumetric efficiency were it not 
for leakage valve resistance and heating during suction, but due to this may 
be very different from it. 

3. The true volumetric efficiency is 

fl.(true)- (^L.P.Cap.) 

(displacement) 

In problems of design or prediction it is necessary either to find dimensions, 
speeds and power necessary to give certain actual results, or with given dimen- 
sions and speeds to ascertain the probable power and capacity or other 



158 ENGINEEEING THERMODYNAMICS 

characteristics of actual performance. Since it is impossible to obtain actual 
performance identical with the hypothetical, and since the former cannot be 
computed, the most satisfactory method of estimate is to perform the computa- 
tions on the hypothetical cycle, as is» explained in previous sections of this chap- 
ter, and then to apply to these results factors which have been found by 
comparing actual with hypothetical performance on existing machines as nearly 
like that under discussion as can be obtained. This necessitates access to data 
on tests performed on compressors in which not only indicator cards are taken 
and speed recorded, but also some reliable measurement of gas compressed. 
The following factors or ratios will be found of much use, and should be 
evaluated whenever such data is to be had : 

= Eyjirue ) = (true L. P. Cap.) , . 

1 E„ (hypothetical) (hypothetical L. P. Cap.)* 

_ .JE* p (true) (true L . P. Cap.) (9oa\ 

^(apparent) (apparent L. P. Cap.)' 

true I.H.P. true m.e.p. rooK\ 

hypothetical I. H. P. hypothetical m.e.p.' 



Then 



true work per cu.ft. gas, supplied 



hypothetical work per cu.ft. gas, supplied 

I.H.P. 



true 



true I.H.P. per cu.ft. gas supplied (L. P. Cap.) 



hypothetical I.H.P. per cu.ft. supplied , , . , I.H.P. 



(226) 



(L. P. Cap.) 

_ true m.e.p. . true L. P. Cap. _es 
hypothetical m.e.p. ' hypothetical L. P. Cap. e\ 

This ratio can be used to convert from hypothetical work per cubic foot 
gas supplied to probable true work per cubic foot. 

Multi-stage Compressors are subject in each stage to all of the characteristics 
described for single stage to a greater or less extent. Valve resistance, friction 
and inertia affect the intake and discharge lines; heat transfer and leakage 
influence the form of compression and reexpansion lines, and the true capacity 
of the cylinder is made different from the apparent due to leakage, pressure 
and temperatures changes. 

In addition to these points it is useful to note one special way in which the 
multi-stage compressor differs from the single stage. The discharge of the first 
stage is not delivered to a reservoir in which the pressure is constant, but a 
receiver of limited capacity. The average rate at which gas is delivered to 
the receiver must equal the average rate at which it passes to the next cylinder. 
The momentary rate of supply and removal is not constantly the same, however. 



WORK OF COMPRESSORS 



159 



This causes a rise or fall of pressure. It is evident that this pressure fluctuation 
is greatest for a small receiver. Very small receivers are not, however, used 
on gas compressors due to the necessity of cooling the gas as it passes from one 
stage to the next. To accomplish this a large amount of cooling surface must 
be exposed, requiring a large chamber in which it can be done. Thus, it is 
seen that the hypothetical cycles assumed for multi-stage compressors do not 
truly represent the actual cycle, but the difference can never be very great, 
due to the large size of receiver which must always be used. 

Another way in which the performance of this multi-stage compressor 
commonly differs from assumptions made in the foregoing discussions is in 



Del. Pr 




Fig. 38. — Effect of Loss of Intercooling in Two-stage Compressors on Receiver-pressure and 
Work Distribution in the Two Cylinders. 

regard to intercooling. It seldom occurs that the gases enter all stages at 
the same temperature. In the several stages the temperature of the gases 
will depend on the amount of compression, on the cooling surface and on the 
amount and temperature of cooling water. The effect of variations in tem- 
oerature upon the work and receiver pressures will be taken up later. It may 
oe noted now, however, that if all cooling water is shut off, the gas passes from 
Dne cylinder to the next without cooling, there is no decrease in volume in the 
-ece iver. For simplicity take the case of zero clearance, two-stage (Fig. 38). 
UtABCDEF be the cycle for perfect intercooling. AB and KD are the low- 
pressure capacities of the first and second stages respectively. If now, inter- 
cooling ceases, the gas will no longer change volume in the receiver. The 
eceiver gas, in order to be made sufficiently dense to occupy the same 



160 ENGINEERING THERMODYNAMICS 

volume (KD) as it did before / must be subjected to a greater pressure in the 
first stage. The new receiver line will be K'D f . The work of the first 
stage will therefore be ABD'K 1 ', of the second stage K'D'GF, and the total 
work in the new condition is greater than when intercoooling was perfect by an 
amount represented by the area DCGE. 

In the case where clearance is considered, the effect is the same, except that 
the increasing receiver pressure, increasing the ratio of compression of the first 
stage, causes the volumetric efficiency of the first stage to become less, and 
hence lessens the capacity of the compressor. The effect on work per unit of 
capacity is the same as without clearance. 

The question as to how many stages should be used for a given compressor 
is dependent upon the ratio of compression largely, and so is due, first, to con- 
siderations of economy, which can be understood from the foregoing sections; 
second, for mechanical reasons, to avoid high pressures in large cylinders; third, 
for thermal reasons, to avoid such high temperatures that the lubrication of 
the cylinders would be made difficult, or other dangers, such as explosions, 
involved. 

Practice varies very widely as to the limiting pressures for single, two, 
three or four-stage compressors. Air compressors of a single stage are com- 
monly used for ratios of compression as high as 6 or 7 (75 to 90 lbs. gage). For 
ratios greater than these, two-stage compressors are used, especially for larger 
sizes, up to ratios of 34 to 51 (500 to 750 lbs. gage). Some three-stage com- 
pressors are used for ratios as low as 11 or 14 (150 or 200 lbs. gage), although 
installations of this nature are rare, and are warranted only when power is costly 
and the installation permanent and continuously used to warrant the high 
investment cost. As a minimum ratio for three stages, 11 (150 lbs. gage) 
is used for large units, while a few small units compress as high as 135 or even 
170 atmospheres (2000 or 2500 lbs. gage). A notable use for the four-stage 
compressor is for charging the air flask of automobile torpedoes used by the 
various navies, which use pressures from 1600 to 3000 lbs. per square inch 
(110 to 200 atmospheres). These require special design of valves, cylinders 
and packings to withstand the extremely high pressures, small clearances, and 
special precautions against leakage, due to the great loss of volumetric efficiency 
and economy that would otherwise result. 

20. Work at Partial Capacity in Compressors of Variable Capacity. It 
is seldom that a gas compressor is run continuously at its full capacity. If 
the duty of the compressor is to charge storage tanks, it may be made to run 
at its full capacity until the process is completed and then may be stopped 
entirely, by hand. Even where the compressed gas is being used continuously 
it is common practice to have a storage reservoir into which the compressor may 
deliver. This enables the compressor to deliver a little faster or slower than 
the demand for a short period without a great fluctuation pressure in the 
reservoir. For many purposes hand regulation is not sufficient or is too 
expensive, hence the demand for automatic systems of capacity regulation. 
These systems may be classified in a general way in accordance with the method 



i 



WORK OF COMPRESSORS 161 

of driving. Some methods of power application permit of speed variations while 
others require constant speed. The former provides in itself a means of regulat- 
ing capacity within certain limits, while, if the compressor must run at constant 
speed, some additional means of gas capacity control must be provided. 

Compressors driven by an independent steam engine, or steam cylinders 
constituting part of the same machine may be made to run at any speed required 
within a very wide range and still kept low enough for safety. If driven 
by gear, belt, rope, chain or direct drive from a source of power whose speed 
is constant, the speed of the compressor cannot be varied. Electric motor, 
gas-engine, oil-engine or water-power drives are subject to only limited speed 
alteration and may, therefore, be placed in the constant speed class. 

Regulation of Capacity by Means of Speed Change. If the speed of the com- 
pressor is decreased below normal : 

1. Displacement of piston is decreased in proportion to the speed. 

2. Mean effective pressure, as to hypothetical considerations, is the same, 
but due to the decrease of velocities in gas passages, the frictional fall of pres- 
sure during inlet and delivery is not so great, and hence the mean effective pres- 
sure ,is not quite so great. If the compressor is multi-stage, since a smaller 
quantity of gas is passing through the intercooler, it is probable that the inter- 
cooling is more nearly perfect, thus decreasing the mean effective pressures in 
the succeeding stages. 

3. The volumetric efficiency is changed, due first to the fact that leakage 
is about the same in total amount per minute as at full speed, but the total 
quantity of gas being less, leakage is a larger percentage of the total; second, 
the inertia of gases in the supply pipe, as well as their friction, has been decreased. 
The former tends to decrease vulmetric efficiency, while the latter may tend 
to increase or decrease it. It may be expected that the true volumetric efficiency 
will be somewhat greater at fractional speed than at full speed. 

For any compressor there is a speed of maximum economy above and below 
which the economy is less, though it may be that this most economical speed is 
greater than any speed of actual operation. 

It is not desirable at this point to discuss the effect of speed variation upon 
the economy of the engine or other motor supplying the power. The reasoning 
above applies to the term economy as applied to the compression effect obtained 
per unit of power applied in the compression cjdinder. It might be noted here, 
however, that the decrease of speed has little effect upon the mechanical efficiency 
of the compressor as a machine, since frictional resistance between solid parts 
remains nearly constant, and, therefore, power expended in friction will vary 
as the speed, as does approximately also the power to drive the compressor. 
The ratio of frictional power to total may then be expected to remain nearly 
constant. 

Regulation of Capacity at Constant Speed may be accomplished in a number 
of ways : 

1. Intermittent running; 

2. Throttling the supply to compressor; 



162 ENGINEEKING THEKMODYNAMICS 

3. Periodically holding open or shut the intake valve; 

4. Closing intake valve before end of intake stroke, or holding intake valve 

open until compression stroke has been partially completed; 

5. Large clearance; 

6. Variable clearance, 

The first necessitates some means for stopping and starting the compressor, 
which is simple with electric drive, and may be accomplished in other cases by 
means of a detaching clutch or other mechanical device. The pressure in the 
reservoir is made to control this stopping and starting device by means of a 
regulator. This arrangement is made to keep the pressure in the reservoir 
between certain fixed limits, but does not maintain a constant pressure. The 
economy of compression in this case is evidently the same as at full speed 
continuous running, provided there is no loss in the driving system due to 
starting and stopping, which may not be the case. This method of 
regulation is used mainly for small compressors in which inertia is not 
great, such as supply the air brakes on trolley cars. The sudden change 
of load on the driving machinery would be too great if large compressors 
were arranged in this way. 

If the compressor whose capacity is regulated by intermittent running is 
multiti-stage, the constant supply of water to the intercoolers while the compres- 
sor is stopped will lower the temperature of the cooling surface, causing more 
nearly perfect intercooling when the compressor is started. Leakage, on the 
other hand, will permit the loss of pressure to a greater or less extent in the 
receivers while the compressor is stationary, which must be replaced after 
starting before effective delivery is obtained. 

Throttling the gas supply to the compressor has certain effects that may 
be studied by referring to Fig. 39, which represents the hypothetical cycles most 
nearly approaching this case. In order to reduce from the full-load low-pres- 
sure capacity, AB, to a smaller capacity, AE, the supply pressure is decreased 
by throttling to the pressure of B', such that B' and E lie on the same adiabatic. 
The work area A'B'EA is entirely used up in overcoming the throttle resistance 
and is useless friction, so that economy is seriously reduced by this method 
of regulation. Such compressors may use almost as much power at partial 
as at full capacity. 

It is easily seen that this method of regulation would be undesirable, its 
only advantage being simplicity. 

The effect of throttling upon a multi-stage compressor may be illustrated as 
in Fig. 40 , by considering the two-stage compressor cycle without clearance, 
ABCDEF. The ratio of compression of the first cylinder is determined with 

perfect intercooling by the ratio of displacements P c = P b lj~ ) . When the 

supply pressure is throttled down to P/, the new receiver pressure will be 

Pc' = Pb [jyJ > a pressure much lower than P c . Hence the receiver pressure 

is decreased, less work done in the first stage, and far more than half the work 



WORK OF COMPRESSORS 



163 



of compression done in the second stage. If best-receiver pressure existed at 
normal capacity, it does not exist in the throttled condition. 

The intake valve may be held wide open or completely closed during one 
or more revolutions, thereby avoiding the delivery of any gas during that period. 
If the intake valve is held wide open, the indicator card would be as shown in 
Fig. 41A in full lines, ABCD, the dotted lines showing the cycle performed when 



I 


D 


C' 


c 




















\ 






































\ 
\ 
\ 

\ 
\ 

\ 


















\ 




\ 
\ 
\ 

\ 

\ 
















\ 


\ 


\ 
\ 

\ 
\ 
\ 

\ 

\ 


















v 




\ 
\ 
\ 
\ 
\ 
N 
\ 


E 








B 






N 
1 \ 
1 \ 

\ 
i \ 








^^ 














N 


A' 






"^-. 


-^_ 


B' 


















5< L.P.Cap Throttled > 




ip at Full 








^ 


D 


> 





Fig. 39. — Effect of Throttling the Suction of One-stage Compressors, on Capacity and 

Economy. 



normal operation is permitted. With the inlet valve open in this way there is 
a loss of power due to friction of the gas in passage during both strokes, measured 
by the area within the loop. 

Closing the inlet valve and holding it shut will give an indicator card of the 
form EFG, Fig. 415, which will be a single line retraced in both directions 
except for probable leakage effects. If leakage is small, there will be but little 



164 



ENGINEEEING THERMODYNAMICS 



area enclosed between the lines. At a high speed this might be expected to 
incur less lost power than the former plan. 

Certain types of compressors are made with an intake valve controlled 
by a drop cut-off, much like the steam valve of the Corliss engine. The effect 
of this is to cut off the supply of gas before the end of the stroke, after which 
time the gas must expand hypothetically according to the adiabatic law. The 
return stroke causes it to compress along the same line continued up to the 
delivery pressure, as indicated by the line FEG, Fig. 41C. There is little work 



p 


F 


" 














































\ 






















\ 






















\ 






















\ 


\ 
















H 






\ 


C 




















K 

1 


\ 




















1 

1 
1 , 


^ 


^&s 












H 
A 






ID 

i 


•fl 




^ 


"^i: 




R 










l 
1 


^ 




\.. 










A' 






1 

l_ 

1 










^-^ 


B' 


I ) 












<- D 2 ->l 
















Oi 













Fig. 40. — Effect of Throttling Multi-stage Compressors on Receiver-pressure, Work Distri- 
bution, Capacity and Economy. 



lost in the process, none, if the line is superimposed as in the figure, and hence 
the process is the same as if only the cycle AEGD were performed. 

The same quantity of gas might have been entrapped in the cylinder by 
holding the intake valve open until the end of the stroke and on the return till 
the point E, Fig. 41D, was reached, then closing it. The same compression 
line EG will be produced. The line AB will not coincide with BE, due to 
friction of the gas in passages, and hence will enclose between them a small 
area representing lost work, which may be no larger than that lost in the process 
EFE, 



WORK OF COMPRESSORS 



165 



If such an automatic cut-off were applied only to the first stage of a multi- 
stage compressor, the effect would be to lower receiver pressures as in the throt- 
tling process. To avoid this, the best practice is to have a similar cut-off to act 
on the supply to all of the stages. If this is properly adjusted, the receiver 
pressures can be maintained the same as at full load. An additional advantage 
of this system is that even if the compressor is to be used for a delivery pres- 
sure for which it was not originally designed, the relative cut-offs may be so 
adjusted as to give and maintain best-receiver pressure. 



















P 


F 
















1 — — 
1 


\ 

\ 
\ 














) 




\ 

\ 
\ 
















\ 
\ 

\ 
















\ 
\ 














\ 


\ 

\ 


\ 














\ 














\ 
\ 
\ 




X 

« 


V 














\ 

\ 


V 










\ 
\ 


=rr — 




D 


*^^ 


»^._ 


(Su 


?. Pr. 








"'^ 


--.^ 






A 


\?n 




B 






C 




F^ 












n 



B 



P| D ^G ,,C 





\ 


-m w 
















\ 




\ 
\ 














\ 


v 


> 


N 










\ 




\ 


NE 


> 


\ 


*^ 


B 




A V 


t 












F 










(A) 













D 


G 


c 


















\ 


^ 


















\ 




\ 














\ 


\ 




\ 

N 


\. 












\ 




\ 


E 




\ 


-^ 










s 




V. 










•■ R 




a" 


























(B) 













Fig. 41.— Control of Compressor Capacity by A. Open Inlet Valve; B. Closed Inlet Valve; 
C. Suction Cut-off; D. Delayed Suction Closure. 

Since the low-pressure capacity per cycle of a compressor involves clearance 
and ratio of compression as two of its variables, it is possible to change capacity 
by changing either the clearance or the ratio of compression. 



i_ 
(L. P. Cap.) =DE v = D(l+c-cR p ' ). 



.... (227) 

Assuming that clearance is a fixed amount and not zero, it is evident that an 
increase in the ratio of compression decreases the capacity, and when it has 



166 



ENGINEERING THERMODYNAMICS 



reached a certain quantity will make the capacity zero. If the clearance is 

i_ 

large, making the coefficient of R p * large in the equation the effect of a change 
in that factor is increased. Fig. 42 indicates the hypothetical performance 
of a compressor with large clearance. When the pressure of delivery is low 
(say P c ) the capacity is large, AB. The cycle is then ABCD. An increase 
of the delivery pressure to P c ' changes the cycle to A' BCD' and the low pres- 
sure capacity is A'B. If the compressor is delivering to a receiver from which 
no gas is being drawn, the delivery pressure will continue to rise and the capacity 



















D 


\c' 
















\ \ 
















\ \ 
















\ \ 
















\ \ 
















\ \ 
















\ \ 
















\ N 
















\ 














D 


\ 


\c 














; ' \ 
















\ \ 
















\ \ 


















\ > 














\ 


v 














\ 


\ 


















\ 


^^ 














A 




A' 




B 




















s 





Fig. 42. — Variation of Compressor Capacity with Rise of Delivery Pressure, Fixed Clearance, 

Pressure for Zero Delivery. 

to decrease till the capacity approaches zero as the delivery pressure approaches 
the pressure P e as a limit. 



1+c 

(limiting del.pr.) = (sup.pr.) ( 



(228) 



When the limiting condition has been reached and the capacity has become 
zero, the compression and reexpansion lines coincide and enclose zero area 
between them; hence, the mean effective pressure and the indicated horse- 
power are zero, for the hypothetical case. Leakage will prevent a perfect 
coincidence of the lines and cause some power to be required in addition to that 
of friction. 

Such a simple method of regulation as this is used for some small com- 
pressors driven constantly from some source of power used primarily for 
other purposes. When it is not necessary to have a constant delivery pressure, 



WORK OF COMPRESSORS 



167 



but only to keep it between certain limits, this may be made use of, especially 
if the limits of pressure are quite wide. 

The expression for low-pressure capacity Eq. (227), suggests the possi- 
bility of decreasing capacity by the increase of clearance. The effect of this is 
shown in Fig. 43. The original compression cycle (full capacity) is shown by 
ABCD, with a clearance volume of cD, so that the axis of zero volume is OP. 
Increasing the clearance to c'D causes a smaller volume CD to be delivered 
and due to the more sloping re-expansion DA', a smaller volume of gas is 
taken in, A'B. 

It has been shown in previous sections that clearance has no effect upon the 
economy of a compressor so far as hypothetical considerations are regarded. 
In practice it is found that a slight loss of economy is suffered at light load, 





p 




J 




C 




c 




















\ 

\ 

\ 




\ 


\ 
\ 
\ 






















\ 

\ 

1 


\ 




\ 
\ 

\ 
\ 
\ 
























\ 

\ 
\ 

\ 
\ 






\ 

\ 
\ 
\ 
\ 
\ 






















\ 
\ 

\ 


\ 




\ 


N \ 
\ 
\ 

•x 


















1 




\ 
\ 
\ 

\ 

\ 










"^^\. 
















\ 




N 
N 


V 






v 


>>X. 


*^^^^ 












JA 






iA' 












B 


0' 















1 














~r^ 


cB 


/ ; 






.P. Cap Part I 


-. 






<- 




l< 




-> 

> 














-D 







Fig. 43. — Variation of Compressor Capacity by Changing Clearance. 

as might be expected, due to greater leakage per unit of capacity. The addi- 
tional clearance is provided in the form of two or more chambers connected to the 
clearance space of the compressor by a passage in which is a valve automatically 
controlled by the receiver pressure. 

In the multi-stage compressor, decreasing the capacity of the first stage by 
an increase of its clearance would evidently permit a decrease of receiver pres- 
sures unless the capacity of each of the various stages is decreased in the same 
proportion. Eq. (132) gives the condition which must be fulfilled to give best 
receiver pressure for a two-stage compressor. 



Di 



RJ 



A 



1 + Cl — CiR p 2s 



1+C2 — C2R V 2& 



168 ENGINEEEING THERMODYNAMICS 

Since D\> Z>2, and R v remain fixed, for any chosen value of clearance of the 
first stage, a, the clearance of the second stage, C2, to give best-receiver 
pressure can be found, 

, = »». . (229) 

(R p 2s-l)R p 2 S D 2 

For every value of first-stage clearance there is a corresponding clearance of 
second stage that will give best-receiver pressure, found by this equation. Sim- 
ilar reasoning can be applied to three- or four-stage compressors. 

21. Graphic Solution of Compressor Problems. In order to obviate the 
necessity of working out the formulas given in this chapter each time a prob- 
lem is to be solved, several of them have been worked out for one or more 
cases and results arranged to give a series of answers graphically. By the 
use of the charts made up of these curves many problems may be solved 
directly and in many others certain steps may be shortened. A description of 
each chart, its derivation and use is given in subsequent paragraphs. 

Chart, Fig. 44. This chart gives the work required to compress and deliver 
a cubic foot of (sup.pr.) air or the horse-power to compress and deliver 1000 cu.ft. 
of (sup.pr.) air per minute if the ratio of pressure (del.pr.)-^ (sup.pr.), the value 
of s and the (sup.pr.) are known, and compression occurs in one stage. The 
work or H.P. for any number of cubic feet is directly proportional to number 
of feet. The curves are dependent upon the formulas, Eq. (31),' for the case 
when s = l, and Eq. (51) for the case when s is not equal to 1. These formulas 
are: 

Eq. (31), W per cu.ft. = 144 (sup.pr) log e R p ; 
" (51), W per cu.ft. = 144--*— (sup.pr.) \R v ~ r --l\. 

These equations are difficult to solve if an attempt is made to get a relation 
between the work and ratio of pressures. This relation may, however, be 
worked out for a number of values of pressure ratios and results plotted to 
form a curve by which the relation may be had for any other ratio within 
limits. This has been done in this figure in the following manner: 

On a horizontal base various values of R v are laid off, starting with the value 
2 at the origin. The values for work were then found for a number of values 
of R p with a constant value of (sup.pr.) and s. A vertical work scale was 
then laid off from origin of R v and a curve drawn through the points found 
by the intersection of horizontal lines through values of work, with vertical 
lines through corresponding values of R v . The process was then repeated for 
other values of s and curves similar to the first, drawn for the other values 
of s. From the construction so far completed it is possible to find the work per 
cubic foot for any pressure ratio and any value of s for one (sup.pr.) by pro- 



WORK OF COMPRESSORS 



1G9 



u 

< 


a 








Cis 


PU 


Ci 


<-> Oi 


CO 


O 3 


<» 


-55 to 


Cs 




Ki 


8^' 


ss 


h 


fH 




O 


s 


<4-H 


8 ° 


CO 




crt 


c 


o 


CO 


CD 


£S 


U 


iJ 


2 










S t 




"8 o 


Ph 




s^ntosqy *m -Tbg aad 'sqi (• jj -dus) 



170 ENGINEERING THERMODYNAMICS 

jecting up from the proper value of R v to the curve of value of s and then hor- 
izontally to the scale of work. It will be noted from these formulas, however, 
that the work may be laid off on the horizontal base and a group of lines drawn 
so that the slope of the line equals ratio of work for any supply pressure to that 
for the (sup.pr.) originally used. For convenience, in order that the group of 
s curves and the latter group may be as distinct as possible, the origin of the 
latter group is taken at the opposite end of the base line. If from the point 
for work originally found, a projection is made horizontally to the proper 
(sup.pr.) curve, the value for work with this (sup.pr.) will be found directly 
below. It will be noted that from point of intersection of the vertical from the 
R v value with the s curve, it is only necessary to project horizontally far enough 
to intersect the desired (sup.pr.) curve, and since no information of value will 
be found by continuing to the work scale for the original (sup.pr.) this is omitted 
from the diagram. 

In brief, then, the use of this chart consists in projecting upward from the 
proper value of R p to the proper s curve, then passing horizontally to the value of 
(sup.pr.) and finally downward to the work scale. As an example of use of the 
curve Ex. 2 of Section (8) may be solved directly. This is to find the work 
to compress 1000 cu.ft. of free air from 1 to 8J atmospheres adiabatically. 
On the curve project upward from R p = 8.5 to curve of s = 1.406, then over 
to 14.7 (sup.pr.) curve and down to read work = 6,300,000 as found, for exam- 
ple, by use of formulas in Section (8). 

Chart, Fig. 45. This gives the work required to compress and deliver a 
cubic foot of (sup.pr.) air or the horse-power to compress and deliver 1000 cu.ft. 
of (sup.pr.) air per minute if the ratio of pressures, the value of s and (sup.pr.) 
are known and if compression occurs in two stages with best-receiver pressure 
and perfect intercooling. The work or H.P. for any other number of cubic feet 
may be found by multiplying work per foot by the -number of feet. The 
method of arriving at this chart was exactly the same as that for one stage. 

As an example of the use of the chart, Example 2 of Section (9) may be 
solved directly. This problem calls for the work to compress 5 cu.ft. of free 
air from 1 to 8 J atmospheres adiabatically in two stages. Project upward from 
R p = 8.5 to curve s = 1.406, then over to 14.7 curve and down to read 5320 ft.-lbs. 
per cubic foot, which is same as found from the formula in Section (9). 

Chart, Fig. 46. This chart gives the work necessary to compress and deliver 
a cubic foot of (sup.pr.) air or horse-power to compress and deliver 1000 cu.ft. 
of (sup.pr.) air per minute, if the ratio of pressures, the value of s, and the 
(sup.pr.) are known and if the compression occurs in three stages with best- 
receiver pressures and perfect intercooling. The work or horse-power for any 
other number of cubic feet may be found by multiplying the work for one 
foot by the number of feet. 

As an example of use of this chart, Example 2 of Section (13) may be 
solved directly byit. This calls for the horse-power to compress 100 cu.ft. free 
air per minute adiabatically in three stages from 15 lbs. per square inch abs. 
to 90 lbs. per square inch gage. From R p = 7, project to curve of s = 1.4 then 



WORK OF COMPRESSORS 



171 



n=s v 



n=s 




o 



U 

ft 

>> 

° 8 



^ M 



HH aT 



8 £ 
2 I 

* 3 



W 



o 

a 



•sqv'ui -bg J9d -sqi (-aj-dng) 



172 



ENGINEERING THERMODYNAMICS 



ii ii 

wwu) m m cq 




S3 S3 8 8 8 §3 8 S3 8-8 
•sqy "ai *nbs aod aiy (Md 'dns) 



WORK OF COMPRESSORS 173 

over to (sup.pr.) = 15 and down, and the horse-power will be found to be 13.6 
as before by use of formulas. 

Chart, Fig. 47. This chart is for finding the (m.e.p.) of compressors. In the 
case of multi-stage compressors with best-receiver pressure and perfect inter- 
cooling, the (m.e.p.) of each cylinder may be found by considering each cylinder 
as a single-stage compressor, or the (m.e.p.) of the compressor referred to the 
L.P. cylinder may be found. 

The chart depends on the fact that the work per cubic foot of (sup.pr.) gas 

is equal to the (m.e.p.) for the no clearance case and that the (m.e.p.) with 

clearance is equal to the (m.e.p.) for no clearance, times the volumetric 

efficiency. Diagrams 1, 2 and 4 are reproductions of Figs. 45, 46 and 47 to a 

smaller scale and hence need no explanation as to derivation. Their use may 

be briefly given. From the proper ratio of pressures project upward to the 

proper curve, then horizontally to the (sup.pr.) and downward to read work per 

cubic feet of (sup.pr.) gas. 

The volumetric efficiency diagram was drawn in the following manner: 

i 

From Eq. (65) vol. eff. = (1 -\-c — cR p s ), showing that it depends upon three 

variables, R P , c and s. A horizontal scale of values of R p was laid off. Values 
i^ 

of R p s were found and a vertical scale of this quantity laid off from the same 

origin as the R v values. Through the intersection of the verticals from various 

values of R P with the horizontals drawn through the corresponding values of 
i 

(R P ) S for a known value of s, a curve of this value of s was drawn. In a similar 

way curves of other values of s were drawn. From the construction so far 

l 

completed it is possible to find the value of (Rp) s by projecting upward from any 

l 

value of R p to the curve of $ and then horizontally to the scale of (Rp) s . Values 

i 
of volumetric efficiencies found for various clearances and the values of (Rp) s 
are laid off on a horizontal base, with the origin at the opposite end of scale 
from that of R v values, in order that clearance curves and s curves might be 

as distant as possible. These clearance curves were drawn through the inter- 

i 

section of horizontals through the (Rp) s values and of verticals through the 

vulmetric efficiency values corresponding to them for the particular clearance 

in question. 

To find volumetric efficiency then it is merely necessary to project from value 

of Rp to the correct s curve, then across to the proper clearance and finally 

i 

down to volumetric efficiency. As the value of (Rp) s is not desired, the hori- 
zontal projection is carried only to the intersection with the clearance curve 
and not to the edge of the diagram. To find the (m.e.p.) for single stage, the 
work per cubic foot is found from the diagram and then the volumetric efficiency, 
both as described above. The product is (?n.e.p.) 

For multi-stage compressors with perfect intercooling and best-receiver 
pressure, as stated above, the (m.e.p.) of each cylinder may be found, considering 



174 



ENGINEERING THERMODYNAMICS 




91 81 



9 7 8 9 10 11 

Ratio of Pressures 

70 63 5b 49 42 3S 28 

Work per Cu. Ft. of (Sup. Pr.) Gas-H44 



\ 



Fig. 47.— Mean Effective Pressure of Compressors, One-, Two-, and Three-stages. 



WORK OF COMPRESSORS 



175 



<tt HK) 




2 


3 


4 


5 6 7 8 9 10 11 
Ratio of Pressure^ 


12 


13 


14 


15 


yi 


84 


77 


70 63 56 49 42 35 28 
Work per Cu. Ft. of (Sup. Pr. ) Gas-KL44 


21 


14 


7 






Fig. 47. — Mean Effective Pressure of Compressors, One-, Two-, and Three-stages. 



176 ENGINEERING THERMODYNAMICS 

each to be a single-stage compressor and remembering that (1 rec.pr.) becomes 
(sup.pr.) for second stage, and (del.pr.) for first stage and that (2 rec.pr.) 
becomes (sup.pr.) for third stage, (del.pr.) for second stage. The (m.e.p.) reduced 
to low-pressure cylinder is found by taking work per cubic feet of (sup.pr.) 
gas and multiplying by volumetric efficiency of low-pressure cylinder. 

To illustrate the use of this curve the example of Section (16) may be 
solved. Projecting upward from the pressure ratio of 9.35 to the line of s = 1.4 
and then over to (sup.pr.) = 15 in diagram 4, since compression is three stage 
and from 15 lbs. per square inch to 140 lbs. per square inch, work per cubic 
foot or (m.e.p.), is found for no clearance to be 37.8 abs. per square inch. Since 
best-receiver pressure assumed is 31.6, which gives a ratio of 2.1 for the low- 
pressure cylinder. From diagram 3, by projecting upward from R p = 2.1 and 
over to the 5 per cent clearance line volumetric efficiency is 96.5. The product 

gives (m.e.p.)reduced to low-pressure cylinder and is 36.5. From the 



33,000 

formula, horse-power is found to be 358 as before. 

Chart, Fig. 48. As mentioned in Section 18, there is one (sup.pr.), which 

for a given (del.pr.) will give the maximum work of compression. The chart, 

Fig. 48, originated by Mr. T. M. Gunn, gives a graphical means of finding this 

value of (sup.pr.) when the (del.pr.), clearance and value of s are known. It also 

gives on the right-hand of the chart a means for finding the (m.e.p.) for this 

condition. The figure was drawn by means of Eqs. (214) and (218). For the 

value of s = 1 the ratio of (del.pr.) to (sup.pr.) was found for cases of clearances 

from per cent to 15 by means of Eq. (218) by trying values of this ratio which 

/del.pr. \ c /del.pr. \ 

would fulfill the condition of the equation, log^ gup pr J =1_ i+^\ vsup .p r . y /- 

For values of s not 1, Eq. (214) was used, and a set of values of R p found 
for the values of s = 1.4 and 1.2 by trial, the correct value of R p being that which 
satisfied the equation, 



del.pr. \ s s r s— 1 /del.pr. \« .1 

sup.pr./ 1+c L s \sup.pr./ J 

As an example the work for the case where s = 1.2 and c = 10 per cent is given. 
Try R p = 2.6, then, R„T^ = J^h+.l-.lx j|x2.6- 833 l 



= 1.091(1. 1-. 01667X2.218), 
= 1.161 






WORK OF COMPRESSORS 



177 




GJnssQJd ifJOAnsa 03. (/d *8 *ni) jo ox^ts^ mnxnxx'BH 




o 

02 

s 

O 
Pi 

03 



c3 

1 

"5 

a» 

1 

Sh 

<g 

O 

I 



^ao^ TanmixBH joj #y 'samssay jo or^Ba; 



178 ENGINEERING THERMODYNAMICS 

R p = ( 1. 161) 6 =2.45, which shows the value of 2.6 to be incorrect. For a second 
trial take 2.45, and then, 

&,*, = 1.091(1. 1-.01667X2.45- 833 ), 
-1.1627 

= (1.1617)6=2.458, 
which is sufficiently close. Therefore the value of R v for s = 1.2 and clearance = 
10 per cent, is 2.45; that is, maximum work will occur for a given (del.pr.) 

when (sup.pr.) is 245 times the (del.pr.) 

When the values for R v had been obtained a horizontal axis of values of s 
and a vertical one of R v values, were laid off and the points for clearance curves 
laid off to their proper values referred to these axes. Through points as plotted 
the clearance lines were drawn. The right-hand diagram was plotted in a 
similar manner from Eqs. (213) and (217), for s not equal to 1 and equal to 
1 respectively. 
. The latter formula was rearranged in the form 



/m.e.p. \ _ 
\del.pr. / 



_log eJ R p 



the last term being found from curve of Fig. 45. The value of R v for each 

value of the clearance was taken from the left-hand diagram, and substituted 

/ m.e.p. \ 
in the above expression to obtain! , ', J for the case of s = 1. Eq. (213) was 

put in form 



/ m.e.p. \ 
\del.pr. / 



(R^-1)E V , 



{s-l)R 2 

and values of R v for each value of the clearance found in the left-hand diagram 

were substituted, together with E„ values from Fig. 45 and the value of ( , ', ' ) 

found for each case of clearance when s = 1.4. When the points for s = l and 
s = 1.4 had been found, a horizontal axis of values of s and a vertical one of 
values of R v were laid off, and points for the clearance curves plotted as for 
the left-hand diagram and the curves drawn in. 

To find the (sup.pr.) to give maximum work for any (del.pr.) it is only 
necessary to project from the proper value of s to the proper clearance curve, 
and then horizontally to read the value of R v . The (del.pr.) divided by this 
gives the (sup.pr.) desired. To obtain the (m.e.p.) project upward from the 

value of s to the clearance curve, then horizontally to read the ratio ( , \ ' ) 

The (del.pr.) times this quantity gives the m.e.p. 

As an example of the use of this chart let it be required to find the (sup.pr.) 
for the case of maximum work for 9X12 in. double-acting compressor running 
200 R.P.M., having 5 per cent clearance and delivering against 45 lbs. per square 
inch gage. Also the horse-power. Compression such that s = 1.3. 



WORK OF COMPRESSORS 



179 



Projecting from the value 1.3 for s on the left-hand diagram to the line of 

60 
5 per cent clearance find R v to be 2.8, hence (sup.pr.) =2g = 21.4 lbs. per square 

inch absolute = 6.4 lbs. per square inch gage. Again, projecting from value 1.3 

(m.e.p.) 



for s on right-hand diagram to line of 5 per cent clearance find that;-r-r- n 
hence fm.e.p). = 23 and 



.383, 



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Note; Solid Lines = 3 Stage; Broken Lines =2 Stage 

Fig. 49.— Relative Work of Two- and Three-stage Compressors Compared to Single Stage. 

Chart, Fig. 49. This chart is designed to show the saving in work done in 
compressing and delivering gases by two-stage or three-stage compression with 
best-receiver pressure and perfect intercooling over that required for com- 
pressing and delivering the same gas between the same pressures in one stage. 



180 ENGINEERING THERMODYNAMICS 

The chart was made by laying off on a horizontal base a scale of pressure 
ratios. From the same origin a scale of work for two or three stage divided by 
the work of one stage was drawn vertically. For a number of values of R p 
the work to compress a cubic foot of gas was found for one, two and three stage 
for each value of s. The values found by dividing the work of two or three 
stage by the work of single stage were plotted above the proper R p values and 
opposite the proper ratio values and curves drawn through all points for one 
value of s. To find the saving by compressing in two or three stages project 
from the proper R v value to the chosen s curve for the, desired number of stages, 
then horizontally to read the ratio of multi-stage to one-stage work. This value 
gives per cent power needed for one stage that will be required to compress 
the same gas multi-stage. Saving by multi-stage as a percentage of single 
stage is one minus the value read. 

To illustrate the use of this chart, find the per cent of work needed to 
compress a cubic foot of air adiabatically from 1 to 8| atmospheres in two 
stages compared to doing it in one stage. From examples under chart Nos. 
44 and 46 it was found that work was 6300 ft.-lbs. and 5320 ft.-lbs. respec- 
tively, for one- and two-stage compression, or that two stage was 84.5 per cent 
of one stage. From R p , 8 J project up on Fig. 49 to s = 1.406 for two stage and 
over to read 84.6 per cent, which is nearly the same. 

Chart, Fig. 50. This chart, designed by Mr. T. M. Gunn, shows the 
economy compared to isothermal compression. 

The chart was drawn on the basis of the following equation: 

^ ,. ,, i\ m.e.p. isothermal (no clearance) 

Economy (isothermal) = — — - — r-^ - — = — - 

m.e.p. actual +E V actual 

(sup.pr.) logeRp 



S s— 1 

—_-^ (sup.pr.) (R P ~T-1) 

logeRp 



(R p —-1) 



s-1 



Values of this expression were worked out for each exponent, for assumed values 
of R p . A scale of values of R p was laid off horizontally and from the same 
origin a vertical scale of values of the ratio of isothermal to adiabatic. The 
results found were then plotted, each point above its proper R p and opposite its 
ratio value. Curves were then drawn through all the points found for the 
same value of s. In a similar way a set of curves for two stage and a set for 
three stage were drawn. 

This chart is also useful in obtaining the (m.e.p.) of the cycle if the (sup.pr.) 
and the volumetric efficiency of the cylinder be known. A second horizontal 
scale laid off above the R p scale shows the (m.e.p.) per pound of (sup.pr. for) 
the isothermal no-clearance cycle. This is found to be equal to log e R p , since 



WORK OF COMPRESSORS 



181 



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REFERRED TO ISOTHERMAL AS STANDARD. 
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182 ENGINEEEING THERMODYNAMICS 

the (m.e.p.) for no clearance is equal to the work per cubic foot of (sup.pr.) gas, 
which, in turn, for the isothermal case is (sup.pr.) log e R p or log e R p when 
(sup.pr.) = 1. 

Knowing the ratio of pressures, economy compared to isothermal can be 
found as explained above. Also knowing R p the (m.e.p.) per pound initial is 
found from the upper scale. 

Since the latter quantity is assumed to be known, by multiplying it by 
factor just found there is obtained (m.e.p.) isothermal. Since volumetric efficiency 
is assumed known, all the factors are known for the first equation given above 
which, rearranged, reads 

, \ , , m.e.p. isothermal (no clearance) 

(m.e.p.) actual = ■ -r- : — -, -^ — t, , 

(economy isothermal)^- th v 

Chart, Fig. 51. This chart is drawn to give the cylinder displacement for a 
desired capacity, with various values of R p , s and clearance. From the formula 
Eq. (64) 

i_ 

(L. P. Cnp.)=D(l-\-c-cR p s ). 

The right-hand portion of the diagram is for the purpose of finding values 

of (Rp) s for various values of R v and s, and is constructed as was the similar 
curve in Fig. 45. The values of the lower scale on the left-hand diagram give 

X 
values of D = (L. P. Cap.)^-(l+c — cR p s ), where capacity is taken at 100 cu.ft., 

this scale was laid out and the clearance curves points found by solving the 

i_ 
above equation for various values of (R p ) s for each value of c. To obtain the 
displacement necessary for a certain capacity with a given value of R p , c and 
s, project upward from R p to the proper s curve across to the c curve and down to 
read displacement per hundred cubic feet. Also on the left-hand diagram are 
drawn lines of piston speed, and on left-hand edge a scale of cylinder areas 
and diameters to give displacements found on horizontal scale. To obtain 
cylinder areas or approximate diameters in inches project from displacement to 
piston speed line and across to read cylinder area or diameter. Figures given 
are for 100 cu.ft. per minute. For any other volume the displacement and 
area of cylinder will be as desired volume to 100 and diameters will be as 
Vdesired volume to 100. 

As an example, let it be required to find the low-pressure cylinder size for a 
compressor to handle 1500 cu.ft. of free air per minute. Receiver pressure to 
be 45 lbs. per square inch gage and (sup.pr.) to be atmosphere. Piston speed 
limited to 500 ft. per minute. Compression to be so that s = 1.4 and clear- 
ance =4 per cent. Projecting upward from R p = 4: tos = 1.4, across to c = 4%, 
and down to piston speed = 500, find the diameter of a cylinder for 100 cu.ft. 
per minute is 6.3. For 1500 cu.ft. diameter will be as Vl5X6,3 = 3.9X6.3 = 24 
ins. 






WORK OF COMPRESSORS 



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184 ENGINEERING THERMODYNAMICS 



GENERAL PROBLEMS ON CHAPTER II. 

Prob. 1. One hundred cubic feet of H 2 S are compressed from 15 lbs. per square 
'nch absolute to 160 lbs. per square inch absolute. 

(a) Find work done if compression occurs isothermally in a no-clearance one-stage 
compressor; 

(6) Adiabatically in a two-stage, no-clearance compressor; 

(c) Adiabatically in two-stage compressor each cylinder having 5 per cent clearance ; 

Prob. 2. Air is being compressed in three plants. One is single-stage, the second is 
two-stage, and the third is three-stage Considering the compressors to have no clear- 
ance and 1000 cu.ft. of free air compressed adiabatically per minute from atmosphere 
to 150 lbs. per square inch gage, what will be the horse-power required and cylinder 
sizes in each case? 

Prob. 3. A two-stage compressor with 5 per cent clearance n the h'gh and 3 per 
cent in the low-pressure cylinder is compressing air from 14 lbs. per square inch ab- 
solute to 125 lbs. per square inch gage. What is the best-receiver pressure and what 
must be the size of the cylinders to handle 500 cu.ft. of free air per minute? 

Prob. 4. A manufacturer gives a capacity of 970 cu.ft. for a 22ix24 in. single- 
stage compressor running at 142 R.P.M. when working pressures are 50 to 100 lbs. per 
square inch gage. What would be the clearance for each of these pressures assuming 
s = 1.4? 

Prob. 5. The card taken from a single-stage compressor cylinder showed an appar- 
ent volmetric efficiency of a 95 per cent and a mean value for s of 1.3. What is the 
clearance and what would be the (m.e.p.) for the ratio of pressures of 6? 

Prob. 6. A compressor with double-acting cylinder 12x14 ins., having 6 per cent 
clearance, is forcing air into a tank. Taking the volumetric efficiency as the mean of 
that at the start and the end, how long w 11 it take to build up 100 lbs. per square inch 
gage pressure in a tank of 1000 cu.ft. capacity if the speed is 100 M.R.P. and compres 
sion is isothermal, and how long wi 1 it take to do it if the compression is adiabatic and 
the air in the tank does not cool during filling? What is the maximum attainable pressure? 

Prob. 7. It is desired to supply 1000 cu.ft. of air per minute at a pressure of 80 lbs. 
per square inch gage. A two-stage compressor is to be used, the clearance in low pres- 
sure of which is 3 per cent. What must be the displacement of the low-pressure cylin- 
der and what will be the horse-power of the compressor? 

Prob. 8. The low-pressure cylinder of a compressor is 18x24 ins. and has a clear- 
ance of 4 per cent. The receiver pressure is 60 lbs. per square inch absolute. The high- 
pressure cylinder has a 5 per cent clearance. What must be its diameter, stroke being 
same as low, so that compressor will operate at its designed receiver pressure? 

Prob. 9. The discharge pressure of a two-stage compressor is 120 lbs. per square 
inch absolute and the supply pressure is 15 lbs. per square inch absolute The compressor 
is 10x16x12 ins. The clearance in the low-pressure cylinder is 3 per cent. What 
must be the clearance in the high-pressure cylinder for the machine to operate at best- 
receiver pressure? Compression is adiabatic. 

Prob. 10. If the clearance in the high-pressure cylinder of Prob. 9 were reduced 
to 3 per cent, would the receiver pressure increase or decrease, how much and why? 

Prob. 11. If the discharge pressure in Prob. 9 fell to 100 lbs. per square inch absolute, 
what would be the new best-receiver pressure and why? Would the original clearance 
allow the new best-receiver pressure to be maintained? 



PROBLEMS ON CHAPTER II 185 

Prob. 12. The discharge pressure for which a 20^x32^x24 in. compressor is 
designed, is 100 lbs. per square inch gage, supply pressure being 14 lbs. per square inch 
absolute. The d'scharge pressure is raised to 125 lbs. per square inch gage. The 
clearance on the high-pressure cylinder can be adjusted. To what value must it be 
changed to enable the compressor to carry the best-receiver pressure for the new 
discharge pressure? Low-pressure clearance is 5 per cent at all times and com- 
pression being adiabatic. 

Prob. 13. A manufacturer builds his 15J X25^X18 'in. compressors with low-pres- 
sure cylinders of larger diameter for high altitude work. What would be the diameter 
of a special cylinder for this compressor to work at an altitude of 10,000 ft. and what 
would be the horse-power per cubic foot of low-pressure air in each case? 

Prob. 14. A three-stage compressor has 4 per cent clearance in all the cylinders. The 
low-pressure cylinder is 34x36 ins., delivery pressure 200 lbs. per square inch gage, 
supply pressure 14 lbs. per square inch absolute. What must be the size of the other 
cylinders for the machine to operate at best-receiver pressure. 

Prob. 15. The cylinders of a two-stage compressor are given as 10| and 16{ ins., 
the stroke being 12 ins. The machine has a capacity of 440 cu.ft. per minute at 160 
R.P.M., the supply pressure is 14 lbs. per square inch absolute and the delivery 
pressure 100 lbs. per square inch gage. What is the clearance of each cylinder? 

Prob. 16. To perform the required useful refrigerating effect in an ice plant, it is 
necessary to compress 250 cu.ft. of ammonia vapor per minute from 30 lbs. per square 
inch gage to 150 lbs. per square inch gage. What must be the size of the compressor 
to handle this at 75 R.P.M. if it be double acting and the clearance be 5 per cent? 

Prob. 17. The maximum delivery pressure of a,type of compressor is controlled by 
making the clearance large so that the volumetric efficiency will decrease as the pressure 
rises and become zero at the desired pressure. What must be the clearance for a single- 
stage compressor where the supply pressure is 14 lbs. per square inch absolute and the 
maximum delivery pressure 140 lbs. per square inch absolute? What will be the volu- 
metric efficiency of the same machine at a delivery pressure of \ the maximum? At j? 

Prob. 18. A three-stage compressor has a clearance of 5 per cent in each cylinder. 
What must be the cylinder ratios for the best-receiver pressures when the machine is 
compressing to 170 lbs. per square inch gage from atmosphere? 

Prob. 19. Show why it was very essential to keep the clearance low in cylinders of 
three-stage compressor used for compressing air for air-driven cars, where the delivery 
pressure carried was 2500 lbs. per square inch, by assuming numerical data and 
calculating numerical proof. 

Prob. 20. With water falling 150 ft. and used to compress air directly, how many 
cubic feet of air could be compressed per cubic foot of water? 

Prob. 21. Air is compressed from atmosphere to 150 lbs. per square inch absolute, 
isothermally in one stage. How much more work would be required per cubic foot if 
compression were adiabatic? How much of this excess would be saved by compressing 
two stage? Three stage? 

Prob. 22. 150 I.H.P. is delivered to theair cylinders of a 14i X22l xl8 in. compres- 
sor, running at 120 R.P.M. The supply pressure is 15 lbs. per square inch absolute. 
The volumetric efficiency as found from the indicator card is 95 per cent. What was 
the discharge pressure? 

Prob. 23. The clearance in the high-pressure cylinder of a compressor is 5 per cent, 
which allows the compressor to run with the best-receiver pressure for a discharge of 
100 lbs. per square inch absolute when the compressor is at sea-level. What would the 



186 ENGINEERING THERMODYNAMICS 

clearance be if the discharge pressure were kept the same and the altitude were 10,000 
ft. to keep the best-receiver pressure? 

Prob. 24. How many cubic feet of supply-pressure air may be compressed per minute 
from 1 to 8 atmospheres absolute by 100 horse-power if the compression in all cases is 
adiabatic? 

(6) Three stage, no clearance; 

(c) Two stage with 5 per cent clearance; 

(d) Single stage with 5 per cent clearance; 
(a) Two stage, no clearance 

Prob. 25. The capacity of a 14ix22ixl4 in. compressor when running at 140 
R.P.M. is said to be 940 cu.ft. for working pressures of 80 to 100 lbs. per square inch 
gage and atmospheric supply at sea level. Check these figures. 

Prob. 26. What horse-power would be required by an 181x301x24 in. compressor 
operating at 100 P. P.M. and on a working pressure of 100 lbs. per square inch gage if 
the clearance in low-pressure cylinder is 4 per cent? What would be the capacity? 

Prob. 27. By means of water jackets on a compressor cylinder the vaue for s of com- 
pression curve in single-stage machine is lowered to 1.3. Compare the work to com- 
press 1000 cu.ft. of air from 1 to 8 atmospheres absolute with this condition with that 
required for isothermal and adiabatic compression. 

Prob. 28. What must be the size of cylinders in a three-stage compressor for com- 
pressing gas from 50 lbs. per square inch absolute to 600 lbs. per square inch absolute 
when s equals 1.3 and each cylinder has 3 per cent clearance. Compressor to run at 
100 R.P.M. to be double acting and handle 1000 cu.ft. of gas per hour? 

Prob. 29. How many cu.ft. of free air could be compressed per minute with an avail- 
able horse-power of 1000 H.P. from atmosphere to 150 lbs. per square inch gage; (a) 
if compression is isothermal; (b) if compression be single-stage adiabatic; (c) if com- 
pression be three-stage adiabatic? 

Prob. 30. A single-stage compressor is compressing air adiabatically at an altitude 
of 6000 ft. to a pressure of 80 lbs. per square inch gage. The cylinder has 2 per cent 
clearance. What must be the s'ze of the cylinder to compress 2000 cu.ft. of free air per 
minute if the piston speed is limited to 600 ft. per minute? What, if the clearance be 
zero? 

Prob. 31. What would be the size of the two-stage compressor for same data as in 
Prob. 30? 



CHAPTER III 

WORK OF PISTON ENGINES. HORSE POWER AND CONSUMPTION OF 
PISTON ENGINES USING STEAM, COMPRESSED AIR, OR ANY OTHER 
GAS OR VAPOR UNDER PRESSURE. 

1. Action of Fluid in Single Cylinders. General Description of Structure 
and Processes. The most commonly used class of engines is that in which 
the operation is dependent on the pushing action of high-pressure fluids on 
pistons in cylinders, and this includes all piston steam engines of the recip- 
rocating or straight-line piston path group as well as the less common 
rotary group, having pistons moving in curved and generally circular paths. 
In these same engines there may be used compressed air as well as steam, 
and equally as well the vapors of other substances or any other gases, without 
change of structure, except perhaps as to proportions, providing only that the 
substance to be used be drawn from a source of supply under high pressure, 
be admitted to the cylinder, there used and from it discharged or exhausted 
to a place of lower pressure. This place of lower pressure may be the open 
air or a closed chamber; the used fluid may be thrown away and wasted or 
used again for various purposes without in any way affecting the essential 
process of obtaining work at the expense of high-pressure gases or vapors. 
It is evident that, regarding a piston as a movable wall of a cylinder, when- 
ever a fluid acts on one side with greater pressure than on the other, the 
piston will move toward the lower pressure end of the cylinder, and in so 
moving can exert a definite force or overcome a definite resistance, measured 
by the difference in pressure on the two sides and the areas exposed to the 
pressure. It is not so evident, but just as true, that the piston may be made 
to move from one end of the cylinder to the other when the average pressure 
on one side is greater than the average pressure on the other, and also do work 
even if the excess of pressure should reverse in direction during the stroke 
and provided only some energy storage device is added. In the common 
steam or compressed-air engine this is a flywheel with the usual connecting 
rod and crank mechanism, uniting the reciprocating piston movement with 
the continuous rotary movement of the flywheel mass. In certain forms 
of pumps the energy is stored in extra cylinders at times of excess and given 
out at times of deficiency in the path of the piston, so that its motion from 
end to end of cylinder may not be interrupted even if the pressure on the 
driving side should fall below that on the resisting side, assuming, of course, 
the average pressure for the whole stroke to be greater on the driving side 
than on the resisting side. 

187 



188 ENGINEERING THERMODYNAMICS 






It appears, therefore, that piston movement in engines of the common 
form and structure, and the doing of work by that movement is not a 
question of maintaining a continuously greater pressure on one side than 
on the other. On the contrary, the process is to be studied by examination 
of the average pressure on the driving side and that on the resisting side, 
or by comparing the whole work done on one side with the whole work 
done on the other side by the fluid. The work done by the fluid on one 
side of a piston may be positive or negative, positive when the pressures are 
assisting motion, negative when they are resisting it. It is most con- 
venient to study the action of fluids in cylinders by considering the 
whole action on one side from the beginning of movement at one end to 
the end of movement at the same point, after the completion of one complete 
forward and one complete return stroke. All the work done by the pressure 
of the fluid on the forward stroke on the side of the piston that is apparently 
moving away from the fluid is positive work, all the work done by the pres- 
sure of the fluid on the same side of the piston during the return stroke is 
negative, and for this stroke the side of the piston under consideration is 
apparently moving toward the fluid or pushing it. 

For the complete cycle of piston movement covering the two strokes 
the work done on one side is the algebraic sum of the forward stroke 
work, considered positive, and the back stroke work, considered negative 
During the same time some pressures are acting on the other side of the 
piston, and for them also there will be a net work done equal to the cor 
responding algebraic sum. The work available for use during the complete 
two strokes, or one revolution, will be the sum of the net work done by the 
fluid on the two sides of the piston during that time, or the algebraic sum of 
two positive and two negative quantities of work. Methods of analysis of the 
work of compressed fluids in cylinders are consequently based on the action in 
one end of a cylinder, treated as if the other end did not exist. 

Just how the high-pressure fluid from a source of supply such as a boiler 
or an air compressor is introduced into one end of a cylinder, how it is treated 
after it gets there, and how expelled, will determine the nature of the varia- 
tion in pressure in that end that acts on that side of the piston, and these 
are subjects to be studied. To determine the work done in the cylinder 
end by the fluid, it is necessary to determine laws of pressure change with 
stroke, and these are fixed first by valve action controlling the distribution 
of the fluid with respect to the piston and second by the physical properties 
of the fluid in question. 

It is necessary that the cylinder be fitted with a valve for getting fluid 
into a C3dinder, isolating the charge from the source of supply and getting it 
out again, and it may be that one valve will do, or that two or even more 
are desirable but this is a structural matter, knowledge of which is assumed 
here and not concerned with the effects under investigation. The first step 
in the process is, of course, admission of fluid from the source of supply to the 
cylinder at one end, which may continue for the whole, or be limited to a part 



WORK OF PISTON ENGINES 189 

of the stroke. When admission ceases or supply is cut off before the end 
of the stroke there will be in the cylinder an isolated mass of fluid which 
will, of course, expand as the piston proceeds to the end. Thus the forward 
stroke, considering one side of the piston only, always consists of full pressure 
admission followed by expansion, the amount of which may vary from zero 
to a very large amount; in fact the final volume of the fluid after expansion 
may be hundreds of times as great as at its beginning, when supply was cut 
off. 

At the end of this forward stroke an exhaust valve is opened, which 
permits communication of the cylinder with the atmosphere in non-con- 
densing steam and most compressed-air engines, or with another cylinder, 
or with a storage chamber, or with a condenser in the case of a steam engine 
in which the pressure approximates a perfect vacuum. If at the moment of 
exhaust opening the cylinder pressure is greater or less than the back pressure, 
there will be a more or less quick equalization either up or down before the 
piston begins to return, after which the return or exhaust stroke will proceed 
with some back-pressure resistance acting on the piston, which is generally 
though not always constant. This may last for the whole back stroke or 
for only a part, as determined by the closure of the exhaust valve. When 
the exhaust valve closes before the end of the return stroke the unexpelled 
steam will be trapped and compressed to a pressure depending partly on the 
point of the stroke when closure begins and the pressure at the time, and 
partly on the clearance volume of the cylinder into which the trapped steam 
is compressed. Of course, at any time near the end of the stroke the admission 
valve may be opened again, and this may occur, 1st, before compression is 
complete, which will result in a sudden pressure rise in the cylinder before 
the end of the stroke to equalize it with the source of supply; 2d, just at the 
end of the stroke, which may result in a rise or a fall to equalize, or no 
change at all, depending on whether compression has raised the cylinder 
pressure not quite to supply pressure, or to something greater than it, or to 
a value just equal to it; 3d, after the end of the stroke, which will result in 
a reexpansion of the steam previously under compression, and then a sudden 
rise. It may be said in general that in cylinders there are carried out with 
more or less variation the following processes: 

Forward stroke, constant-pressure admission followed by expansion. 

Back stroke, constant-pressure exhaust followed by compression, while at 
both ends of the stroke there may or may not be a vertical line on a pressure- 
volume diagram representing a const ant- volume change of pressure. 

These processes will result in a cycle of pressure- volume changes which will 
be a closed curve made of more or less accurately defined phases, and the work 
of the cycle will be the area enclosed by the cyclic curve. Of course, there are 
causes of disturbance which make the phases take on peculiar characteristics. 
For example, the valve openings and closures may not take place as desired 
or as presupposed with respect to piston positions; leakage may occur, steam 
may condense during the operations in the cylinder, and T%ater of condensation 



190 



ENGINEERING THERMODYNAMICS 



may evaporate; the resistance through valves will always make the cylinder 
pressure during admission less than in the supply chamber and greater during 
exhaust than the atmosphere or than in exhaust receiver or condenser and may 
through the valve movements make what might have been a constant-pres- 
sure straight line become a curve. There will, by reason of these influences, 
encountered in real engines, be an almost infinite variety of indicator cards 
or pressure-volume cycles for such engines. 



































Ai 






B 
















B 




A' 
























A-B STEAM ADMITTED 
B-C " EXPANDED 
C-D '■ EXHAUSTED 
D-A » COMPRESSED 




























































ACTUAL CARD FROM 
CORLISS ENGINE 




















L 


c 










^ 


A 




A 


























D 






























C 














































































D 


















C 










Fig. 52. — Diagram to Indicate Position of Admission, Cut-off, Release, Compression on 

Engine Indicator Card. 



The various points of the stroke at which important events occur, 
important in their pressure-volume significance, have names, as do also the 
lines between the points, and these names are more or less commonly accepted 
and generally understood as follows: letters referring to the diagram Fig. 52. 

Point Names: Events of Cycle. 

A. Admission is that point of the stroke where the supply valve is 

opened. 

B. Cut-off is that point of the stroke where the supply valve is closed. 

C. Release is that point of the stroke where the exhaust valve is opened. 

D. Compression is that point of the stroke where the exhaust valve 

is closed. 
Names of Lines, or Periods: 

4-J5. Admission or steam line joins the points of admission and cut-off. 

B-C. Expansion line joins the points of cut-off and release. 

C-D. Exhaust line joins the points of release and compression if there 

is any, or admission if there is not. 
D-A. Compression line joins the points of compression and admission. 



WORK OF PISTON ENGINES 



191 



By reason of the interferences discussed, these points on actual indicator 
cards may be difficult to locate, one line merging into the next in so gradual 
a manner as to make it impossible to tell where the characteristic point lies, 
as will be apparent from Fig. 53, in which is reproduced a number of actual 
indicator cards. In such cases equivalent points must be located for study 









Fig. 53. — Actual Steam Engine Indicator Cards Showing Distortion of Lines and Uncertain 

Location of Characteristic Points. 



These same terms, which it appears sometimes refer to points and some- 
times to lines, are also used in other senses, for example, cut-off is com- 
monly used to mean the fraction of stroke completed up to the point of 
cut-off, and compression that fraction of stroke remaining incomplete at the 
point of compression, while compression is also sometimes used to express 
the pressure attained at the end of the compression line. In general, there 



192 



ENGINEERING THERMODYNAMICS 



is nothing in the use of these words to indicate just which of the various 
meanings is intended except the text, and experience will soon eliminate most 
of the possible chances of confusion. 

Prob. 1. Draw a card in which admission and exhaust are late. Draw a card in 
which there is no compression, in which compression is very early, so that compression 
pressure is equal to admission pressure. Draw a card with per cent cut-off, and cut- 
off = 100 per cent. Draw cards with same cut-off but with varying initial pressures. 

Prob. 2. The following diagrams, Fig. 54, are reproductions of indicator cards 
actually taken from engines. Explain what features are peculiar to each and if 
possible give an explanation of the cause. 







No. 8 



Fig. 54. — Indicator Diagram from Steam Engines with Improperly Set Valve Gear. 

For example, in No. 1 a line of pressure equalization between the end of the com- 
pression line and the beginning of the admission line inclines to the right instead of 
being perpendicular, as in a perfect diagram. This is due to the fact that admis- 
sion does not occur until after the piston has begun to move outward, so that pres- 
sure rise does not occur at constant volume, but during a period of increasing volume. 

2. Standard Reference Cycles or PV Diagrams for the Work of Expansive 
Fluids in a Single Cylinder. Simple Engines. To permit of the derivation 
of a formula for the work of steam, compressed air, or any other fluid in a 



WORK OF PISTON ENGINES 193 

cylinder, the various pressure volume, changes must be denned algebraically. 
The first step is, therefore, the determination of the cycle or pressure-volume 
diagram representative of the whole series of processes and consisting of a 
number of well-known phases or single processes. These phases, ignoring all 
sorts of interferences due to leakage or improper valve action, will consist of 
constant-pressure and constant-volume lines representing fluid movement 
into or from the cylinder, combined with expansion and compression lines 
representing changes of condition of the fluid isolated in the cylinder. These 
expansion and compression lines represent strictly thermal phases, laws for 
which will be assumed here, but derived rigidly later in the part treating of 
the thermal analysis; however, all cases can be represented by the general 
expression 

PV S = C, 

in which the character of the case is fixed by fixing the value of s. For all 
gases and for vapors that do not contain liquid or do not form or evaporate 
any during expansion or compression, i.e., continually superheated, the exponent 
s may have one of two characteristic values. The first is isothermal expansion 
and compression, and for this process s is the same for all substances and 
equal to unity. The second is for exponential expansion or compression and 
for this process s will have values peculiar to the gas or superheated vapor 
under discussion, but it is possible that more than one substance may have the 
same value, as may be noted by reference to Section 8, Chapter I, from 
which the value s = 1.406 for air and s = 1.3 for superheated steam or ammonia 
adiabatically expanding are selected for illustration. 

When steam or any other vapor not so highly superheated as to remain 
free from moisture during treatment is expanded or compressed in cylinders 
different values of s must be used to truly represent the process and, of course, 
there can be no isothermal value, since there can be no change of pressure of 
wet vapors without a change of temperature.. For steam expanding adiabatically 
the value of s is not a constant, as will be proved later by thermal analysis, so 
that the exact solution of problems of adiabatic expansion of steam under 
ordinary conditions becomes impossible by pressure-volume analysis and can 
be handled only by thermal analysis. However, it is sometimes convenient 
or desirable to find a solution that is approximately correct, and for this a 
sort of average value for s may be taken. Rankine's average value is s = 1 . 1 1 1 = y>- 
for adiabatic expansion of ordinarily wet steam, and while other values have 
been suggested from time to time this is as close as any and more handy than 
most. The value s = 1.035+.14X(the original dryness fraction), is given by 
Perry to take account of the variation in original moisture. 

Steam during expansion adiabatically, tends to make itself wet, the 
condensation being due to the lesser heat content by reason of the work done; 
but if during expansion heat be added to steam originally just dry, to keep it so 
continuously, as the expansion proceeds, it may be said to follow the saturation 
law of steam, for which s = 1.0646. This is a strictly experimental value found 



194 



ENGINEERING THERMODYNAMICS 



A 



s=l 



Exp. 



,*-S=l 



■Exp 













\ 


-s=-l 


-Exp. 








D 


V 










V 


*- 



























P- 










p- 










p- 




























-s=l 


Exp. 


-,-i 


\—s- 


=1 

-Exp 








VJ— s=: 


> 






* 


-CI 


E 








-* * 


-ci 


F 








G \ 








->* 


. V— Exp. 








































H^ 




' — ■ 












































V 












V 










V 










V 













-,( 


■Cl 










r-s= 


i 

L 








V-* — Cpmp. 








r 1 







£ 


J 




















\ 


-s=l 


M 










V 


'Comp. 
























V 




p 












II s " 1 








c 4 

3 = 1 


1 *\ 


-Exp. 






NjjXv 

















Exp. 











r 








r 










\ 


Q 








Comp. 




















Pp 


di 




















\\- 


=1 
Exp. 






--1- 




S 




S= 




3omp. 








c 









Comp 



Exp. 



Fig. 55. — Standard Reference Cycles or Pressure-volume Diagrams for Expansive Fluid 

in Simple Engines. 



WORK OF PISTON ENGINES 



195 



by studying the volume occupied by a pound of just dry steam at various 
pressures, quite independent of engines. 

Direct observation of steam engine indicator cards has revealed the fact 
that while, in general, the pressure falls faster at the beginning of expansion 
and slower at the end than would be the case if 8 = 1, yet the total work is 
about the same as if s had this value all along the curve. This law of expansion 
and compression, which may be conveniently designated as the logarithmic law, 
is almost universally accepted as representing about what happens in actual 
steam engine cylinders. Later, the thermal analysis will show a variation of wet- 
ness corresponding to s=l, which is based on no thermal hypothesis what- 
ever, but is the result of years of experience with exact cards. Curiously 







a 




















a 
















p 








V 


1 

-Exp.PV=C 






p 










_Exp. 


PVS= 


c 








e 






\ 


V 


b 




























CYCLE 1 \* 

SIMPLE ENGINE LOGARITHMIC- 
EXPANSION. 
ZERO CLEARANCE 










e CYCLE 2 ^\ 

SIMPLE ENGINE EXPONENTIAL 
EXPANSION. 
ZERO CLEARANCE 














c 








c 


















































d 
















d 












a 














V 






a 














V 


p 












-Exp. 


PV= 








p" 




f 






v — 


Exp. 


?VS= 


3 










/ 


Comi 


).PV= 


c 




6 














-Com 


J.PV8 


= c 














\ CYCLE 3 ^^ 










~\ 


CYCLE 4 X 


Isb 










\expansion and compression. 
^k with clearance 




c 




X 


.EXPANSION AND COMPRESS 
\ WITH CLEARANCE 


ION. 




c 












1 










S 






















| d 














d 







Fig. 56. — Simple Engine Reference, Cycles or PV Diagrams. 

enough, this value of s is the same as results from the thermal analysis of con- 
stant temperature or isothermal expansion for gases, but it fails entirely to 
represent the case of isothermal expansion for steam. That s = l for isothermal 
gas expansion and actual steam cylinder expansion is a mere coincidence, a 
fact not understood by the authors of many books often considered standard, 
as in them it is spoken of as the isothermal curve for steam, which it most 
positively is not. This discussion of the expansion or compression laws indicates 
that analysis falls into two classes, first, that for which 8=1, which yields a 
logarithmic expression for work, and second, that for which s is greater or 
less than one, which yields an exponential expression for work, and the former 
will be designated as the logarithmic and the latter as the exponential laws, for 
convenience. 



196 



ENGINEERING THERMODYNAMICS 






The phases to be considered then may be summed up as far as this analysis 
is concerned as: 

1. Admission or exhaust, pressure constant, P = const. 

2. Admission or exhaust, volume constant, V = const. 

3. Expansion, PV = const., when s = l. 

4. Expansion, PV S = const., when s is greater or less than 1. 

5. Compression, PV = const., when s = l. 

6. Compression, PV S = const., when s is greater or less than 1. 

Considering all the possible variations of phases, there may result any or 
the cycles represented by Fig. 55. These cycles have the characteristics indicated 
by the following table, noting the variation in the law of the expansion or 
compression that may also be possible. 



Cycle. 


Clearance. 


1 
Expansion. 


Compression. 


A 
B 
C 
D 


Zero 
Zero 
Zero 
Zero 


Zero 

Little 

Complete 

Over-expansion 


Zero 
Zero 
Zero 
Zero 


E 
F 
G 
H 


Little 
Little 
Little 
Little 


Zero 

Little 

Complete 

Over-expansion 


Zero 
Zero 
Zero 
Zero 


I 
J 
K 
L 


Little 
Little 
Little 
Little 


Zero 

Little 

Complete 

Over-expansion 


Little 
Little 
Little 
Little 


M 

N 

P 


Little 
Little 
Little 
Little 


Zero 

Little 

Complete 

Over-expansion 


Complete 
Complete 
Complete 
Complete 


Q 
R 

S 
T 


Little 
Little 
Little 
Little 


Zero 

Little 

Complete 

Over-expansion 


Too much 
Too much 
Too much 
Too much 



It is not necessary, however, to derive algebraic expressions for all these 
cases, since a few general expressions may be found involving all the variables 
in which some of them may be given a zero value and the resulting expression 
will apply to those cycles in which that variable does not appear. The result- 
ing cycles, Fig. 56, that is is convenient to treat are as follows: 

SIMPLE ENGINE REFERENCE CYCLE OR PF DIAGRAMS 

Cycle 1. Simple Engine, Logarithmic Expansion without Clearance. 
Phase (a) Constant pressure admission. 

" (b) Expansion PV = const, (may be absent). 

" (c) Constant volume equalization of pressure with exhaust (may be absent). 

" (d) Constant pressure exhaust. 

" (e) Constant (zero) volume admission equalization of pressure with supply. 






WORK OF PISTON ENGINES 



197 



Cycle II. Simple Engine, Exponential Expansion without Clearance. 
Phase (a) Constant pressure admission. 

f< (6) Expansion P V s = const, (may be absent). 

11 (c) Constant- volume equalization of pressure with exhasut (may be absent). 

11 (d) Constant-pressure exhaust. 

" (e) Constant (zero) volume admission equalization of pressure with supply. 

Cycle III. Simple Engine, Logarithmic Expansion and Compression with Clearance. 
Phase (a) Constant pressure admission. 

" (b) Expansion PV = const, (may be absent). 

11 (c) Constant volume equalization of pressure with exhaust (may be absent). 
" (d) Constant pressure exhaust. 
" (e) Compression PV = const, (may be absent). 

" (/) Constant volume admission, equalization of pressure with supply (may 
be absent). 

Cycle IV. Simple Engine, Exponential Expansion and Compression with Clearance. 
Phase (a) Constant pressure admission. 

" (b) Expansion PV S = const, (may be absent). 

" (c) Constant volume equalization of pressure with exhaust (may be absent). 
" (d) Constant pressure exhaust. 
" (e) Compression PV S = const, (may be absent). 

" (/) Constant admission, equalization of pressure with supply (may be 
absent). 



p 


^ 




- 7-D 








B 




















A 


















(in. pr ) 






























































































































































^.c 


(rel.pr.) 








































































E 




























n 


(bk.pr.) 


















































n 














H 






G 












F 














' 





Fig. 57. — Work of Expansive Fluid in Single Cylinder with No Clearance. 
Expansion for Cycle I. Exponential for Cycle II. 



V 

Logarithmic 



3. Work of Expansive Fluid in Single Cylinder without Clearance. Loga- 
rithmic Expansion, Cycle I. Mean Effective Pressure, Horse-power and 
Consumption of Simple Engine. Referring to the diagram, Fig. 57, the net 
work, whether expansion.be incomplete, perfect, or excessive, is the sum of 






198 ENGINEEEING THERMODYNAMICS 

admission and expansion work less the back-pressure work, or by areas, 
net work area, ABODE = admission work area ABFG 

or algebraically, 



+ expansion work area FBCH 



— back-pressure work area GEDH 



W = P b V b +P b V b \og e j±-P d V d , 



(230) 



Work of cycle in foot-pounds is 



'I l+log e ^j-P d V d (a) 



P c V c (l+log e y)-P d V d (6) 

n+iog e ^}-p d v d (c) 

i l+\og e ^j-P d V d (d) 



(231) 






As V d or V c represent the whole displacement, the mean effective pressure 
will be obtained by dividing Eqs. (231), by V d or V c , giving, 



M.E.F. = P c (l+\og e ^-P d (a) 



= P e (l+\og,^)-F a (6) 

m.e.p. = pil+log e ^) -p d (c) 

= p e ( l+lo&y^ Z )~Va (d) 
V e \ (V b 



= Pb[ l+\0ge 



VJ\V, 



Pa (e) 



(232) 



Similarly, dividing the Eq. (232) by the volume of fluid admitted, V b , will 
give the work per cubic foot, which is a good measure of economy, greatest 
economy being defined by maximum work per cubic foot, which, it may be noted, is 
the inverse of the compressor standard. 



Work per cu i t . supplied = P b i 1 + log e ~ j — P d ~ (a 



l+log e g-P d g (6) 



(233) 



WORK OF PISTON ENGINES 199 

According to. Eq. (19), Chapter I, the piston displacement in cubic feet 

per hour per I.H.P. is for z = l,- ( — l , and this multiplied by the fraction of 

whole displacement occupied in charging the cylinder or representing admission, 
which is 

Va 0F V c > 

will give the cubic feet oj high pressure fluid supplied per hour per I.H.P. f 
hence 



Cu.ft.supplied per hr. per I.H.P. = i^vZ? ( fl ) 

(m.e.p.) V c 

_ 13,750 R 
"(m.e.p.) X P 6 W 



(234) 



Introducing a density factor, this can be transformed to weight of fluid. If 
then ^i is the density of the fluid as supplied in pounds per cubic foot, 



Lbs. fluid supplied per hr. per I.H.P. = 7 — l -X^XBi (a) 

(m.e.p.) V c 

13/750 £ ^ 

(m.e.p.) P b 



. . (335) 



All these expressions, Eqs. (230) to (235), for the work of the cycle, the 
mean effective pressure, work per cubic feet of fluid supplied, cubic feet and 
pounds of fluid supplied per hour per I.H.P., are in terms of diagram point 
conditions and must be transformed so as to read in terms of more generally 
defined quantities for convenience in solving problems. The first step is to 
introduce quantities representing supply and back pressures and the amount 
of expansion, accordingly : 

Let (in.pr.) represent the initial or supply pressure p b expressed in pounds 

per square inch; 

(rel.pr.) represent the release pressure p c , in pounds per square inch; 

(bk.pr.) represent the back pressure p d , in pounds per square inch; 

" R v represent the ratio of expansion defined as the ratio of largest to 

smallest volume on the expansion line ( ^ ) , or I ~ J which is, of 

course, equal to the ratio of supply to release pressure ( — ) , when 

the logarithmic law is assumed; 
D represent the displacement in cubic feet which is V d or V c when no 
clearance is assumed: 



200 



ENGINEERING THERMODYNAMICS 






Let Z represent the fraction of stroke or displacement .completed up to 
cut-off so that ZD represents the volume Vb admitted to the cylinder. 

1 



In this case when clearance is zero, Z 



Rv 



Work of the cycle in foot-pounds 



W 



= 144 (in.pr. 



)I±^-(bk.p,)]z) (a)) 
144[(rel.pr.)(l+log e R v ) - (bk.pr. )]Z) (b) J 



(236) 



m.e.p. = (rel.pr.) (1 +k>g e R v ) — (bk.pr.) (a) 
'l-\-\og e R v 



= (in.pr.) 



R^ 



-(bk.pr.) (6) 



(in.pr.)z(l+log e j) - (bk.pr.) (c) 



Work per cu.ft. supplied = 144[(in.pr.) (l+log e R v ) — (bk.pr.)iiV] (a) 

(bk.pr.) 



Cu.ft. supplied per hr. per I.H.P. = 



= 144 [ (in.pr. ){l-\-\og e ~ 
13,750 1 



(&) 



m.e.p. R^ 



13,750 „ 

— - r Zj 

(m.e.p.) 



(a) 



(237) 

(238) 
(239) 



Lbs. supplied per hr. per I.H.P. 



13,750 S 



1 (a) 



(m.e.p.) R v 
(m.e.p.) 



(240) 



The indicated horse-power may be found by multiplying the work of the 
cycle, Eq. (236), by the number of cycles performed per minute n and divid- 
ing the product by 33,000. 

Dn |~/. x/1+loge.Rv- 



or 



I.H.P. = 



I.H.P. = 



229.2 

Dn(m.e.p.) 
229.2 



I (in.pr.) I - 



Rv 



(bk.pr.) 



]■ 



(241) 



(242) 



In any of these expressions where Rv is the ratio of greatest to smallest volume 



during expansion, either R P , ratio of greater to smaller pressures, or — , the 

Li 



WOEK OF PISTON ENGINES 



201 



reciprocal of the cut-off, may be substituted, since the expressions apply only 
to the logarithmic law, and clearance is assumed equal to zero. When 
clearance is not zero; it is shown later that the cut-off as a fraction of stroke 
is not the reciprocal of Rp or R v . 

These expressions are perfectly general, but convenience in calculation 
will be served by deriving expressions for certain special cases. The first of 
these is the case of no expansion at all, the second that of complete expansion 
without over-expansion. This latter gives the most economical operation from 
the hypothetical standpoint, because no work of expansion has been left 
unaccomplished and at the same time no negative work has been introduced 
by over-expansion. 



A 


























B 


ln.pr/ 




































































































• 


























































































































































D l 


)K. pr.) 


E 






















































1 
|H 





Fig. 58. — First Special Case of Cycles I and II. Expansion = zero. Full Stroke Admission. 

First Special Case. If there is no expansion, together with the above assump- 
tion of no clearance, the diagram takes the form (Fig. 58), and 

IF=144Z)[(in.pr.)- (bk.pr.)] (243) 

m.e.p. = (in.pr.) — (bk.pr.) (244) 

Work per cu.ft. supplied = 144[(in.pr.) — (bk.pr.)] (245) 

Cu.ft. supplied per hr. per LH.P.= T . \ 3?7 ^ 1 r (246) 

(m.pr.) — (bk.pr.) v J 

Lbs. supplied per hr. per I.H.P. = T . ^-^ — .. ....... (247) 

(m.pr.) — (bk.pr.) v J 



202 



ENGINEERING THERMODYNAMICS 



Second Special Case. When the expansion is complete without over-expan- 
sion, no clearance, the points C and D, Fig. 59, coincide, and (rel.pr.) = (bk.pr.), 

hence R v = Rp = ,' p ~ = y. This value of cut-off, Z, is known as best cut-off, as 

it is that which uses all the available energy of the fluid by expansion. 



W= 144D(in.pr.) ^|^ = 144Z) (bk.pr.) log e R P . 



Ri 



(m.e.p.) = (in.pr.) 



log e Ri 



Ri 



Work per cu.ft. supplied = 144(in.pr.) log 6 R P . 



(248) 

(249) 
(250) 



A 




B (in 


Pr) 


























I 




























j\ 




























| \ 






















• 


















































































































E. 


























D(i 




























! 






|h 






















1 
If 



(bk.pr) 



Fig. 59. — Second Special Case of Cycles I and II. Complete Expansion Without Over 

Expansion Case of Best Cut-off. 



Cu.ft. supplied per hr. per I.H.P. = 
Lbs. supplied per hr. per I.H.P. = 



13,750 



(in.pr.) log e Rp 
13,7503i 



(in.pr.) log e R P 



(251) 
(252) 



Example 1. Method of calculating diagrams. Fig. 57 and Fig. 59. 
Assumed data for Fig. 57. 

Pa =P„ =90 lbs. per sq.in. abs. V a = V e =0 cu.ft. 
Pa =Pe = 14 lbs. per sq.in. abs. V c = V d = 13.5 cu.ft. 

F 6 = 6cu.ft. 



WORK OF PISTON ENGINES 203 

To obtain point C: 

V b 6 

P c =P b X— =90 X—— =40 lbs. per sq.in. abs. 
V c 13.5 

Assumed data for Fig. 59. 

P a =P b =90 lbs. per sq.in. V a = V e =0 cu.ft. 
Pa=Pe = 14 lbs. per sq.in. Va = 13.5 cu.ft. 

To obtain point B: 

V, = 7 d X^ = 13.5 X^ =2.1 cu.ft. 

Example 2. A simple double-acting engine admits steam at 100 per square inch 
absolute for i stroke, allows it to expand to the end of the stroke and then exhausts it 
against a back pressure of 5 lbs. per square inch absolute. If the engine has no 
clearance, a 7 x9-in. cylinder and runs at 300 R.P.M., what is the horse-power and steam 
consumption when steam is expanding according to the logarithmic law? Note: 
1 cu.ft. steam at 100 lbs. per square inch absolute weighs .2258 lb. 

From Eq. (237 6 ), 

m.e.p. = (in.pr.)( 1 -(bk.pr.) 



a00 (1+1 : g ' 4) -5 = 100 (1+1 / 86) -5=54.7: 
4 4 



(m.e.p.)Lan 54.7 X. 75x38.5x600 
*~ 33,000 " 33,000 " 



or directly from Eq. (242) 



IHp = Dn(m.e.p.), 



229.2 
2X600X54.7 



229.2 



=28. 



Lbs. steam per I.H.P. = — x — 

m.e.p. Ry 



13750 .2258 

X— 7— =14.13. 



54.7 

Therefore, steam per hour used by engine = 14.15x28 =396 lbs. 

Prob. 1. A steam engine has a cylinder 12x18 ins. with no clearance. It runs at 
200 R.P.M. and is double-acting. If the steam pressure be fixed at 100 lbs. per 
square inch absolute, and the back pressure at 10 lbs. per square inch abs., show how 
the horse-power and steam consumption will vary as cut-off increases. Take cut-off 
from i to f by eighths. Plot. 



204 ENGINEERING THERMODYNAMICS 

Prob. 2. Two engines of the same size and design as above are running on a steam 
pressure of 100 lbs. per square inch absolute, but one exhausts through a long pipe to 
the atmosphere, the total back pressure being 20 lbs. per square inch absolute, 
while the other exhausts into a condenser in which the pressure is but 3 lbs. per 
square inch absolute. If the cut-off is in each case |, how will the I.H.P. and steam 
used in the two cases vary? 

Prob. 3. By finding the water rate and the horse-power in the two following cases, 
show the saving in steam and loss in power due to using steam expansively. A pump 
having a cylinder 9 Xl2 ins. admits steam full stroke, while an engine of same size admits 
it but I of the stroke; both run at the same speed and have the same back pressure. 

Prob. 4. Steam from a 12x24 in. cylinder is exhausted at atmospheric pressure 
(15 lbs. per square inch absolute) into a tank, from which a second engine takes steam. 
Neither engine has clearance. The first engine receives steam at 100 lbs. per square 
inch absolute and the cut-off is such as to give complete expansion. The second engine 
exhausts into a 24 in. vacuum and its cut-off is such that complete expansion occurs in 
its cylinder. Also the cylinder volume up to cut-off equals that of the first cylinder 
at exhaust. If the stroke is the same in both engines and the speed of each is 200 
R.P.M., what is the diameter of the larger cylinder, the total horse-power developed, 
the total steam used, and the work per cubic foot of steam admitted to the first 
cylinder, the water rate of each engine and the total horse-power derived from each 
pound of steam? 

Prob. 5. The steam pressure for a given engine is changed from 80 lbs. per square 
inch gage to 120 lbs. per square inch gage. If the engine is 12x16 ins., running 250 
R.P.M. with a fixed cut-off of 25 per cent and no clearance, the back pressure being 
15 lbs. per square inch absolute, what will be the horse-power and the water rate in 
each case? 

Note: 1 cu.ft. of steam at 80 and 120 lbs. weighs .215 and .3 lb. respectively. 

Prob. 6. By trial, find how much the cut-off should have been shortened to 
keep the H.P. constant when the pressure was increased and what effect this would 
have had on the water rate. 

Prob. 7. A certain type of automobile engine uses steam at 600 lbs. per square 
inch absolute pressure. The exhaust is to atmosphere. For a cut-off of | and no 
clearance, what would be the water rate? 

Note: for 600 lbs. ^=1.32. 

Prob. 8. Engines are governed by throttling the initial pressure or shortening the 
cut-off. The following cases show the effect of light load on economy. Both engines, 
12x18 ins., running at 200 R.P.M. , with 125 lbs. per square inch absolute. Initial 
pressure and back pressure 'of 10 lbs. per square inch absolute. The load is sufficient 
to require full steam pressure at J cut-oif for each engine. Load drops to a point 
where the throttle engine requires but 50 lbs. per square inch absolute initial pressure 
with the cut-off still fixed at J. What is the original load and water rate, and new 
load and water-rate for each engine? 

Note: d for 125 lbs. absolute = .279 and for 50 lbs. =.117 lb. 

Prob. 9. The guarantee for a simple engine 18x24 ins., running at 200 R.P.M., 
states that the I.W.R. when cut-off is i will not exceed 15 lbs. if the initial pressure be 
100 lbs. per square inch gage, and back pressure 5 lbs. per square inch absolute. 
If engine has no clearance, see if this would be possible. 

4. Work of Expansive Fluid in Single Cylinder without Clearance. Expo- 
nential Expansion Cycle II. Mean Effective Pressure, Horse-power and 



WORK OF PISTON ENGINES 



205 



Consumption of Simple Engines. Referring to the diagram, Fig. 57, the 
work is given by the same areas as for Cycle I, but its algebraic expression 
is different because s is greater than 1 and an exponential expansion results on 
integration instead of a logarithmic one. 

In general, from Eq. (13a), Section 7, Chapter I, 

W = P b V b + P f^(^y~ 1 -lj-P d V d (253) 

Putting this in terms of initial conditions by the relations 

= (Jf) S = R s v and V c = V d = V b R v . 



Also 



there results 



P c V c -P b V b [yJ -£-^1, 



W=P b V b + 
= P b V b [] 



P b V b 



(s-l)Rv s - 
1 



tiRvS-i-V-PtVoRi 



-i(fli 



(s-l)R v s 
Rv \s-l 



1) 



P d V,R v . 

(a) 



(s-l)Rv 
= 144D \z (in.pr.) S "^f'~ 1 - (bk.pr.)l 



pr.)] (6) 

(c) 



which is the general equation for work of this cycle. 
Dividing by V b , the volume of fluid supplied, 



Work per cu.ft. supplied =P\ 



1 



H s-1 (s-VRv 8 - 1 
1 



PaRv (a) 



= 144(in.pr.) f — - (g _ W - i] ~ (bk.pr.)« F (6) 



(254) 



(255) 



Similarly, the mean effective pressure results from dividing the work by the 
displacement, Vd=V b Rv 

P b 



M.E.P. 



or 



Rv\s-1 (s-l)R 
(in.pr.) 



Rv 8 - 1 ) ' 



^ \r--i~ J^m?^) " (bk ' pr ° 



= Z(in.pr.) 



-Z 8 - 1 
s-1 



(bk.pr.) 



(a) 

(o) 
(c) 



(256) 



206 



ENGINEERING THERMODYNAMICS 



First special case of no expansion, when R v = l, results the same diagram 
as in the previous section, Fig. 58, and exactly the same set of formulas. 

Second special case when the expansion is complete without over-expansion, 
is again represented by Fig. 59 and for it 



P d = P> 



Whence 



or 



W 



\vj Rv s 



Work for complete expansion is 

•s£=*) (a) , 



TT=P ft 7 6 — T (l 
s—1 



144- 



Rv 



/D \ 

which is the general equation for the work of V b or ( tt ) cubic feet of fluid 

when the economy is best or for best cut-off. 

The work per cubic foot of fluid supplied for this case of complete expansion 
gives the maximum value for Eq. (255) and is obtained by dividing Eq. (257) 

by V b or (§- 



Max. work per cu.ft. supplied = Pv~^t ( 1 — „ s-1 I 



144(in.pr.)- 



s-\\ l Ry^ 1 ) 



(a) 
(b) 






(258) 



which is the general equation for maximum work per cubic foot of fluid supplied. 
The expression for mean effective pressure becomes for this case of best cut-off, 



or, 



M.E.P.= 

(m.e.p.) = 



Pt, s 



Rvs-l 

(in.pr.) 
Rv 



R 



h) 



(«) 



i 1 -*^; (6) 



(259) 



It is convenient to note that in using Eqs. (257), (258) and (259) it may be 
desirable to evaluate them without first finding Rv Since 



Rv 



v b 



i_ i^ 

= Rp 



(c)s = 1.111 and (m.e.p.) =-°-(^^- _ *„ )-!( =47.5. 



WORK OF PISTON ENGINES 207 

this substitution may be made, and 

Example. Compare the horse-power and the steam consumption of a 9x12 in. 
simple double-acting engine with no clearance and running at 250 R.P.M. when initial 
pressure is 100 lbs. per square inch absolute and cut-off is J, if 

(a) steam remains dry and saturated throughout expansion, 

(6) remains superheated throughout expansion, and 

(c) if originally dry and suffers adiabatic expansion. 

Back pressure is 10 lbs. per square inch absolute. 

(in.pr.)/ s 1 \ 

(m.e. P .)=- ^ (jzi- (s . w -i j -( bk -p r -)- 

For case (.).-! .0646 and (m.e.p.) JJ (^f - ^4"*" ) " 10 =48 ' 6; 

(6) 8=1.3 and (m.e.p.) =^i|-^-^ -10=43.5; 

'( 
4 \ .111 .111 X4" n 

T n D / s r (m.e.p.)1.0x63.6x500 nnr 
I.H.P. = (m.e.p.)Lan = 33 0QQ = .965 m.e.p. 

/. I.H.P. for case (a) =46.9, 

(b) =42.0, 

(c) =45.8. 

From Eq. (240), lbs. steam per hour per I.H.P. =, 13 ' 750 x Jl 

m.e.p. i?« 

.*. For case (a) steam per hr. =46.9 X — ■ — X— ; 

48.6 4 

(6) steam per hr. =42 X ' r X—; 
4:3.0 4 

a /x . . JEO 13,750 Si 

(c) steam per hr. = 45.8 X — - — X— . 
47.5 4 

Prob. 1. On starting a locomotive steam is admitted full stroke, while in running 
the valve gear is arranged for f cut-off. If the engine were 18 X30 ins., initial pressure 
150 lbs. per square inch absolute, back pressure 15 lbs. per square inch absolute, what 
would be the difference in horse-power with the gear in normal running position and 
in the starting position for a speed of 20 miles per hour with 6-ft. driving wheels? Con- 
sider the steam to be originally dry and expanding adiabatically. What would be the 
difference in steam per horse-power hour for the two cases and the difference in total 
steam? Clearance neglected. 

Prob. 2. Consider a boiler horse-power to be 30 lbs. of steam per hour; what must be 
the horse-power of a boiler to supply the following engine? Steam is supplied in a super- 



208 ENGINEERING THERMODYNAMICS 

heated state and remains so throughout expansion. Initial density of steam = .21 lbs. per 
cubic foot. Engine is 12x20 ins., double-acting, 200 R.P.M., no clearance, initial 
pressure 125 lbs. per square inch absolute, back pressure a vacuum of 26 ins. of 
mercury. Cut-off at maximum load f , no load, rV. What per cent of rating of boiler 
will be required by the engine at no load? 

Prob. 3. While an engine driving a generator is running, a short circuit occurs 
putting full load on engine, requiring a f cut-off. A moment later the circuit-breaker 
opens and only the friction load remains, requiring a cut-off of only ^. The engine 
being two-cylinder, double-acting, simple, 12 Xl8 ins., running at 300 R.P.M., and having 
no clearance, what will be the rate at which it uses steam just before and just after 
circuit-breaker opens if the steam supplied is at 125 lbs. per square inch absolute and is 
just dry, becoming wet on expanding, and back pressure is 3 lbs. per square inch 
absolute? 

Prob. 4. A pumping engine has two double-acting steam cylinders each 9x12 ins. 
and a fixed cut-off of J. It runs at 60 R.P.M. on 80 lbs. per square inch absolute steam 
pressure and atmospheric exhaust. Cy finder is jacketed so that steam stays dry 
throughout its expansion. How much steam will it use per hour? Neglect clearance. 

Prob. 5. If an engine 10x14 ins. and running 250 R.P.M. has such a cut-off that 
complete expansion occurs for 90 lbs. per square inch absolute initial pressure, and at 
atmospheric (15 lbs. absolute) exhaust, what will be the horse-power and steam used per 
hour, steam being superheated at all times, and what would be the value for the horse- 
power and steam used if full stroke admission occurred? 

Prob. 6. The steam consumption of an engine working under constant load is 
better than that of a similar one working under variable load. For a 16 X24 ins. engine 
running at 250 R.P.M. on wet steam of 125 lbs. per square inch absolute and atmospheric 
exhaust, find the horse-power and steam used per horse-power per hour for best con- 
dition and by taking two lighter and three heavier loads, show by a curve how steam 
used per horse-power per hour will vary. 

Prob. 7. For driving a shop a two cylinder single-acting engine, 6x6 Ins., running 
at 430 R.P.M., is used. The cut-off is fixed at \ and intitial pressure varied to control 
speed. Plot a curve between horse-power and weight of steam per hour per horse- 
power for 20, 40, 60, 80, 100, 120 lbs. per square inch absolute initial pressure and 
atmospheric exhaust. Steam constantly dry. Clearance zero. 

Note: *i for above pressures equals .05, .095, .139, .183, .226, and .268 lbs. per 
cubic foot, respectively. 

Prob. 8. Taking the loads found in Prob. 7, find what cut-off would be required 
to cause the engine to run at rated speed for each load if the initial steam pressure 
were 100 lbs. per square inch absolute, and the back pressure atmosphere, and a plot 
curve between horse-power and steam used per horse-power hour for this case. 

Prob. 9. For working a mine hoist a two-cylinder, double-acting engine is used 
in which compressed air is admitted f stroke at 125 lbs. per square inch absolute and then 
allowed to expand adiabatically and exhaust to atmosphere. If the cylinders are 18 X24 
ins. and speed is 150 R.P.M., find the horse-power and cubic feet of* high pressure air 
needed per minute. 

5. Work of Expansive Fluid in Single Cylinder with Clearance. Loga- 
rithmic Expansion and Compression; Cycle III. Mean Effective Pressure, 
Horse-power, and Consumption of Simple Engines. As in previous cycles, 
the net work of the cycle is equal to the algebraic sum of the positive work 



WORK OF PISTON ENGINES 



209 



done on the forward stroke and the negative work on the return stroke. By 
areas, Fig. 60, this is 



Work area = JABN+NBCW- WDEO - OEFJ. 
Expressed in terms of diagram points this becomes 



W = 



P b {V b -V a )+P b VAog e ^ 

Vb 
-P d {V d -V c )-P e V e \0& 



Vr 



(260) 



(CD) 


l 


-Volun 


le 




B 


















All 






J ia.pr.) 




-X}& 






Lu A 






Ql 


























































































I 


F 
























































^C(r 


\ 






























1 


\ 




























i 


V 


























D( 


IE 




























i 




1 












^ 












1 

jW 


i 

i 


J 


lc 


) 








N 


~~ D 












i 


-H 


<-D^I 


























V 



G- 



Fig. 60. — Work of Expansive Fluid in Single Cylinder with Clearance. Logarithmic Expan- 
sion for Cycle III. Exponential for Cycle IV- 



Expressing this in terms of displacement, in cubic feet D ; clearance as a frac- 
tion of displacement, c; cut-off as a fraction of displacement, Z; compres- 
sion as a fraction of displacement, X; initial pressure, in pounds per square 
inch (in.pr.), and exhaust or back pressure, in pounds per square inch (bk.pr): 



P 6 = 144(in.pr.); 

Pa = 144 (bk.pr.); 

(V b -V a )=ZD; 

V c= D(l+c) = l+c 
V b D(Z+c) Z+c' 



(V d -V e ) = (l-X)D; 
V b = D(Z+c); 
V e = D(X+c). 
V e D(X+c) X+c 



Vr 



Dc 



210 



ENGINEERING THERMODYNAMICS 



Whence 
Work in ft .-lbs. per cycle is 



W 



144D { (in.pr.) [z+(Z+c) log* ^g] 
-(bk.pr.) [(l-Z) + (Z+c)log e ^J 



(261) 



From Eq. (261), the mean effective pressure, pounds per square inch, follows 
by dividing by 144D: 



(m.e.p.); 



(in.pr.) Z+(Z+c) log e -^-j—. . . (mean forward press.) 
-(bk.pr.)r(l-Z) + (X+c)loge— , ^- C .1 (mean bk.pr.) 



- (262) 



This is a general expression of very broad use in computing probable mean 
effective pressure for the steam engine with clearance and compression, or for 
other cases where it is practicable to assume the logarithmic law to hold. Fig. 
117, at the end of this chapter, will be found of assistance in evaluating this 
expression. 

Indicated horse-power, according to expressions already given, may be 
found by either of the following equations : 

T tt p _ (m.e.p.)La n_ 144(m.e.p.)Dn _(m.e.p.)Dn 
33,000 " 33,000 " 229.2~ ' 

where L is stroke in feet, a is effective area of piston, square inches, n is the 
number of cycles performed per minute and D the displacement, cubic feet. 

It might seem that the work per cubic foot of fluid supplied could be 
found by dividing Eq. (261) by the admission volume, 



(V b -V a ,) 

but this would be true only when no steam is needed to build up the pressure 
from F to A. This is the case only when the clearance is zero or when com- 
pression begins soon enough to carry the point F up to point A, i.e., when by 
compression the pressure of the clearance fluid is raised to the initial pressure. 
It is evident that the fluid supplied may perform the two duties: first, 
building up the clearance pressure at constant or nearly constant volume, 
and second, filling the cylinder up to cut-off at constant pressure. To measure 
the steam supplied in terms of diagram quantities requires the fixing of the 
volume of live steam necessary to build up the pressure from F to A and adding 
it to the apparent admission volume (V b — V a )» This can be done by producing 



WORK OF PISTON ENGINES 211 

the compression line EF to the initial pressure Q, then LQ is the volume that 
the clearance steam would have at the initial pressure and QA the volume of 
live steam necessary to build up the pressure. The whole volume of steam 
admitted then is represented by QB instead of AB or by (V b —V a ) instead of 
by (V b —V a ), and calling this the supply volume, 



But 



Hence 



(Sup.Vol.) = (F 6 -F a .) 



Va = vF^ = ^! bk - pr ; } = d+c)D; bk ^ 

P tt (m.pr.) (m.pr.) 



(Sup.Vol.)=i>[(Z+c)-(Z-hc)|g^], .... (263) 

which is the cubic feet of fluid admitted at the initial pressure for the dis- 
placement of D cubic feet by the piston. Dividing by D there results 

(Sup.Vol.)_ (z+c) _ (X+c) (bk : prO j (264) 



D K ' ' N ' '(in.pr.)' 

which is the ratio of admission volume to displacement or cubic feet of live 
steam admitted per cubic foot of displacement. 

Dividing the work done by the cubic feet of steam supplied gives the 
economy of the simple engine in terms of volumes, or 

W 
Work per cu.ft. of fluid supplied 



(Sup.Vol.) 



(in.pr.)|z+(Z+c)log e M -(bk.pr.)r(l-Z) + (Z+c)log e ^±~ C l 
= 144 L ^±n L^ LA. (265) 

(Z+c)-(X+c)^r7 

It is more common to express economy of the engine in terms of the weight 

of steam used per hour per horse-power or the " water rate," which in more 

general terms may be called the consumption per hour per I.H.P. 

Let di be the density or weight per cubic foot of fluid supplied, then the 

weight per cycle is (Sup.Vol.) di, and this weight is capable of performing 

W 
W foot-pounds of work or (Sup.Vol.) di lbs. per minute will permit of 5^7^ 

horse-power. But (Sup.Vol.) £1 lbs. per mintue corresponds to 60 (Sup.Vol.) 
di lbs. per hour, whence the number of pounds per hour per horse-power is 

60(Sup.Vol.)3i 
IF/33,000 "' 



212 ENGINEERING THERMODYNAMICS 

which is the pounds consumption per hour per I.H.P., whence 

n +• • iu v, tttd 60X33,000( Sup.Vol.)^i /tuiQ l 

Consumption m lbs. per hr. per I.H.P.= '- ^-^ •— ■, . . . (266) 

which is the general expression for consumption in terms of the cubic feet of 
fluid admitted per cycle, §i initial density, and the work per cycle. 

As work is the product of mean effective pressure in pounds per square foot, 
(M.E.P.,) and the displacement in cu.ft. or W = (M.E.P.)D, or in terms of 
mean effective pressure pounds per square inch IF =144 (m.e.p.)D, the 
consumption may also be written 

n +• • ik u TTTT3 60X33,000(Sup.Vol.)3i 

Consumption in lbs., per hr. per I.H.P. = ., : . , — ^-^ — - — 

144(m.e.p.)D 



13,750 (Sup.Vol.)tf 



(m.e.p.) D 

= ^U +c) _ (x+c) fe)U (267) 

(m.e.p.) L (m.pr.) J v ' 

which gives the water rate in terms of the mean effective pressure, cut-off, 
clearance, compression, initial and back pressures and initial steam density. 
It is sometimes more convenient to introduce the density of fluid at the back 
pressure ^2, which can be done by the relation (referring to -the diagram), 



whence 



This on substitution gives 



<?2 = 



(in.pr.) 
(bk.pr.) 


V e 


dl 


, (bk.pr 


) 





(in.pr.) * 



Consumption in lbs., per hr. per I.H.P 

13,750 



(m.e.p.) 



(Z+c)di-(X+c)d 2 .\ (268) 



Since the step taken above of introducing ^2 has removed all pressure or 
volume ratios from the expression, Eq. (268) is general, and not dependent 
upon the logarithmic law. It gives the consumption in terms of mean effective 
pressure, cut-off, clearance, compression and the density of steam at initial 
and back pressure, which is of very common use. 

It cannot be too strongly kept in mind that all the preceding is true only when 
no steam forms from moisture water during expansion or compression or no steam 
condenses, which assumption is known to be untrue. These formulae are, there- 
fore, to be considered as merely convenient- approximations, although they 



WORK OF PISTON ENGINES 



213 



are almost universally used in daily practice. (See the end of this chapter for 
diagrams by which the solution of this expression is facilitated.; 

Special Cases. First, no expansion and no compression would result in 
Fig. 61. For it 



W= 144£>[(in.pr.) - (bk.pr.)]. 
(m.e.p.) = (in.pr.) — (bk.pr.) . . 



(269) 
(270} 



p 


A 












>ply-V 


olume 


















1 [ 
























B 


(in.pr.) 


1 


2 






























1 

! 

i 








































































































































cDi 


F" 


































1 
































N 


1 




























D 






I 
































K 


1 

1 














-[)— 



















W V 

Fig. 61. — First Special Case of Cycles III and IV. Expansion and Compression both Zero, 

but Clearance Finite. 

The volume of fluid supplied per cycle is QB, or from Eq. (263) it is 

(Sup.Vol.)=c[l+c-c|g^] (271) 



Consumption in lbs. per hr. per I.H.P.= 



13,750 



L (in.pr.) J (21 



(272) 



(in.pr.) — (bk.pr.) 
or in terms of initial and final densities, 

Consumption in lbs. per hr. per I.H.P.= ' ^n^ ^+^i-^] (273) 

The second special case is that of complete expansion and compression, as 
indicated in Fig. 62. Complete expansion provides that the pressure at the 



214 



ENGINEERING THERMODYNAMICS 



end of expansion be equal to the back pressure, and complete compression that 
the final compression pressure be equal to the initial pressure. 



Here 



and hence 



Vc = Ve == 1+c = X+c = (in.pr.) 
V b V a Z+c c (bk.pr.)' 



and 



Vc—V in.pr. , , ^i ■ in. 



»h>-<m)l 



p 

1 


A*— 


-7F)H 


2b 
























(in.pr.) 










[ 


























CD 


r 






i\ 


































! N 


\ 
















j 




| 














\ 




























\ 


































































































G 






























c 


(bk.pr.) 








1 1 — 






















1 
































1 

jw 




K 


j 




JN 











u 















Fig. 62. — Second Special Case of Cycles III and IV. Perfect Expansion and Perfect 
Compression with Clearance. 



Hence by substitution 



1 I (bk.pr.) J 



from which 



Again, 



ZD 



Z-H+c) ( ^)-c 



(in.pr.) 
(bk.pr.) ' 

(in. pr. 



(274) 



v v cd\(^)-i\ r/ . , -. 
= Ve-Vg _ LVbk.pr./ J [/ln.prA 1 

V c -Va D lAbk-prJ 'J 



. . (275) 



WORK OF PISTON ENGINES 215 

Eq. (274) gives the cut-off as a fraction of the displacement necessary to give 
complete expansion, while Eq. (275) gives the compression as a fraction of dis- 
placement to give complete compression, both in terms of clearance, initial 
pressure and back pressure, provided the logarithmic law applies to expansion 
and compression. 

Substitution of the values given above in Eq. (261) gives, after simplifi- 
cation, 

TF=144D[(l+c)(bk.pr.)-c(in.pr.)]lo ge |^ ) . . . (276) 

(m.e.p.) = [(l+ C )(bk.pr.)-c(in.pr.)]Ioge^!^- ) (277) 

In this case the volume supplied is exactly equal to that represented by the 
admission line AB, and is equal to 

(Sup.Vol.)=ZZ) (278) 

Hence, the consumption, in pounds fluid per hour per I.H.P. in terms of 
initial density, is 

Consumption in lbs. per hr. per I.H.P. =- — - Zdi, 



but 



(bk^) 
(m.pr.) 



.p. r . [ (bk.pr.) "L (in.pr. ) r N1 

(m.prj^l+cj^^-cjlofr^j. (m.pr.) log 



'(bk.pr.)' 
hence 

Consumption in lbs. fluid per hr. per I.H.P. = ? ^ r-. . . (279) 

/• m (m.pr.) 

(m - pr - } l0& (bEpTo 

This last equation is interesting in that it shows the consumption (or water 
rate, if it is a steam engine) is independent of clearance, and dependent only 
upon initial density, and on the initial and final pressures. 

An expression may also be easily derived for the consumption in terms of 
initial and final density, but due to its limited use, will not be introduced here. 

Example 1. Method of calculating Diagrams Fig. 60 and Fig. 62. 
Assumed data for Fig. 60: 

Pa =Pa =Pb =90 lbs. per square inch abs. V a = Vf = .5 cuit. 
P g —P e =Pd = 14 lbs. per square inch abs. Va = V c = 135 cu.ft. 
P/ = 50 lbs. per square inch abs. Vb =6 cuit. 



216 ENGINEERING THERMODYNAMICS 

« 

To obtain point C: 



To obtain point E: 



To obtain point Q : 



Vh 6 

Pe =P b X— =90 X—— =40 lbs. per sq.in. abs. 

V c lo.O 



Ve = V f X?f = .5X^ = 1. 78 cuit. 



F fl = F / X^ = .5x^ = .278 cuit. 



Intermediate points from B to C and E to Q are found by assuming volumes 
and computing the corresponding pressures by relation P x V x =PbV b or P x V x =P e V e . 

V .5 

Clearance is — — ^=- =~ =3.8 per cent, 

V d — Va lo 

Cut-off is — — =— - =42.3 per cent, 

Vd~ Va lo 
n . . Ve-Va 1.28 no 

Compression is =—r- =9.9 per cent. 

Vd~ Va lo 

Assumed data for Fig. 62. 

Pa =Pb =90 lbs. per square inch absolute. V a = .5 cuit. 
P e = Pc= 14 lbs. per square inch absolute. Vd = 13.5 cuit. 



To obtain point B. 



To obtain point E: 



V b = Vc~ = 13.5 X^ =2.11 cuit. 



V e = VaX^ = .5x~=S2m.ft. 

Pe 14 



Intermediate points from B to C and from A to E are to be found by assuming 
yarious volumes and finding the corresponding pressures from relation P x V x =PaV a or 
P x V x =PbV b . 

Example. 2. What will be the horse-power of, and steam used per hour by the 
following engine: 

(a) cut-off 50 per cent, compression 30 per cent, 

(6) complete expansion and compression, 

(c) no expansion or compression. 

Cylinder, 12xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial pressure 
85 lbs. per square inch gage, back pressure 15 lbs. per square inch absolute, and 
logarithmic expansion and compression, 



WORK OF PISTON ENGINES 217 

Note: d for 85 lbs. gage = .23, <n for 15 lbs. absolute = .038 cu.ft. 
(a) From Eq. (262) 

(m.e.p.) = (in.pr.) \Z + (Z+c) log c |±^J -(bk.pr.) |~(1 -X) + (X+c)log —"^1 , 

= 100[.5 + (.5+.07)log e -|±:^]-i5[(l-.3) + (.3 + .07) log, *+^] 

= 86-20=66 lbs. sq.in. 

IHP = (m.e.p . ) La n ^6 6X1.5 X113.1X400 _ 

33,000 33,000 ~ =135 * 

From Eq. (267) steam per hour per I.H.P. in pounds is 

= 1 ^ 5 °[(.5 + .07)-(.3 + .07)x^ jx.23=25 1bs. 



Hence total steam per hour =25x135 =3380 lbs. 
(b) From Eq. (277) 

(m.e.p.) =[(1 + C )(bk. pr.) - C (in. pr.)] log e (^^ 

\bk.pr. 

=[(l+.07) Xl5-.07xl00]log e 6.67 = 17.2 lbs. sq.in. 



17.2X1.5X113.1X400 

33,000 60A ' 



From Eq. (279) 



Steam per I.H.P. per hour = 13^50^ = 13^750x^23 

(m.pr.)log/ in - pr ^ 10 °X1.9 



^bk.pr. 

Total steam per hour = 16.^X35.4 =588 lbs. 

(c) FromEq. (270) (m.e.p.) = (in.pr.) -(bk.pr.) =100-15 =85 lbs. sq.in. 

85X1.5X13.1X400 
.. l.xl.Jr. = — = 1 74 k 

33,000 1/4 '°- 

From Eq. (273) 

Steam per I.H.P. per hour - gg[(l +c)h _ c a 2 ] _1M50h.o7x.23_.o7x.o38] =35.4. 

Total steam per hour = 174.5x35.4 =6100 lbs. 

Prob 1. What will be the horse-power and water rate of a 9 X 12 in. simple engine 
having 5 per cent clearance, running at 250 R.P.M. on 100 lbs. per square inch abso- 



218 ENGINEERING THERMODYNAMICS 



lute initial pressure and 5 lbs. per square inch absolute back pressure when the cut-off is 
-§-, -J-, and I, expansion follows the logarithmic law, and there is no compression? 

Note : 8 for 10O lbs. absolute = .23, 8 for 5 lbs. absolute = .014. 

Prob. 2. Will a pump with a cylinder 10x15 ins. and 10 per cent clearance give the 
same horse-power and have the same water rate as a pump with cylinders of the same 
size but with 20 per cent clearance, both taking steam full stroke? Solve for a case 
of 125 lbs. per square inch absolute initial pressure, atmospheric exhaust and a speed of 
50 double strokes. No compression. 

Note: 8 for 125 lbs. absolute = .283, 8 for 15 lbs. absolute = .039. 

Prob. 3. Solve the above problem for an engine of the same size, using steam expan- 
sively when the cut-off is | and R.P.M. 200, steam and exhaust pressure as in Prob. 2 
and compression zero. 

Prob. 4. Two engines, each 9x15 ins., are running on an initial pressure of 90 lbs. 
per squarennch absolute, and a back pressure of one atmosphere. One has no clearance, 
the other 8 per cent. Each is cutting off so that complete expansion occurs. The speed 
of each is 200 and neither has any compression. What will be the horse-power and 
water rate? 

Note: 8 for 90 lbs. =.24, 8 for 15 lbs. = .039. 

Prob. 5. By finding the horse-power and water rate of a 12x18 in. double-acting 
engine with 8 per cent clearance, running at 150 R.P.M. on an initial steam pressure 
of 90 lbs. per square inch absolute and atmosphere exhaust for a fixed cut-off of £ and 
variable compression from to the point where the pressure at the end of compression 
is equal to 125 per cent of the initial pressure, plot the curves between compression and 
horse-power, and compression and water rate to show the effect of compression on 
the other two. 

Note: 8 for 90 lbs. =.21, 8 for 15 lbs. = .039. 

Prob. 6. A steam engine is running at such a load that the cut-off has to be -f at 
a speed of 150 R.P.M. The engine is 14 X20 ins. and has no clearance. Initial pressure 
100 lbs. per square inch absolute and back pressure 5 lbs. per square inch absolute. 
What would be the cut-off of an engine of the same dimensions but with 10 per cent 
clearance under similar conditions? 

Prob. 7. The steam pressure is 100 lbs. per square inch gage and the back 
pressure is 26 ins. of mercury vacuum when the barometer is 30 ins. For a 16x22 
in.engine with 6 per cent clearance running at 125 R.P.M., cut-off at ^ and 30 per cent 
compression, what will be the horse-power and the water rate? Should the steam 
pressure be doubled what would be the horse-power and the water rate? If it should 
be halved? 

Note: 8 for 100 lbs. gage = .2017, 8 for 26 ins. Hg.=.0058. 

Prob. 8. While an 18x24 in. simple engine with 4 per cent clearance at speed 
of 150 R.P.M. is running with a \ cut-off and a compression of \ on a steam pressure of 
125 lbs. per square inch gage, and a vacuum of 28 ins. of mercury, the condenser fails 
and the back pressure rises to 17 lbs. per square inch absolute. What will be the change in 
the horse-power and water rate if all other factors stay constant? What would the new 
cut-off have to be to keep the engine running at the same horse-power and what 
would be the water rate with this cut-off? 

Note: 8 for 125 lbs. gage = .315, 8 for 28 in. Hg.=.0029, 8 for 17 lbs. absolute 
= .0435. 

Prob. 9. Under normal load an engine has a cut-off of f , while under light load 
the cut-off is but A. What per cent of the steam used at normal load will be used 



WOEK OF PISTON ENGINES 



219 



at light load for the following data? Cylinder 10x18 ins.; 7 per cent clearance; 200 
R.P.M.; initial pressure 120 lbs. per square inch gage; back pressure 2 lbs. per square 
inch absolute; compression at normal load 5 per cent; at light load 25 per cent. 
Note: 8 for 120 lbs. gage =.304, S for 2 lbs. absolute = .0058. 

6. Work of Expansive Fluid in Single Cylinder with Clearance ; Exponential 
Expansion and Compression, Cycle IV. Mean Effective Pressure, Horse- 
power and Consumption of Simple Engines. As pointed out in several places, 
the logarithmic expansion of steam only approximates the truth in real engines 
and is the result of no particular logical or physically definable hypothesis as to 
the condition of the fluid, moreover its equations are of little or no value 
for compressed air or other gases used in engine cylinders. All expansions 
that can be defined by conditions of physical state or condition of heat, 
including the adiabatic, are expressible approximately or exactly by a definite 
value of s, not unity, in the expression PV S = const. All these cases can 
then be treated in a group and expressions for work and mean effective 
pressure found for a general value of s, for which particular values belonging 
to, or following from any physical hypothesis can be substituted. The area 
under such expansion curves is given by Eq. (13) Chapter I, which applied to 
the work diagram, Fig. 60, in the same manner as was done for logarithmic 
expansion, gives the net work: 



■ . (280) 



S J. 


1- 


-(- 


1) ] 


(area JABCWJ) 


-P«{V d -V e )-^ 

s —1 


[©" 


-1 


(area WDEFJW ) 


lucing the symbols, 








P 6 = 144(in.pr.), 






V b = D(Z+c). 


P d = 144(bk.pr.), 






V e = D(X+c). 


(V i -V a )=ZD, 






(VA Z+c 
\VJ 1 + c 


(V d -V e )=D(l-X), 








(V e \ X+c 
\Vf) c ' 



-(bk.pr.)[(l-X)+f±^[ 



X+c 



'-'}} 



(281) 



Eq. (281) gives the work in foot-pounds for D cubic feet of displacement in a 
cylinder having any clearance c, cut-off Z, and compression X, between two 



220 ENGINEERING THERMODYNAMICS 

pressures, and when the law of expansion is PV S = const, and s anything except 
unity, but constant. 

The mean effective pressure, pounds per square inch, is obtained by divid- 
ing the expression for work by 144Z), giving 



(m.e.p.) = (in.pr.) { Z+—-^ 



1 — (ttt ) f (mean for'd pr.) 

(bk.pr.) { (1-X)+J±- C [(^ p) Sl -l] ) (mean bk.pr.) 



(282) 



which is the general expression for mean effective pressure for this cycle. 

It was pointed out in Section (5) that the cubic feet of fluid admitted at the 
initial pressure was not represented by AB, Fig. 60, but by QB, and the same 
is true for this case, so that the 

'Sup.Vol.) = 7 6 -F ff . 

But when the expansion and compression laws have the form PV s = c 



JPJ \in.pr 

Whence 



l 



(Sup.Vol.)=Z)[(Z+c)-(X+c)(g^) s I (283) 

Eq. (283) gives for these cases what was given by Eq. (263) for the logarithmic 
law, the cubic feet of fluid supplied at the initial pressure for the displacement 
of D cubic feet in terms of cut-off, clearance, compression and the pressures. 
From this, by division there is found the volume of high pressure fluid per 
cu.ft. of displacement 



(Sup.Vol.) =(z+c) _ (x+c) /bki>r. r (2g4) 



D \in.pr 

The consumption is given by the general expression already derived, 
Eq. (34), from which is obtained, 



Consumption lbs. per hr. per I.H.P. 



iW (Z+c) _ (Z+c) fey~L (285) 

(m.e.p.) L '\m.pr. / J 



WORK OF PISTON ENGINES 221 

Eq. (285) gives the water rate or gas consumption in terms of mean effective 
pressure, initial and back pressure, cut-off, clearance, compression and initial 
fluid density. Introducing the density at the back pressure by the relation, 



P q V q s = P e V e s ; 

7 e _Si_ / in.pr. \ s 
y a ~S2~\bk.pr.; ; 



there results 

Consumption lbs. per hr. per I.H.P. = ^ 50 |~(Z+c) h - (X+c) B 2 1 , . (286) 

which is identical with Eq. (268) and is, as previously observed, a general expres- 
sion, no matter what the laws of expansion and compression, in terms of 
mean effective pressure, cut-off, clearance, compression and the initial and 
final steam density. 

The first special case of full admission, no compression might at first thought 
appear to be the same as in the preceding section, where the logarithmic law 
was assumed to hold, and so it is as regards work and mean effective pressure, 
Eq. (269) and (270), but referring to Fig. 61 it will be seen that since the expo- 
nential law is now assumed instead of the logarithmic, the point Q will be dif- 
ferently located (nearer to A than it was previously if s is greater than 1), and 
hence the supply volume QB is changed, and its new value is 

i 
(Sup.Vol.)=D[l+c-c(g^) 7 ] (287) 

Hence, consumption stated in terms of initial density of the fluid h, is 
Consumption lbs. per hr. per I.H.P. 

13,750 



(in.pr.) — (bk 



.FihMSf] 8 « 



Stated in terms of initial and final densities, the expression is as before, Eq. (273), 
The second special case, complete expansion and compression is again repre- 
sented by Fig. 62. From the law of expansion it is evident that 



i 
Ay 



222 ENGINEEKING THERMODYNAMICS' 

or stated in symbolic form, 

whence 

z ^+iwsf-° «»> 

Again referring to Fig. 62, 



X= 






c Kfer- cD 



Whence 

i 



X = c 



impr i y_ I 
bk.pr./ J 



(290) 



Eq. (289) gives the cut-off as a fraction of displacement necessary to give 
complete expansion, and (290), the compression fraction to give complete com- 
pression, both in terms of clearance, initial and back pressures, and the exponent 
s, in the equation of the expansion or compression line, PV S = const. 

The work of the cycle becomes for this special case, by substitution in Eq. 
(281), 

W-1W> (in.p,)^[(l + c)(^)--c][l-(^:)^], . (29!) 

and the mean effective pressure, lbs. per sq.in., is 

Ke.p.) = (in.p,)^[(l + c)(^:)^ C ][l-(gg)^. . . . (202) 



The volume of fluid supplied is, 

(Sup.Vol.)=ZD, (293) 



hence 



1 o 7Kf) 

Consumption, lbs. per hr. per I.H.P.= — '■ — Zl\. 

m.e.p. 



but 
Z 



M^) ; -1 



s— 1 L \m.pr. / J L \m.pr. / J 



^->M>-(my i 






WOEK OF PISTON ENGINES 223 



whence 

Consumption lbs. fluid per hr. per I.H.P. is, 

13,750XBi 



s-l t 



(294) 



. s I" /bk.prA « 1 

m.pr.) 1-f- — £- ) 

s-lL Vin.pr./ J ■ 



the expression for smallest consumption (or water rate if steam) of fluid for 
*the most economical hypothetical cycle, which may it be noticed, is again in- 
dependent of clearance. 

The expressions for work and mean effective pressure are not, however, 
independent of clearance, and hence, according to the hypothetical cycles here 
considered, it is proved that large clearance decreases the work capacity of a 
a cylinder of given size, but does not affect the economy, provided complete 
expansion and compression are attained, a conclusion similar to that in regard 
to clearance effect on compressor capacity and economy. Whether the actual 
performance of gas or steam engines agrees with this conclusion based only 
on hypothetical reasoning, will be discussed later. 

Example 1. What will be the horse-power of and steam used per hour by the 
following engine: 12xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial 
pressure 85 lbs. per square inch gage, back pressure 15 lbs. per square inch absolute, 
and expansion such that s = 1.3. 

(a) cut-off =50 per cent; compression = 30 per cent; 

(b) complete expansion and compression; 

(c) no expansion or compression; 

Note: 8 for 85 lbs. gage =.18; for 15 lbs. absolute. = .03. 
(a) From Eq. (282) 



-^H+sK-?)*"-']; 

--K MSI ! -KKD "-]) —■*- * 



59.8X1.5X113.1X400 
IJLR= 33,-000 =122 ' 



From Eq. (286) 
Steam per hour per I.H.P. 



13,750 [" 
(m.e.p.) [ 



(Z+c)*! -(X+c)B 2 l =^^[(.57) X.18-C37) X.03] =20.9 lbs. 



.-. Steam per hour = 122 X20.9 = 2560 



224 ENGINEERING THERMODYNAMICS 

(b) From Eq. (292) 



1 s-l 



'-'Hi..^[« + ..||) r -.][ 1 -(^)'] 1 



TTTT) 26.2X1.5X113.1X200 _ 
LH - P -= 3000 =64 



From Eq. (294) steam used I.H.P. per hour is, 



(in.pr.) 



13,750Bi 13,750 X.18 <fl K „ 
^j- = — j- = 16.5 lbs., 



hence total steam per hour = 16.5 X 64 = 1060 lbs. 

(c) From Eq. (270) which holds for any value of s, m.e.p. =100 — 15 =85 lbs. sq.in. 

tttt, 85X1.5X113.1X400 

and I.H.P. = ^r^: = 174.5, 

33,000 

From Eq. (288) steam per I.H.P. hour 

and total steam per hour = 174.5X24.5 =2475 lbs. 

Prob. 1. Two simple engines, each 12 Xl8 ins., with 6 per cent clearance are operated 
at \ cut-off and with 20 per cent compression. One is supplied with air at 80 lbs. 
per square inch gage, and exhausts it to atmosphere; the other with initially dry steam 
which becomes wet on expansion and which also is exhausted to atmosphere. For a speed 
of 200 R.P.M. what is the horse-power of each engine and the cubic feet of stuff supplied 
per horse-power hour ? 

Prob. 2. A crank-and-fly wheel two-cylinder, double-acting, pumping engine is 
supplied with dry steam and the expansion is such that it remains dry until exhaust. 
The cylinder size is 24x36 ins., cut-off to give perfect expansion, clearance 5 per cent, 
compression to give perfect compression, initial pressure 50 lbs. per square inch ab- 
solute, back pressure 5 lbs. per square inch absolute. What is the horse-power and 
water rate? What would be the horse-power and water rate of a full-stroke pump of the 
same size and clearance but having no compression, running on the same pressure range 
and quality of steam. 

Note: 8 for 50 lbs. absolute = .12, 8 for 15 lbs. absolute = .0387. 

Prob. 3. Should the cylinder of the following engine be so provided that the 
steam was always kept dry, would there be any change in the horse-power developed as 






WORK OF PISTON ENGINES 225 

compared with steam expanded adiabatically, and how much? Cylinder 20x24 ins., 
initial pressure 125 lbs. per sq. in. gage, back pressure 26 ins. vacuum, standard barom- 
eter, clearance 6 per cent, cut-off |, compression 10 percent, and speed 125 R.P.M. 
Prob. 4. What will be the total steam used per hour by a 20x28-in. double-acting 
engine running at 150 R.P.M. if the initial pressure be 125 lbs. per square inch absolute, 
back pressure one atmosphere, clearance 8 per cent, compression zero, for.cut-off 
i, £, f, and 2, if steam expands adiabatically and is originally dry and saturated? 
Note: 6 for 125 lbs. absolute = .283, 8 for 15 lbs. absolute =.0387. 
Prob. 5. An engine which is supplied with superheated steam is said to have an 
indicated water rate of 15 lbs. at I cut-off and one of 25 lbs. at | cut-off. See if 
this is reasonable for the following conditions: engine is 15x22 ins., 7 per cent clearance, 
no compression, initial pressure 100 lbs. per square inch gage., back pressure 28-in. 
vacuum, barometer 30 ins. and speed 180 R.P.M. 

Note: 8 for 100 lbs. gage = .262, for 28 in. Hg = .0029. 

Prob. 6. The tank capacity of a locomotive is 4500 gals. The cylinders are 
18X36 ins., initial pressure 200 lbs. per square inch gage, exhaust atmospheric, cut-off \, 
clearance 7 per cent, speed 200 R.P.M., no compression. The stream is dry at 
start and expansion adiabatic, how long will the water last if 40% condenses during 
admission? 

Note: 8 for 200 lbs. gage =.471, 8 for 15 lbs. absolute = .0387. 
Prob. 7. To drive a hoist, an air engine is used, the air being supplied for \ 
stroke at 80 lbs. per square inch gage expanded adiabatically and exhausted to atmos- 
sphere. If the clearance is 8 per cent and there is no compression how many cubic 
feet of air per horn' per horse-power will be needed? What, with complete compression? 
Prob. 8. A manufacturer rates his 44x42-in. double-acting engine with a speed 
of 100 R.P.M. at 1000 H.P. when running non-condensing, initial pressure 70 lbs. 
per square inch gage and cut-off \. No clearance is mentioned and nothing said about 
manner of expansion. Assuming s = 1.0646, find on what clearance basis this rating is 
made. 

Prob. 9. The water supply of a town is supplied by a direct-acting non-condensing 
pump with two cylinders, each 24x42 ins., with 10 per cent clearance, and no com- 
pression, initial pressure being 100 lbs. per square inch gage. What must be the size of the 
steam cylinder of a crank-and-flywheel pump with 6 per cent clearance to give the same 
horse-power on the same steam and exhaust pressures with a cut-off of |? Speed in 
each case to be 50 R.P.M. 

Note: 8 for 100 lbs. gage =.262, 8 for 15 lbs. abs.=.0387. 

7. Action of Fluid in Multiple-expansion Cylinders. General Description 
of Structure and Processes. When steam, compressed air, or any other high 
pressure working fluid is caused to pass through more than one cylinder in 
series, so that the exhaust from the one is the supply for the next, the engine 
is, in general, a multiple-expansion engine, or more specifically, a compound 
when the operations are in two expansion stages, triple for three, and quadruple 
for four stages. It must be understood that while a compound engine is one 
in which the whole pressure-volume change from initial to back pressure takes 
place in two stages, it does not necessarily follow that the machine is a two- 
cylinder one, for the second stage of expansion may take place in two cylinders, 
in each of which, half of the steam is put through identical operations; this 



226 ENGINEERING THERMODYNAMICS 

would make a three-cylinder compound. Similarly, triple-expansion engines, 
while they cannot have less than three may have four or five or six cylinders 
Multiple expansions engine, most of which are compound, are of two classes 
with respect to the treatment and pressure-volume changes of the steam, first 
without receiver, and second, with receiver. A receiver is primarily a chamber 
large in proportion to cylinder volumes, placed between the high- and low-pres- 
sure cylinders of compounds or between any pair of cylinders in triple or 
quadruple engines, and its purpose is to provide a reservoir of fluid so that 
the exhaust from the higher into it, or the admission to the lower from it, will 
be accomplished without a material change of pressure, and this will be accom- 
plished as its volume is large in proportion to the charge of steam received 
by it or delivered from it. With a receiver of infinite size the exhaust line of 
a high-pressure cylinder discharging into it will be a constant-pressure line, 
as will also the admission line of the low-pressure cylinder. When, however, 
the receiver is of finite size high-pressure exhaust is equivalent to increasing 
the quantity of fluid in the receiver of fixed volume and must be accompanied 
by a rise of pressure except when a low-pressure cylinder may happen to be 
taking out fluid at the same rate and at the same time, which in practice never 
happens. As the receiver becomes smaller in proportion to the cylinders, the 
pressure in it will rise and fall more for each high-pressure exhaust and low- 
pressure admission with, of course, a constant average value. The greatest 
possible change of pressure during high-pressure exhaust and low-pressure admis- 
sion would occur when the receiver is of zero size, that is when there is none at all, 
in which case, of course, the high-and low-pressure pistons must have synchronous 
movement, both starting and stopping at the same time, but moving either in 
the same or opposite directions. When the pistons of the no-receiver compound 
engines move in the same direction at the same time, one end of the high- 
pressure cylinder must exhaust into the opposite end of the low; but with 
oppositely moving pistons, the exhaust from high will enter the same end of the 
low. It is plain that a real receiver of zero volume is impossible, as the connect- 
ing ports must have some volume and likewise that an infinite receiver is equally 
impracticable, so that any multiple-expansion real engine will have receivers 
of finite volume with corresponding pressure changes during the period when a 
receiver is in communication with a cylinder. The amount of these pressure 
changes will depend partly on the size of the receiver with respect to the cylin- 
ders, but also as well, on the relation between the periods of flow into receiver, 
by high-pressure exhaust and out of it, by low-pressure admission, which 
latter factor will be fixed largely by crank angles, and partly by the settings 
of the two valves, relations which are often extremely complicated. 

For the purpose of analysis it is desirable to treat the two limiting cases 
of no receiver and infinite receiver, because they yield formulas simple enough to 
be useful, while an exact simple solution of the general case is impossible. These 
simple expressions for hypothetical cases which are very valuable for estimates 
and approximations are generally close to truth for an actual engine especially 
if intelligently selected and used. 



WORK OF PISTON ENGINES 227 

Receivers of steam engines may be simple tanks or temporary storage 
chambers or be fitted with coils or tubes to which live or high-pressure steam 
is supplied and which may heat up the lower pressure, partly expanded steam 
passing from cylinder to cylinder through the receiver. Such receivers are 
reheating receivers, and as noted, may heat up the engine steam or may evapo- 
rate any moisture it might contain. As a matter of fact there can be no 
heating of the steam before all moisture is first evaporated, from which 
it appears that the action of such reheating receivers may be, and is quite 
complicated thermally, and a study of these conditions must be postponed 
till a thermal method of analysis is established. This will introduce no 
serious difficulty, as such reheating receivers assist the thermal economy 
of the whole system but little and have little effect on engine power, likewise 
are now little used. Reheating of air or other gases, as well as preheating 
them before admission to the high-pressure cylinder is a necessary practice, 
when the supply pressure is high,' to prevent freezing of moisture by the gases, 
which get very cold in expansion if it be carried far. This is likewise, however, 
a thermal problem, not to be taken up till later. 

Multiple-expansion engines are built for greater economy than is possible 
in simple engines and the reasons are divisible into two classes, first mechanical, 
and second thermal. It has already been shown that by expansion, work is 
obtained in greater amounts as the expansion is greater, provided, of course, 
expansion below the back pressure is avoided, and as high initial and low 
back pressures permit essentially of most expansion, engines must be built 
capable of utilizing all that the steam or compressed gas may yield. If 
steam followed the logarithmic law of expansion, pressure falling inversely 
with volume increase, then steam of 150 lbs. per square inch absolute 
expanding to 1 lb. per square inch absolute would require enough ultimate 
cylinder space to allow whatever volume of steam was admitted up to cut-off 
to increase 150 times. This would involve a valve gear and cylinder structure 
capable of admitting T ^ = .0067 of the cylinder volume. It is practically 
impossible to construct a valve that will accurately open and close in this 
necessarily short equivalent portion of the stroke. This, however, is not the 
worst handicap even mechanically, because actual cylinders cannot be made 
without some clearance, usually more than 2 per cent of the displacement and 
in order that any steam might be admitted at all, the clearance in the example 
would have to be less than .67 per cent of the total volume, which is quite 
impossible. These two mechanical or structural limitations, that of admission 
valve gear and that of clearance limits, supply the first argument for multiple- 
expansion engines, the structure of which is capable of utilizing any amount 
of expansion that high boiler pressure and good condenser vacuum make 
available. For, if neglecting clearance, the low-pressure cylinder had ten 
times the volume of the high, then the full stroke admission of steam to the high 
followed by expansion in the low would give ten expansions, while admission 
to the high for r \ of its stroke would give 15 expansions in it, after which this 
final volume would increase in the low ten times, that is, to 150 times the original 



228 ENGINEERING THERMODYNAMICS 

volume, and cylinder admission of ^ or 6.7 per cent is possible with ordinary 
valve gears, as is also an initial volume of 6.7 per cent of a total cylinder 
volume, even with clearance which in reasonably large engines may be not 
over 2 per cent of the whole cylinder volume. 

It is evident that the higher the initial and the lower the back pressures 
the greater the expansion ratio will be for complete expansion, and as in steam 
practice boiler pressures of 225 lbs. per square inch gage or approximately 240 
lbs. per square inch absolute with vacuum back pressures as low as one or even 
half a pound per square inch are in use, it should be possible whether desirable 
or not, to expand to a final volume from 250 to 500 times the initial in round num- 
bers. This is, of course, quite impossible in simple engine cylinders, and as it 
easy with multiple expansion there is supplied another mechanical argument for 
staging. Sufficient expansion for practical purposes in locomotives and land 
engines under the usually variable load of industrial service is available for even 
these high pressures by compounding, but when the loads are about constant, 
as in waterworks pumping engines, and marine engines for ship propulsion, 
triple expansion is used for pressures in excess of about 180 lbs. gage. 

Use of very high initial and very low back pressures will result in simple 
engines, in a possibility of great unbalanced forces on a piston, its rods, pins 
and crank, when acting on opposite sides, and a considerable fluctuation in tan- 
gential turning force at the crank pin. Compounding will always reduce the 
unbalanced force on a piston, and when carried out in cylinders each of which 
has a separate crank, permits of a very considerable improvement of turning 
effort. So that, not only does multiple expansion make it possible to utilize 
to the fullest extent the whole range of high initial and low back pressures, 
but it may result in a better force distribution in the engine, avoiding shocks, 
making unnecessary, excessively strong pistons, and rods and equalizing turn- 
ing effort so that the maximum and minimum tangential force do not depart 
too much from the mean. 

The second or thermal reason for bothering with multiple-expansion com- 
plications in the interest of steam economy is concerned with the prevention 
of steam loss by condensation and leakage. It does not need any elaborate 
analysis to show that low-pressure steam will be cooler than high-pressure 
steam and that expanding steam in a cylinder has a tendency to cool the 
cylinder and piston walls, certainly the inner skin at least, so that after 
expansion and exhaust they will be cooler than after admission; but as 
admission follows exhaust hot live steam will come into contact with cool 
walls and some will necessarily condense, the amount being smaller the less 
the original expansion; hence in any one cylinder of a multiple-expansion 
engine the condensation may be less than a simple engine with the same range 
of steam pressures and temperatures. Whether all the steam condensation 
during admission added together will equal that of the simple engine or not 
is another question. There is no doubt, however, that as the multiple expan- 
sion engine can expand usefully to greater degree than a simple engine, and 
so cause a lower temperature by expansion, that it has a greater chance to 



WORK OF PISTON ENGINES 229 

reevaporate some of the water of initial condensation and so get some work 
out of the extra steam so evaporated, which in the simple engine might have 
remained as water, incapable of working until exhaust opened and lowered 
the pressure, when, of course, it could do no .good. It is also clear that steam 
or compressed-air leakage in a simple engine is a direct loss, whereas in a 
compound high-pressure cylinder leakage has at least a chance to do some 
work in the low-pressure cylinder. The exact analysis of the thermal reasons 
for greater economy is complicated and is largely concerned with a study of 
steam condensation and reevaporation, but the fact is, that multiple-expan- 
sion engines are capable of greater economy than simple. The thermal analysis 
must also consider the influence of the reheating receiver, the steam-jacketed 
working cylinder, and the use of superheated steam, their effects on the pos- 
sible work per pound of steam and the corresponding quantity of heat expended 
to secure it, and for air and compressed gas the parallel treatment of pre- 
heating and reheating. 

To illustrate the action of steam in multiple-expansion engines some indi- 
cator cards are given for a few typical cases in Figs. 63 to 66, together 
with the combined diagrams of pressure-volume changes of the fluid in all 
cylinders to the same scale of pressures and volumes, which, of course, makes 
the diagram look quite different, as indicator cards are usually taken to the 
same base length, fixed by the reducing motion, and to different pressure 
scales, to get as large a height of diagram as the paper will permit. Fig. 63 
shows four sets of cards taken from an engine of the compound no-receiver 
type, namely, a Vauclain compound locomotive. In this machine there are 
two cylinders, one high pressure and one low, on each side, the steam from 
the high pressure exhausting directly into the low-pressure cylinder so that 
the only receiver space is made up of the clearance and connecting passages 
between the cylinders. Starting with set A, the cards show a decreasing 
high pressure cut-off of 76 per cent in the case of set A to 54 per cent in the 
case of set D. The letters A, B, C and D refer in each case to admission, cut- 
off, release and compression, the use of primed letters denoting the low-pressure 
cylinder. 

In set A the high-pressure admission line AFB may be considered as made 
up of two parts, the part AF representing pressure rise at constant volume, 
wfrch is the admission of steam to the clearance space at dead center to raise the 
pressure from that at the end of compression to that of boiler pressure. From 
F to B admission occurred at constant pressure, steam filling the cylinder 
volume as the piston moved outward. At B cut-off or closure of the steam 
valve occurred and the steam in the cylinder expanded. At C, release or open- 
of the exhaust valve of the high-pressure cylinder occurred and the admission 
valve of the low-pressure cylinder opened, the steam dropping in pressure until 
the pressure in both high- and low-pressure clearance became equal, and then 
expanding in both cylinders, as the exhaust from the high and admission to 
the low occurred, the exhaust line CD of the high pressure and the admission 
line F'B' of the low pressure being identical except for the slight pressure drop 



230 



ENGINEERING THERMODYNAMICS 







(A) 



(B) 







(C) (D) 

Fig. 63. — Set of Indicator Cards from Vauclain Locomotive Illustrating the No-receiver 

Compound Steam Engine. 



'WOEK OF PISTON ENGINES 231 

in the passages between the high-- and the low-pressure cylinders. At D the 
high-pressure exhaust valve closed and compression of the steam trapped in 
the high-pressure cylinder occurred to point A, thus closing the cycle. From 
point D' in the low-pressure cylinder, which corresponds to D in the high 
pressure, no more steam was admitted to the low-pressure cylinder. What 
steam there was in the low expanded to the point C when the exhaust valve 
opened and the pressure dropped to the back pressure and the steam was 
exhausted at nearly constant back pressure to D', when the exhaust valve closed 
and the steam trapped in the cylinder was compressed to A', at which point 
steam was again admitted and the cycle repeated. 

In set B the cycle of operation is exactly the same as in set A. In set C 
the cycle is the same as in A, but there are one or two points to be especially 
noted, as they are not present in set A. The admission line of the high-pres- 
sure cylinder is not a constant pressure, but rather a falling pressure one, due to 
throttling of the steam, or "wire drawing ," as it is called, through the throttle 
valve or steam ports, due to the higher speed at which this card was taken. 
It will also be noticed that the compression pressure is higher in this case, due 
to earlier closing of the exhaust valve, which becomes necessary with the type of 
valve gear used, as the cut-off is made earlier. In the low-pressure card it will 
be seen that the compression pressure is greater than the admission pressure 
and hence there is a pressure drop instead of rise on admission. In set D the 
peculiarities of C are still more apparent, the compression in high-pressure 
cylinder being equal to admission pressure and above it in the low-pressure 
cylinder. The wire drawing is also more marked, as the speed was still 
higher when this set of cards was taken. 

In Fig. 64, one set of the cards of Fig. 63 is redrawn on cross-section 
paper and then combined. Cards taken from the different cylinders of a 
multiple-expansion engine will in nearly all cases have the same length, the great- 
est that can be conveniently handled by the indicator, and will be to two different 
pressures scales, in as much as that indicator spring will be chosen for each 
cylinder which will give the greatest height of card consistent with safety to 
the instrument. To properly compare the cards they must be reduced to the 
same pressure scale, and also to the same volume scale. As the lengths 
represent volumes, the ratio of the two volume scales will be as that of the 
cylinder volumes or diameters squared. Hence, the length of the high-pressure 
card must be decreased in this ratio or the low increased. As a rule it is found 
more convenient to employ the former method. When the cards have been 
reduced to a proper scale of pressures and volumes the clearance must be 
added to each in order that the true volume of the fluid may be shown. 
The cards may now be placed with these atmospheric lines and zero volume 
lines coinciding and will then appear in their true relation. In this case the 
cylinder ratio was 1.65, the indicator springs 100 lbs. and 70 lbs. respectively 
and clearance 5 per cent in each cylinder. 

The steps in combining the cards were as follows : The zero volume lines were 
first drawn perpendicular to the atmospheric line and at a distance from the end of 



232 



ENGINEEKING THERMODYNAMICS 



the card equal to the length of the card times the clearance. PV axes were laid 
off and a line drawn parallel to the zero-pressure line at a distance above it equal to 
14.7 lbs. to scale of combined diagram. This scale was taken to be that of the 
high-pressure diagram. A number of points A'B'C, etc., were then chosen 
on the low-pressure card, and the corresponding points a'b'c', etc., plotted by 
making the distances of a',b', etc., from the zero- volume line equal to thoce 
of A' ,B' ', etc., and the distances of the new points above the atmosphere .7 
the distances of the original. By joining the points as plotted, the new diagram 
for the low-pressure card was formed. The high-pressure card was then redrawn 







r 




R 












































A] 




p 




















D^ 






































E^ 














G y 






a 




b 


"cV 
























_Fx 








9 










d 




























a (x 










e 






















^ 






e' 


f 












_6^ 




E 




?' 












c' 










d! 










c'^) 




















i— — ■ 












■J 






















V 


s 


rV 










u> 


s 









































Fig. 64. — Diagram to Show Method of Combining the High- and Low-pressure Cylinder 
Indicator Cards of the No-receiver Compound Engine. 



by taking a number of points A, B } C, etc., and plotting new points a, b, c, 

etc., so that the distances of a, b, c, etc., from the zero-volume line were — — the 

distances of A, B, C, etc., while the distances of new points above the atmos- 
pheric line were the same as for the original points. 

In Fig. 65 are shown two cards from a compound steam engine with receiver. 
Diagram A shows the cards as taken, but transferred to cross-section paper for 
ease in combining, and with the zero-volume axis added. On the high-pressure 
card admission occurred practically at constant volume, piston being at rest 
at dead center, at A, bringing the pressures in the cylinder up to the initial 
pressure at B. Admission continued from B to C at nearly constant pressure, 
the piston moving slowly with correspondingly small demand for steam and 
consequently little wire drawing. From C to D the piston is moving more 
rapidly and there is in consequence more wire drawing, admission being no 
longer at constant pressure. At D the steam valve closes and expansion occurs, 
to E, where release occurs, the pressure falling to that in the receiver. From 
F to G exhaust occurs with increase of pressure due to the steam being forced 
into the receiver, (receiver + decreasing H.P. cyl.vol.) while from G to H the 



WORK OF PISTON ENGINES 



233 



pressure falls, due to the low-pressure cylinder taking steam from the receiver 
and consequently volume of receiver, (receiver + increasing L.P. cyl.vol. + 
decreasing H.P. cyl.vol.) increasing. At H exhaust closes, a very slight com- 
pression occurring from H to A . 

On the low-pressure card, admission occurred at A' and continued to B' at 
constant volume, the piston being on dead center as from A to B in high-pressure 
cylinder. From B' to C" admission occurred with falling pressure due to increase 
in receiver volume, (receiver + increasing L.P. cyl.vol.), and from C to D' 
admission still took place, but with less rapidly falling pressure, as high-pres- 
sure cylinder is now exhausting and receiver volume, (receiver + increasing 
L.P. cyl.vol. +decreasing H.P. cyl.vol.) was receiving some steam as well as 





















i 
p 


































}.-> 


i 


























H.P. 


Clear; 
( 


mce-> 


1 




6\ 


N 


I 














F-, 




















\ 


i 














F 








G 




A 

—tL/ 










V 
































a 


9 



























B 

"1 


j 




EHr-- 


4f 


c' 












f£ 












a 4 


i 

i 




v 










e' 














L.P. 


Clear 


nice f 


L. 


-» 


r^ 






® 


















® 


























i 














V 



Fig. 65.— Indicator Cards from a Compound Engine with a Receiver, as Taken and as 

Combined. 

delivering. At D' admission ceased and expansion took place to E' where 
release occurred, the pressure falling to the back pressure and continuing from 
F' to G', where the exhaust valve closed and compression took place to A' 
thus completing the cycle. At H' leakage past the exhaust valve was so great 
as to cause the curve to fall off considerably from H f to A', instead of con- 
tinuing to be a true compression curve, ending at /, as it should have done. 
The combined diagrams are shown in B. 

In Fig. 66 are shown a set of three cards from a triple^expansion pumping 
engine with large receivers and cranks at 12Q°. In diagram A the cards are 
shown with the same length and with different pressure scales as taken, but 
with the zero volume line added and transferred to cross-section paper. On 
the high-pressure card admission occurred at A, causing a constant volume 
pressure rise to B, the piston being at rest with the crank at dead center. From 
B to C admission occurred at nearly constant pressure to C, where steam was 



234 



ENGINEERING THERMODYNAMICS 



cut off and expansion took place to D. At this point release occurred, the 
pressure dropping at constant volume to E with the piston at rest. From E 
exhaust took place with slightly increasing pressure, since the intermediate 
cylinder was taking no steam, the intermediate piston being beyond the point 
of cut-off. The pressure rise is slight, however, due to the size of the receiver, 
which is large compared to the cylinder. At two-thirds of the exhaust stroke, 
point F, the back pressure became constant and then decreased, for at this 
point the speed of the intermediate piston increased and the receiver pressure 
fell. At G exhaust closed and a slight pressure rise occurred to A, due to the 
restricted passage of the closing exhaust valve. On the intermediate card 



















































C 














p 


I, 






















B 




















y 
































































A 
















D 


























G 




F 












E 


























X 










































B' ' 




- C 






































A' 




















a\ 






















L 














Atn 


/' / 






d 
e 


















l(J 
















V 
Atm 






\c> 




















f: 






( 












a' 






































D" 










^d' 


c" 














\ 
















e" 




,a" , 
















d" 

II 

e 




G" 














\g" 





















® ® \ 

Fig. 66. — Indicator Cards from a Triple-expansion Engine with Receiver as Taken and 

Combined. 



admission occurred at A', the pressure rising to B' '. From B' the admission 
was at nearly constant pressure to X while the piston speed was low and then 
at a falling pressure to C". Pressure was falling, since the steam was supplied 
from a finite receiver into which no steam was flowing during intermediate 
admission. At C cut-off occurred and steam expanded to D f , where release 
took place, and the steam was exhausted. As in the case of the high-pressure 
cylinder the back pressure was rising for two-thirds of the stroke, since the 
steam was being compressed into the receiver or rather into a volume made 
up of receiver and intermediate cylinder volume, which is, of course, a decreas- 
ing one, since the cylinder volume is decreasing. At two-thirds of the stroke 
the low-pressure cylinder begins to take steam and the receiver volume is 



WORK OF PISTON ENGINES 235 

now increased, inasmuch as it was made up of the receiver portion of the inter- 
mediate cylinder and a portion of the low-pressure cylinder, and the low- 
pressure cylinder volume increased faster than intermediate decreased for the 
same amount of piston travel. At G' exhaust closed and a slight compression 
occurred to A f , thus completing the cycle. 

On the low-pressure card admission occurred at A" and the pressure rose at 
constant volume to B" , and then admission continued first at constant pressure 
and then falling, as in the intermediate cylinder, to the point of cut-off 
at C" . From here expansion took place to D" '. At this point the exhaust 
valve opened, the pressure fell nearly to back pressure at E'\ and the steam 
was exhausted at practically constant back pressure to G", where the exhaust 
valve closed and there was compression to A" , thus completing the cycle. 
The combined diagram is shown in B. 

Prob. 1. In Fig. 67 are shown six sets of indicator cards from compound engines. 
The cylinder sizes and clearances are given below. Explain the cylinder events and the 
shape of lines for each card and form a combined diagram for each set. 

No. 1. From a four- valve Corliss engine, 26x48 ins., with 3 per cent clearance in 
each cylinder. 

No. 2. From a single-valve engine, 12x20x12 ins., with 33 per cent clearance in 
high-pressure cylinder and 9 per cent in low. 

No. 3. From a four-valve Corliss engine 22x44x60 ins., with 2 per cent clearance 
in the high-pressure cylinder and 6 per cent in low. 

No. 4. From a single-valve engine 18 X30 Xl6 ins., with 30 per cent clearance in the 
high-pressure cylinder and 8 per cent in the low. 

No. 5. From a single- valve engine 11|X 18^X13 ins., with 7 per cent clearance in 
the high and 10 per cent in the low. 

No. 6. From a double-valve engine 14x28x24 ins., with 3.5 per cent clearance in 
the high-pressure cjdinder and 6.5 per cent in the low. 

Prob. 2. In Fig. 68 are shown four sets of indicator cards from triple-expansion 
marine engines. The cylinder sizes and clearances are given below. Explain the cylinder 
events and the shape of the lines of each card and form a combined diagram of each set. 

No. 1. From the engine of a steam-ship, cylinders 21.9x34x57 ins. X39 ins. with 
6 per cent clearance in each and fitted with simple slide valves. 

No. 2. From an engine 20x30x50x48 ins. 

No. 3. From the engine of a steam-ship with cylinders 22x35x58x42 ins., 
assume clearance 7 per cent in each cylinder. 

No. 4. From the engine of the steamer " Aberdeen, " with cylinders 30 X45 X70 X54 
ins., and with 4 per cent clearance in the high, 7 per cent in the intermediate, and 8 per 
cent in the low. 

Prob. 3. In Fig. 69 are shown some combined cards from compound engines. 
Explain the cylinder events and the shape of the lines and reproduce the indicator cards. 

Prob. 4. In Fig. 70 are ahown some combined cards from triple-expansion engines. 
Draw the individual cards and explain the cylinder events and shape of lines. 

8. Standard Reference Cycles or PV Diagrams for the Work of Expansive 
Fluids in Two-cylinder Compound Engines. The possible combinations of 
admission with all degrees of expansion for forward strokes and of exhaust with 



236 



ENGINEEEING THEKMODYNAMICS 



all degrees of compression for back strokes, with and without clearance, in 
each of the two cylinders of the compound engine, that may have any volume 
relation one to the other and any size of receiver between, and finally, any sort 
of periodicity of receiver receipt and discharge of fluid, all make possible a 

rioo 




Fig. 67. — Six Sets of Compound Engine Indicator Cards. 

very large number of cycles. In order that analysis of these conditions of work- 
ing may be kept within reasonable space, it is necessary to proceed as was 
done with compressors and simple engines, concentrating attention on such 
type forms as yield readily to analytical treatment and for which the formulas 



WORK OF PISTON ENGINES 



237 



are simple even if only approximate with respect to actual engines, but, of 
course, keeping in mind the possible value of the formulas, as those that teach 




Fig. 68. — Four Sets of Triple-expansion Engine Indicator Cards. 

no principles or fail to assist in solving problems must be discarded as useless. 
The work that fluids under pressure can do by losing that pressure is no 



238 



ENGINEEEING THERMODYNAMICS 



different in compound than in simple engines, if the fluid has a chance to do 
what it can. Provided the structure is such as will not interfere with the com- 
pleteness of the expansion, and no fluid is wasted in filling dead spaces without 



200 



100- 




Fig. 69. — Combined Diagrams of Compound Engines. 



working, then the work per cubic foot or per pound of fluid is the same for simple, 
compound and triple engines. Furthermore, there is a horse-power equiv- 
alence between the simple and compound, if, in the latter case the steam 



WORK OF PISTON ENGINES 



239 



admitted up to cut-off may be assumed to be acting only in the low-pressure 
cylinder, that is, ignoring the high-pressure cylinder except as it serves as a 




No. 1 



100 



50 




200- 

150 

100- 

50 




100 



50 



No. 3 




No. 4 




No. 5 



Fig. 70. — Combined Diagrams of Triple-expansion Engines. 

cut-off means or meter. This should be clear from a comparison of Figs. 71, 
A and B. In Fig. 71 A, representing the case of the simple engine without 



240 



ENGINEERING THERMODYNAMICS 



clearance and with complete expansion, the volume admitted, AB, expands 
to the back pressure on reaching the full cylinder volume DC, and exhausts 
at constant back pressure, the work represented by the area ABCD. It should 
be clear that no difference will result in the work done if a line be drawn across 
the work area as in Fig. 715, all work done above the line HG to be developed 
in the high-pressure cylinder and that below in the low. This is merely equiv- 
alent to saying that a volume of steam A B is admitted to the high-pressure 
cylinder expands completely to the pressure at G on reaching the full high- 
pressure cylinder volume, after which it exhausts at constant pressure (into 
a receiver of infinite capacity), this same amount being subsequently admitted 
without change of pressure to the low-pressure cylinder, when it again expands 
completely. Thus, it appears that the working of steam or compressed air 



p 


if 


Vol. 


A.dmi 


tted to Cyl 


inder 






P 


^ 


I 

Vol.Admi 


1 
;tedtoH.P.Cyl 


nder 




A 




ti 
















A 


] 




















































\ 












































A 
















B 










\ 


















\ 
















\ 






















olume Adr 


iitte( 


Itol 


.P. Cj 


d. 




C 




















n 


<Z — c- 


^V 


H.P. 


Displ 


icem 


mi 










































p 
















C 


D 
















C 




























■ ^ 






I 


Mspla 


ceme 


nt Vc 






" 




L.P.J 


JlSpiirl 









Fig. 71. — Diagram to Show Equality of Work for Expansion in One-cylinder Simple and in 
Two-cylinder Compound Engines for the Same Rate of Expansion. 

in two successive cylinders instead of one will in no way change the maximum 

amount of work a cubic foot supplied can do, the compounding merely making 

it easier to get this maximum. In simple engine cases, Fig. 71 A, the cut-off in per 

AB 
cent of stroke is 100 Xj^, which is a very small value, leaving but little time to 

open and close the admission valve, whereas in the compound case the per cent 

AB 
cut-off in the high-pressure cylinder is 100 X"^, and in the low-pressure cylinder, 

jjri 

100 X yr^, which are quite large enough ratios to be easily managed with ordinary 

valve gears. 

Compounding does, however, introduce possibilities of loss not present in 
the single-stage expansion, if the dimensions or adjustments are not right, 
which may be classed somewhat improperly as receiver losses, and these are 






WORK OF PISTON ENGINES 



241 




Diagram to Show Correct Low-pressure 
Cut-off for No Receiver Loss. 



of two kinds, one due to incomplete 
i expansion in the high and the other 

to over-expansion there. Thus, in 

Fig. 72, if ABCEFGDA represent a 

combined compound diagram for 
i the case of complete expansion in 

the high-pressure cylinder continued 

in the low without interruption but 

incomplete there, DC represents the 

volume in the low-pressure cylinder 

at cut-off, and at the same time the 

total high-pressure cylinder volume. 
If now, the low-pressure cut-off 

be made to occur later, Fig. 73, 

then the volume that the steam 

would occupy when expansion began F 79 

in the low-pressure cylinder is rep- 
resented by D'C This adjust- 
ment could not, of course, change the high-pressure total volume DC, so 

that at release in the high-pressure 
cylinder the pressure would drop 
abruptly to such a value in the 
receiver as corresponds to filling 
the low pressure up to its cut-off, 
and work be lost equal to area 
CC'C". 

A shortening of the low-pressure 
cut-off will have an equally bad 
effect by introducing negative work 
as indicated in Fig. 74, in which 
the L.P. cut-off volume is reduced 
from DC to D'C- Expansion in the 
high pressure proceeds as before till 
the end of the stroke, at which time 
it has a volume DC greater than 
the low pressure can receive D'C, 
hence the receiver pressure must 
rise to such a value as will reduce 
the volume the required amount, 
introducing the negative work CC'C" . 
Changes of low-pressure cut-off may, 
therefore, introduce negative work, 























A 




B 








































\ 




















\ 




















\ 




















\ 


\ 




















\ 














D 






\ 


c 




















A 




























w 


k 




















L 








D' 








c" 






















E 


G 


















F 






























D 2 








^ 























Fig. 73.-Diagram to Show Effect of Lengthening change the receiver pressure and, 



L.P. Cut-off Introducing a Receiver Loss Due to 
Incomplete High-pressure Cylinder Expansion. 



of course, modifiy the distribution 
of work between high and low, but 



242 



ENGINEERING THERMODYNAMICS 






these same effects might also have resulted from changes of high-pressure 
cut-off or of cylinder ratio. 

For such conditions as have been assumed it seems that compounding 
does not increase the work capacity of fluids, but may make it easier to realize 
this capacity, introducing at the same time certain rather rigid relations between 
cut-offs and cylinder volumes as necessary conditions to its attainment. It 
can also be shown that the same proposition is true when there are clearance and 
compression, that is, in real cylinders and when the receiver is real or not 
infinite in size, or when the exhaust of high and admission of low, are not con- 
stant-pressure lines. The former 
needs no direct proof, as inspec- 
tion of previous diagrams makes 
it clear, but the latter requires 
some discussion. 

A real receiver of finite size 
is at times in communication 
with the high-pressure cylinder 
during its exhaust, and at other 
times with the low-pressure 
cylinder during admission, and 
these two events may take place 
at entirely independent times, be 
simultaneous as to time, or over- 
lap in all sorts of ways. Suppose 
the periods to be independent 
and there be no cylinder clear- 
ance, then at the beginning of 
high-pressure exhaust two sepa- 
rate volumes of fluid come 
together, the contents of both 
the high-pressure cylinder and the 
receiver, and this double volume 
is compressed by the H.P. piston 
into the receiver, in which case the 
high-pressure exhaust would take 
place with rising pressure . Follow- 
ing this will come low-pressure admission, during which the volume of fluid in the 
receiver expands into the low-pressure cylinder up to its cut-off, and if the same 
volume is thus taken out of the receiver as entered it previously, low-pressure 
admission will take place with falling pressure, the line representing it exactly 
coinciding with that for the high-pressure exhaust. Independence of H.P. 
cylinder exhaust and L.P. cylinder admission, as to time, may result in a cycle 
such as is represented in Fig. 75 for the case of no cylinder clearance. On this 
diagram the receiver line is DC, an expansion or compression line referred to a 
second axis of volumes KJ, placed away from the axis of purely cylinder volumes 




Fig. 74. — Diagram to Show Effect of Shortening L.P. 
Cut-off, Introducing a Receiver Loss Due to 
Over-expansion in the High-pressure Cylinder. 



WORK OF PISTON ENGINES 



243 



by the distance LD, equal to the receiver volume to scale. All diagram points 
are referred to the axis A I except those on the line DC. 

This same case of time independence of H.P. exhaust and L.P. admission 
yields quite a different diagram when the cylinder clearance is considered. 
Such a case is represented by the diagram, Fig. 76, which also serves to illustrate 
the effect of incomplete expansion and compression as to equalization of 
receiver with cylinder pressures. At high-pressue release the volume of fluid 
in the H.P. cylinder is ML and its pressure is LR. This is about to come 
into communication with the receiver volume ON from which the low-pres- 
sure cylinder finished taking fluid and which is, therefore, at the same pressure 



p 


K'r 


3 

> 








u 


A 




B 




















1 


S 

a 








IS 






\ 




















o 

> 

O 


is 
d 














\ 




















CD 

3 


a 








2 £ 

'R «g 






\ 


















L 


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JO 

o 











P5 


^ 




\ 


\ 




































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_Re< 


elver. 


-Volui 


ne . 












\ 
















M 




















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G 




















F 




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I 




















H 



Fig. 75. — Diagram to Illustrate Variable Receiver Pressure for the Case of Independent 
High-pressure Exhaust and Low-pressure Admission with Zero Clearance. 

as the L.P. cylinder cut-off KS. The question, therefore, is — what will be 
the pressure at P in both H.P^cylinder and receiver when LM cu.ft. of fluid 
at LR pressure combines with ON cu.ft. at pressure ~KS, and together occupy 
a volume ON+LM. By hypothesis the pressure after mixture is 

(first volume Xits pressure) + (second volume X its pressure) 
sum of volumes 



From this or the graphic construction following, the point P is located. If the 
high-pressure expansion had continued to bring LQ to the receiver pressure 
KS, it would reach it at X. At this hypothetic point there would be a volume 
NX in the H.P. cylinder to add to the volume ON in the receiver at the same 
pressure, resulting in OX cu.ft. This fluid would have a higher pressure at the 
lesser volume of receiver and H.P. cylinder and the value is found by a compres- 



244 



ENGINEEEING THERMODYNAMICS 



sion line through X, XPAT referred to the receiver axis. This same line is also 
the exhaust of the H.P. cylinder from P to A . A similar situation exists at admis- 
sion to the L.P. cylinder as to pressure equalization and location of admission 
line. At the end of the L.P. compression there is in the L.P. cylinder FE cu.ft. 
at pressure EH, to come into communication with the receiver volume CB cu.ft. 
at pressure BG, that at which H.P. exhaust ended. Producing the L.P. com- 
pression line to I, the volume BI is found, which, added to receiver, results 
in no pressure change. An expansion line, referred to the receiver axis through 7, 
fixes the equalized pressure J and locates the L.P. admission line JK, which, it 
must be observed, does not coincide with the H.P. exhaust. 

So far only complete independence of the time of H.P. exhaust and L.P. 
admission have been considered, and it is now desirable to consider the diagram- 













































^ 




l 
1 




\ 


























\ 


^ 






\ 


















r, 










\ x 


\ 




\ 


L 


























B ; 




\ 




\ 


































p\ 



























N 


i 
\ 






<> 

^s^. 


\ 


X 
























\ 








I 


N 


>Q 












Re 


:eiv.er 


Vnln 


me . 


F 


^ 


\ 






1 
I 




























V 






1 

I 


























i 

L 








1 

i 















D G H R S V 

Fig. 76. — Diagram to Illustrate Variable Receiver Pressure for the Case of Independent High- 
pressure Exhaust and Low-pressure Admission with Clearance. 



matic representation of the results of complete coincidence. Such cases occur 
in practice with the ordinary tandem compound stationary steam engine and 
twin-cylinder single-crosshead Vauclain compound steam locomotive. In the 
latter structure both pistons move together, a single valve controlling both 
cylinders, exhaust from high taking place directly into low, and for exactly 
equal coincident time periods. The diameter of the low-pressure cylinder being 
greater than the high, the steam at the moment of release suffers a drop in 
pressure in filling the low-pressure clearance, unless, as rarely happens, the pres- 
sure in the low is raised by compression to be just equal to that at H.P. release. 
After pressure equalization takes place, high-pressure exhaust and low-pres- 
sure admission events are really together a continuation of expansion, the 
volume occupied by the steam at any time being equal to the difference between 



WORK OF PISTON ENGINES 245 

the two cylinder displacements and clearances up to that point of the stroke. 
Before this period of communication, that is, between high-pressure cut-off and 
release, the volume of the expanding fluid is that of the high-pressure displace- 
ment up to that point of the stroke, together with the high-pressure clearance. 
After the period of communication the volume of the expanding fluid is that 
of the low pressure cylinder up to that point of the stroke, together with the low 
pressure clearance, plus the high-pressure displacement not yet swept out, 
and the high-pressure clearance. 

These fluid processes cannot be clearly indicated by a single diagram, 
because a diagram drawn to indicate volumes of fluid will not show the volumes 
in the cylinders without distortion. If there be no clearance, Fig. 77 will 
assist in showing the way in which two forms of diagram for this purpose 
are derived. Referring to Fig. 11 A, the volume AB admitted to the high 
pressure cylinder expands in it until it occupies the whole H.P. cylinder volume 
DC At this point expansion proceeds in low and high together, with decreasing 
volumes in high and increasing in low until the low-pressure cylinder volume is 
attained at E. The line BCE then indicates the pressures and volumes of the 
fluid expanding, but does not clearly show the volume in either cylinder between 
C and E, with the corresponding -pressure. It is certain, however, that when 
the volume in the H.P. cylinder becomes zero the pressure must have fallen 
to a value the same as that in the low when the fluid completely fills it, 
or P f =P e . 

As the high-pressure piston returns, on the exhaust stroke, the low- 
pressure piston advances an equal distance, on its admission stroke, sweep- 
ing through a greater volume than the high pressure, in the ratio of low-pres- 
sure to high-pressure displacements. If at any point of the stroke the volume 
remaining in the high-pressure cylinder be x, and the high- and low-pressure 
displacements be respectively Z>i, and Z>2, then (Di—x) is the volume swept 

out by high-pressure piston, x the volume remaining in it, and -=^-(Di — x) 

the volume swept in by the low-pressure piston. Then the total volume 
still in the two cylinders is, for a point between C and E, 



y=* + g(z> 1 -x)=z) 2 -*(g-i). 

Since the equation of the curve CE is, PV = P C V C = PcDi, the value of V may 
be substituted, giving P D2-X y^-l) = P C P>\, = constant. Dividing by 

~P\ fr~ZTn — x =a new cons ^ an t? so 



D 2 \ 

jj — 1 J this becomes P 



D 2 

— x 



Di ) 



246 



ENGINEERING THERMODYNAMICS 



my 
point on FC will be distant from the new axis LM an amount f -^ jz — )x\ 



that if a new axis LM be laid off on B, GV = I n n ~ ) from the axis GP any 



as the product of this distance and the pressure P, is constant, the curve FC 
is an equilateral hyperbola referred to the axis LM. Therefore Fig. 77B is 







P 

A 






-D 
























P 












L 
















\ 






















A 






t 
























\ 




























x 








1 


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" 


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c 

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of-L-.i-— volumes 

mbi'ned diagra 


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COMBINED DIAGRAM TO ONE SCALE 

Fig. 77. — Diagram to Illustrate the Compound Engine Cycle with No-receiver, and Exact 
Coincidence of H.P. Exhaust and L.P. Admission Periods. No Clearance. 

the pressure-volume representation of the entire cycle of the high-pressure 
cylinder. 

In Fig. 77C is shown the corresponding pressure-volume diagram for the low- 
pressure cylinder, for which it may similarly be shown that the curve DE may be 



WOKK OF PISTON ENGINES 247 

plotted to an axis NO at a distance to the left of the axis GP equal to the same 
quantity, 

GN=~~ D k • • • < 295 > 

JJ2 — L>\ 

These diagrams, 77 A, B and C may be superposed, as in Fig. 77 E, giving 
one form of combined diagram used for this purpose, and the one mo st nearly 
comparable with those already discussed. In this diagr am, t he area ABCFA 
represents the work of the high-pressure cylinder, and DEIHD, the work of 
the low-pressure cylinder. Together, they equal the work of the enclosing 
figure ABEIHA, and hen ce the w ork of the low-pressure cylinder must also be 
represented by the area FCEIHF. 

It is not difficult to show that if a vertic al, CJ, be dropped from the point 
C to the exhaust line HJ, the figure CFHJ, in Fig. 772) is similar to DEIH, 
in Fig. 77E, reversed, but drawn to a different horizontal scale. Here the 
length of the low-pressure diagram is made equal to the length of the high- 
pressure diagram. The two scales of volumes are shown above and below the 
figure. While this appears to be a very convenient diagram, it will be found 
to be less so when clearance and compression are considered. 

It may be noted that since it has been shown that the curves CF and DE 
are of the same mathematical form (hyperbolic) as the expansion line^ CE, 
they may be plotted in the same way after having in any way found the axis. 
The location of this axis may be computed as given above, or may be found 
graphically by the method given in connections with the subject of clearance, 
Chapter I, and shown in Fig. 775. Knowing two points that lie on the curve, 
C and F, the rectangle CDFK is completed. Its diagonal, DK, extended, cuts 
the horizontal axis GV in the point M, which is the base of the desired axis ML. 

If now the axis NO and the figure DEIH, part of which is referred to this 
axis, be reversed about the axis GP, Fig. 77C, NO will coincide with ML, 
Fig 77D and Fig. 78 results. Note that the axis here may be found graphically, 
in a very simple way. Draw the vertical CK from C to the axis GV and 
the horizontal DJ to the vertical IE extended, DC is then the high-pressure 
displacement and DJ the low-pressure displacement. Draw the two diagonals 
DK and JG, extending them to their intersection X. By similar triangles it 
may be shown that a horizontal line, UW, will have an intercept between these 
two lines, JG and DK, equal to the volume of fluid present in the two cylinders 
combined. The intersection X is the point at which this volume would 
become zero if the mechanical process could be continued unchanged to that 
point, and is, therefore, on the desired axis ML extended. T being the inter- 
section of WU with the axis GP, when the volume UT is present in the high- 
pressure cylinder, TW gives the volume in the low-pressure cylinder. 

When clearance and compression are considered, this diagram is changed in 
many respects, and is shown in Fig. 79. The axes OP, OV and OV' are laid out, 
with OZ equal to the clearance and ZK, the displacement, of the high-pressure 



248 



ENGINEERING THERMODYNAMICS 



cylinder, and OQ and QY, clearance and displacement of the low-pressure cylinder 

It is necessary to know high-pressure cut-off, ~=; high-pressure compression 

ZK. 



ler. 



ZE" 
~ZK 



I'Q 

and low-pressure compression, =, in addition to the initial and back 



pressures, in order to lay out the diagram. The drop in pressure CD at release 
is due to the coming together of (volume V c at pressure P c ), with (volume V 3 - 
at pressure P 3 ), If the volume V 3 (measured from axis OP) were decreased 
sufficiently to raise the pressure in the low-pressure clearance to the pressure 




Fig. 78. — Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compound 
Engine at any Point of the Stroke for the Case of No-clearance and Coincidence of 
H.P. Exhaust with L.P. Admission. 

at C, the volume would become V 8f as indicated at the point S, and the volumes 
now combined in the hypothetical condition would occupy the volume SC. 
Increasing this volume to D 7 !), that actually occupied after communication, 
the pressure would fall along the curve SD', which is constructed on KV and KT 
as axes. The pressure of D is then laid out equal to the pressure at D f . To 
find the axis, ML, for the curves DE and D'E', from any convenient point N 
on ZA, draw the line NK extended to X. Extend HG to R, at a height equal 
to that of N, and draw RQX, and through the intersection draw the desired axis 
XML. The fraction of stroke completed at E' in the low pressure at cut-off must 
be equal to that completed at E in the high pressure at compression, and may be 
laid out graphically by projecting up from E to the point U on the line NK and 
horizontally from U to W on the line RQ. Projecting down from W to the 
curve at E' locates the point of effective cut-off in the low-pressure cylinder. 



WORK OF PISTON ENGINES 



249 



After the supply from the high-pressure cylinder has been cut off at E', the 
expansion is that of the volume in the low-pressure cylinder and its clearance, and 
hence the curve E'G is constructed on OP as an axis. 

While in this last case a zero receiver volume has been assumed, there is 
nothing to prevent a receiver volume being interposed between H.P. and L.P. 
so that common expansion takes place with a volume greater than assumed 
by so much as this volume, the. effect of which is to decrease the slope of DE 
and D'E '. Such receivers usually consist of the spaces included in a valve 
body and connecting passages and may be treated generally as increased L.P. 
clearances. 

The most common of all relations between H.P. exhaust and L.P. admis- 
sion is, of course, that of partial coincidence of periods, as it is thus with all cross- 

























w 


P 

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- 


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. 




























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Cu.Ft. 


2 


L.P. pomp? Mt> 


h 




j 

-r > 




2 duV 3 




A 








U- 




U-L. 

T 


P. Cut Off due 


toE 


.P. C 


omp 


-4 H.P. k> 


>^H.P. CutOff j v 

iH.p.i^ I j 


\ 












i 

1 

— |— 




















>l 




Disr 


lacei 


nenl 




' 


\ 










1 


































\ 


\ 
V 


\ i 














































% 







Fig. 79. — Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compound 
Engine at any Point of the Stroke for the Case of Clearance and Coincidence of H.P. 
Exhaust with L.P. Admission. 



compound and triple-expansion engines having separate cranks for the 
individual cylinders. For these there is no simple fixed relation between the 
periods, for, while crank angles are generally fixed in some comparatively simple 
relation, such as 90°, 180° and 270° for compounds and 120° for triples, they 
are sometimes set at all sorts of odd angles for better balance or for better 
turning effort. Even if the angles were known the receiver line would have 
to be calculated point by point. When the H.P. cylinder begins to discharge into 
a receiver for, say, a cross compound with cranks at 90°, steam is compressed 
into the receiver, and so far the action is the same as already considered for 
independence of periods, but at near mid-stroke the low-pressure admission 
opens while high-pressure exhaust continues. This will cause the receiver 
pressure to stop rising and probably to fall until the low pressure cuts off, which 



250 



ENGINEERING THERMODYNAMICS 



may occur before the H.P. exhaust into the receiver ceases. If it does, the receiver 
pressure will again rise. Exact determination of such complex receiver lines is 
not often wanted, and when needed is best obtained graphically point by point. 
Its value lies principally in fixing exactly the work distribution between cylin- 
ders, which is not of great importance except for engines that are to work at 
constant load nearly all the time, such as is the case with city water works 
pumping, and marine engines. While equations could be derived for these 
cases, they are not worth the trouble of derivation, because they are too cumber- 
some, and graphic methods are to be substituted or an approximation to be made. 
Four kinds of approximation are available, as follows, all of which ignore 
partial coincidence of periods: 

1. Receiver pressure constant at some mean value and clearance ignored. 

2. Receiver pressure constant at some constant value and clearance con- 
sidered with compression zero or complete. 

3. Receiver pressure fluctuates between fixed limits as determined by an 
assumed size, clearance ignored. 

4. Receiver pressure fluctuates between fixed limits as determined by an 
assumed size, clearance considered, with compression zero or complete. 

These are not all of equal difficulty in solution, and the one to be used is 
that nearest the truth as to representation of conditions, which is usually the 
most difficult, provided time permits or the information is worth the trouble. 
Quickest work is accomplished with assumption (1) and as this is most often 
used in practical work it indicates that its results are near enough for most 
purposes. 

This discussion leads, therefore, to the analytical study of the following cycles: 

Infinite Receiver, Zero Cylinder Clearance. 
CYCLES V, AND VI (Fig. 80). 

(a) Admission at constant supply pressure to H.P. cylinder. 

(b) Expansion in H.P. cyl. (may be zero) by law PV =c for (V); 
PV S =c for (VI)." 

(c) Equalization of H.P. cyl. pressure with receiver pressure 
at constant volume (may be zero). 

(d) Exhaust into infinite receiver at constant pressure from 
H.P. cylinder. 

(e) Equalization of H.P. cylinder pressure with supply pressure 
at constant zero volume. 

' (/) Admission from receiver at constant receiver pressure to 

L.P. cylinder. 
(g) Expansion in L.P cylinder (may be zero) by law PV=c for 

(V); PV s =c for (VI). 
(h) Equalization of L.P. cylinder pressure with back pressure 

at constant volume (may be zero) . 
(i) Exhaust at constant back pressure for L.P. cylinder. 
(j) Equalization of L.P. cylinder pressure with receiver pressure 

at constant zero volume. 



H.P. Cylinder 

Events 



L.P. Cylinder 
Events 



WORK OF PISTON ENGINES 



251 































































p 






















P 




































a 
































a \ 






















\ 




































\ Cycles VII & VIII 








\ 


Cycles V & VI 
Cycle V PV = C Cycle VI PV*=C 














e 




Acycle VII PV=C Cycle VIII PVs=C 








\ 


















X 




















c 




I 


VT 




























































\ 


































d 


























c 


















-* — 


Rec 


jivei; Vol 


line 










c 


\ 

\ 




















d 










































s 




















/ 






























i 
























./' 














u 




































9 




















I 








h 




























i 








h 






















































































V 




































V 






























































p 
































































a 




















P 
















(i 


























\ 


Cycles IX & X 
Cycle IX PV=C Cycle X PVs=C 














f 




\ Cycles XIII & XIV 








\ 


















\Cycle XIII PV=C Cycle XIV PV S =C 








\ 
































( 




t 




















d 






r 


































\ 
























\ 


































\ 


















\ 






■J 


\ 


















Receivei 


Vol 


um« 


£* 


r 


N> 


*4 


c 




















V 


> 


T 
































; 


^ 























y 


/ 




^c 




N 


■.^ 






























































■^-ii 






















v* 












,\ 




















/ 








c 




















\ 


^ 














i 
























































j 


























V 




































V 






























































p 






















P 








































a 






















a 






































\ 


Cycles XI & XII 
Cycle XI PV=C Cycle XII PVs=C 


/ 


• 


\ 






Cycles XV & XVI 
Cycle XV PV=C Cycle XVI PVS=C 














\ 






\ 














i 




\ 
























A 






































r 


















1 






\ 




































\ 


















\ 


e 




\ 
































; 




c 




















I 




c 


































d 






















v 


\ 


tf. 


































9 






















k \ 


/ 


d 
































l 








ft^ 














\ 












r 


- 






















\ 


k 






j 








i 








\ 


K 








i 






7i 













































































Fig. 



V V 

-Compound Engine Standard Reference Cycles or PV Diagrams. 



252 



ENGINEERING THERMODYNAMICS 



Relations 

between H.P. 

and L.P. Cylinder 

Events 



(1) H.P. exhaust and L.P. admission independent as to time, 

coincident as to representation (except as to length). 

(2) H.P. expansion line produced coincides as to representation 

with L.P. expansion line. 

(3) The length of the constant pressure receiver line up to the 

H.P. expansion line produced is equal to the length of 
the L.P, admission line. 



Finite Receiver, Zero Cylinder Clearance. 
CYCLES VII, AND VIII, (Fig. 80). 
(a) Admission at constant supply pressure to H.P. cylinder. 



H.P. Cylinder 
Events 



L.P. Cylinder 

Events 



(c) 



id) 



(e) 



(I) 



(9) 



(i) 



Expansion in H.P. cylinder (may be zero) by law PV =c 

for (VII); PV s =c for (VIII). 
Equalization of H.P. cylinder pressure with receiver pressure 

at constant volume (may be zero) with a change of receiver 

pressure toward that at H.P. cy Under release (may be 

zero). ► 

Exhaust into finite receiver from H.P. cylinder at rising 

pressure equivalent to compression of fluid in H.P. cylinder 

and receiver into receiver by law PV = c for (VII) and 

PV s =c for (VIII). 
Equalization of H.P. cylinder pressure with supply pressure 

at constant zero volume. 



Admission from receiver to L.P. cylinder at falling pressure 

equivalent to expansion of fluid in receiver into receiver 

and L.P. cylinder together by law PV=c for (VII), PV S ■■ 

for (VIII). 
Expansion in L.P. cylinder (may be zero) by law PV=c 

for (VII); PV s =c for (VIII). 
(h) Equalization of L.P. cylinder pressure with back pressure 

at constant volume (may be zero). 
Exhaust at constant back pressure for L.P. cylinder. 
Equalization of L.P. cylinder pressure with receiver pressure 

at constant zero volume to value resulting from H.P. 

exhaust. 



Relation 

between h.p. and 

L.P. Cylinder 

Events 



r (1) H.P. exhaust and L.P. admission independent as to time, 
coincident as to representation, except as to length. 

(2) H.P. expansion line produced coincides as to representation 

with L.P. expansion line. 

(3) The length of the receiver pressure line up to the H.P. 

expansion line produced is equal to the length of the 
L.P. admission line. 



WOKK OF PISTON ENGINES 



253 



No Receiver, Zero Cylinder Clearance. 
CYCLES IX, AND X, (Fig. 80). 



H.P. Cylinder 

Events 



Both H.P. and L.P. 

Simultaneously 

H.P. Cylinder 

Events 



L.P. Cylinder 

Events 



(a) Admission at constant supply pressure to H.P. Cylinder. 

(b) Expansion in H.P. cylinder (may be zero) by law PV =c 

for (IX); PV s =c for (X). 

(c) Transference of fluid from H.P.to L.P. cylinder with simul- 

taneous continuation of expansion until all fluid is in 
L.P. cylinder and expanded to its full volume by law 
PV=c for (IX); PF s =cfor(X). 

(d) Equalization of H.P. cylinder pressure to the pressure of 

supply. 

(e) Equalization of L.P. cylinder pressure with back pressure 

at constant volume (may be zero). 
(/) Exhaust at constant back pressure for L.P. cylinder. 
(g) Equalization of L.P. cylinder pressure to the pressure in 

H.P. cylinder at the end of its expansion. 



Infinite Receiver, with Cylinder Clearance. 



CYCLES XI, AND XII, (Fig. 80). 



H.P. Cylinder 
Events 



L.P. Cylinder 
Events 



(a) 
(b) 

(c) 

(d) 

(e) 

(/) 

\{g) 
(h) 

(0 

(i) 
(*) 

(l) 



Admission at constant supply pressure to H.P. cylinder. 
Expansion in H.P. cylinder (may be zero) by law PV=c 

for (XI); PV s =c for (XII). 
Equalization of H.P. cylinder pressure with receiver pressure 

at constant volume (may be zero) pressure. 
Exhaust into infinite receiver at constant pressure from H.P. 

cylinder. 
Compression in H.P. cylinder to clearance volume (may be 

zero) by law PV =c for (XI) ; PV S =c for (XII). 
Equalization of H.P. cylinder pressure with supply pressure 

at constant clearance volume, may be zero. 



receiver at constant-receiver pressure to 



Admission from 
L.P. cylinder. 

Expansion in L.P. cylinder (may be zero) by law PV =c 
for (XI); PV s =c for (XII). 

Equalization of L.P. cylinder pressure with back pressure 
at constant volume (may be zero) . 

Exhaust at constant back pressure for L.P. cylinder. 

Compression in L.P. cylinder to clearance volume by law 
PV=c for (XI) ; PV S =c for (XII) (may be zero). 

Equalization of L.P. cylinder pressure with receiver pres- 
sure at constant clearance volume without change of 
receiver pressure (may be zero). 



254 



Relations 

between h.p. and 

L.P. Cylinder 

Events 



ENGINEERING THERMODYNAMICS 






' (1) H.P. exhaust and L.P. admission independent as to time, 
coincident as to representation except as to length. 

(2) L.P. expansion line does not coincide as to representation 

with H.P. expansion line produced by reason of clearance 
influence except in one special and unusual case. 

(3) The length of the constant-receiver pressure line intercepted 

between H.P. compression line and H.P. expansion line 
produced is equal to the same intercept between L.P. 
expansion line and L.P. compression line produced. This 
is equivalent to the condition that the volume taken in by 
low is equal to expelled by the high reduced to the same 
pressure. 



Finite Receiver, with Cylinder Clearance. 



CYCLES XIII, AND XIV, (Fig. 80). 



H.P. Cylinder 

Events 



L.P. Cylinder 
Events 



' (a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV=c 
for (XIII) ; PV S =c for (XIV). 

(c) Equalization of H.P. cylinder pressure with receiver pres- 

sure at constant volume (may be zero) toward that at 
H.P. cylinder release (may be zero). 

(d) Exhaust into finite receiver from H.P. cylinder at rising 

pressure equivalent to compression of fluid in H.P. cylinder 
and receiver into receiver by law PV = c for (XIII) ; PV S =c 
for (XIV). 

(e) Compression in H.P. cylinder to clearance volume (may be 

zero) by law PV=c for (XIII) ; PV S =c for (XIV). 
(/) Equalization of H.P. cylinder pressure with supply pressure 
at constant clearance volume. 

(g) Admission from receiver to L.P. cylinder at falling pressure 
equivalent to expansion of fluid in receiver into receiver 
and L.P. cylinder together by PV = c for (XIII); PV S =c 
for (XIV). 

(h) Expansion in L.P. cylinder (may be zero) by law PV =c 
for (V); PV =cfor(VI). 

{%) Equalization of L.P. cylinder pressure with back pressure 
at constant volume (may be zero). 

(j) Exhaust at constant back pressure for L.P. cylinder. 

(k) Compression in L.P. cylinder to clearance volume by law 
PV=c for (XI); PV s =c for (XII) (may be zero). 

(I) Equalization of L.P. cylinder pressure with receiver pressure 
at constant clearance volume with change of receiver 
pressure in direction of L.P. compression pressure (may 
be zero). 



WORK OF PISTON ENGINES 



255 



Relations 

between h.p. and 

L.P. Cylinder 

Events 



(1) H.P. exhaust and L.P. admission independent as to time, 

representation and length. 

(2) L.P. expansion line does not coincide as to representation 

with H.P. expansion line produced by reason of clearance 
influence except in one special and unusual case. 

(3) The high-pressure exhaust and low-pressure admission lines 

do not coincide as to representation by reason of clearance 
influences. 

(4) There is a relation between the lengths of the L.P. admission 

and H.P. exhaust lines, but not a simple one. 



No Receiver, with Cylinder Clearance. 
CYCLES XV, AND XVI, (Fig. 80). 



H.P. 

Cylinder 

Events 



LP. 
Cylinder 

Events 



(a) Admission at constant-supply pressure to H.P. cylinder. 

(b) Expansion in H.P. cylinder (may be zero) according to law 

PV=c for (XV); PV S =c for (XVI). 

(c) Equalization of pressures in H.P. cylinder after expansion 

with that in L.P. after compression at constant volume 
(may be zero). 
Also f (d) Transference of fluid from H.P. to L.P. cylinder until all 
L.P. \ fluid is in L.P. cylinder and expanded to its full volume 

Event [ by same law as (b). 

(e) Compression in H.P. cylinder to clearance volume (may be 

zero) by law PV =c for (XV) ; PV S =-c for (XVI). 
(/) Equalization of pressure in H.P. cylinder with supply at 
constant-clearance volume (may be zero). 

(g) Expansion in L.P. cylinder may be zero by law PV=c for 

(XV); PV s =c for (XVI). 
(h) Equalization of pressure in L.P. cylinder with back pressure, 

at constant volume (may be zero). 
(i) Exhaust at constant pressure for L.P. cylinder. 
(j) Compression in L.P. cylinder to clearance, may be zero by 

law PV = c for (XV) ; PV S =c for (XVI). 
(k) Equalization of L.P. cylinder pressure with H.P. cylinder 

pressure. 



Cycle XVII, Fig. 81, for the triple expansion is defined in the same way 
as the corresponding case of compounds Cycle V, with appropriate alterations 
in wording to account for a third or intermediate cylinder between high- and low- 
pressure cylinders and an additional receiver. Thus, high- pressure cylinder- 
exhausts into first, and intermediate cylinder into second receiver: inter- 
mediate cylinder receives its supply from first, and low-pressure cylinder from 
second receiver. This being the case, it is unnecessary to write out the cylin- 
der events, noting their relation to the corresponding compound case. 



256 



ENGINEERING THERMODYNAMICS 



9. Compound Engine with Infinite Receiver, Logarithmic Law. No Clear- 
ance, Cycle V. General Relations between Pressures, Dimensions, and Work. 

It must be understood that the diagrams representing this cycle, Fig. 82, 
indicating f A) incomplete expansion and (B) over-expansion in both cylinders, 

P 

























A 




B 




















a 


\ 






















\ 






















X 


















e 




\ 






















\ 


C 
















E 


d 


c 


D f 


















f 




\ 
























L 














3 








\ 


G 












1 




i 




h 


H 




J 














k 








I 




K 





















m 






M 








n 










L 



























Fig. 81. — Triple-expansion Engine Standard Reference Diagram or PV Cycle for Infinite 

Receiver. 



may just as well stand for over, complete or incomplete expansion in all possible 
combinations in the two cylinders. Applying the principles already derived 
for calculating the work areas, 



High-pressure cylinder work 



W H = PMll+log, y) -P d V d 



(296) 



Low-pressure cylinder work 



l + los, J?) 



(297) 



WORK OF PISTON ENGINES 257 

Total work 

W = P b V i) (l+\og e y^j +P e F«(l+lo & ^ -P d V d -P g V g , . . (298) 

pressure being in pounds per square foot, and volumes in cubic feet. 

In theses expressions the receiver pressure P e = Pa is unknown, but determinate 
as it is a function of initial pressure and certain volumes, giving it the value, 

Pe = Pd = PblT} 

is merely satisfying the condition that the point E at which expansion begins 
in the low-pressure cylinder must lie in the expansion line of the high. Sub- 
stituting this value there results 

W = P b V b (l+\o&^+P tf V i ,(l+\o& l |?) -PbV b y-P g V g 

= P»7»[2+loge^+lo&^-^J-P,7,. . (299) 

This is a perfectly general expression for the work of the fluid expanding to 
any degree in two cylinders in succession when the clearance is zero and receiver 
volume infinite, in terms of initial and back pressures, pounds per square foot, 
the volumes occupied by the fluid in both cylinders at cut-off, and at full 
stroke in cubic feet. Dividing this by the volume of the low-pressure cylinder 
V g gives the mean effective pressure referred to the low-pressure cylinder, 
from which the horse-power may be determined without considering the high- 
pressure cylinder at all. Hence, in the same units as are used for P b and P 0l 

(M.E.P. referred to L.P.) = pX & [2+log e ^ c +log e ^-^1 -P g . (300) 

Vg\_ Vb Ve Vej 

Proceeding as was done for simple engines, the work per cubic feet of fluid 
supplied is found by dividing Eq. (299) by the volume admitted to the high- 
pressure cylinder F?,, whence, 

Work per cu.ft. supplied = pj2+log e ^-hlog e ^-pl-P^. . . . (301) 

L "& Ve V ej Vb 

Also applying the consumption law with respect to horse-power, 

Cu.ft. supplied per hour per I.H.P. = -, J - r - — ; — r =? (302) 

*^ (m.e.p. ref. to low) V g 

Lbs. supplied per hr. per I.H.P. =-, , — F — — -, — r ^di. . . . . (303) 

(m.e.p. ref. to low) V g 



258 



ENGINEERING THERMODYNAMICS 



These last five equations, (299), (300), (301), (302), (303), while character 
istic, are not convenient for general use in their present form, but are ren 



A 


B 


















A 

N 
N 

p M 

r (rel.pr.) L 
(atm.pr.) 

GO>k.pr.) 

■ 

L 






I 










•) 








B 


\ 




• 


\ 
















H.T. \ 


<-Z H 


Ds*j 


\ 
















'D 




! 


\ 


C 




INCC 
(rec-.p 


















D 


E ® 


r -/H 

MPLE 


TEE) 


PANS 


ON 




N 


_JJ H, 




E 










\ 










~\ 








> 
























L.P. >v 

\^F 
























INDICATOR CARDS OF 
EQUAL BASE AND 






















M 










~ U L 










HEIGHT 


























K 




) 


1 










V \ 


G (bk.pr.) 

f (rel.pr.)u 

(atm.pr.) 




A 


! 


i 




n 


a.pr.)- 














\ 




















\ 


v 


<-Zh 




\ 
















H.P. \ 

\ 


Dh > 


E 
























\l 


N 






I 


? (r 


ec.pr. 


®o 


VER E 


XPAN 


ilON 


























;l.pr.) 


H 














D H 




C (r« 
























■ — i 


M 










V 










L.P.\ 












V 


N,^ 




























\ 


\J 








L 


L 













INDICATOR CARDS OF 
EQUAL BASE AND 


L_ 


























J H V 

Fig. 82. — Work of Expansive Fluid in Compound Engine with Infinite Receiver, Zero 
Clearance. Cycle V, Logarithmic Expansion and Cycle VI Exponential Expansion. 



dered so by substituting general symbols for initial and back pressures 
displacement, cut-off, and amount of expansion for each cylinder. 



WORK OF PISTON ENGINES 259 

r .et (in.pr.) = initial or supply pressure, pounds per square inch = ^; 

p 
' (rel.pr.)* = release pressure, in H. P. cylinder pounds per square inch = ^; 

p 
" (rel.pr.)/,= release pressure in L.P. cylinder, pounds persquare inch = -^; 

P P 
" (rec.pr.)= receiver pressure, pounds per square incn = J44 = 144 ' 

• i. P ° 
(bk.pr.) = back pressure, pounds per square mch=^g, 

" R H = ratio of expansion in high-pressure cylinder = y ; 

" R L = ratio of expansion in low-pressure cylinder = y ; 

" # r = ratio of expansion for whole expansion = ^; 

" D H = displacement of high-pressure cylinder = V d = V c ; 

" D L = displacement of low-pressure cylinder =V f =V ; 

" R c = cylinder ratio = YT - ~ ; 

Z H = fraction of displacement completed up to cut-off in high-pressure 
" cylinder, so that Z H D H = V b = -jr- ; 

" Z L = fraction of displacement completed up to cut-off in low-pressure 

cylinder, so that Z L D L = V e — tt- 

Substitution of these general symbols in Eqs. (299), (300), (301), (302), and 
(303) gives another set of five equations in useful form for direct substitu- 
tion of ordinary data as follows : 

Work of cycle 

1) 



= 144D*(in.pr.) [^(2+loge fl*+lo fr Rl) ~^\ ~ ^(bk.pr.)^ (a 
= 144D, { (in.pr.)|(2+log e ^+log e \--^) - (bk.pr.) ) (6) 

= 144D L I ( in -P r -)^;( 2+l0 ^ «h+1o& Rl-j£) ~ (bk.pr.) 1 (c) 



. (304) 



(m.e.p.) lbs. per sq.in. referred to L.P. cyl. 

= (in.pr .)^-(2+log e RH+log e Kl-j£) ~ (bk.pr.) (a) 

- (in.pr.)|(2-Mog e ^+log e ^-^) - (bk.pr.) (6) 



(305) 



260 



ENGINEERING THERMODYNAMICS 



Work per cu.ft. supplied 
= 144 



= 144 



1 



(in.pr.) (Wlog e R H +\og e R L ~) - (bk.pr.)fl c #*l 

-(bk.pr.)g] 

1 



(in.pr.) (2+log e ^-+log e y- 

\ Zjh £l &lKc 



(a) 
(b) 






(306) 



Cu.ft. supplied per hr. per I.H.P. 

13,750 



(m.e.p. ref. to L.P.) R H Rc 



(a) 



13,750 Zh J 

(m.e.p. ref. to L.P.) R c { } J 



(307) 



From this, of course, the weight in pounds supplied per I.H.P. results directly 
from multiplication by the density of the fluid. 

To these characteristic equations for evaluating work, mean pressure, 
economy and consumption in terms of the initial and final pressures and cylin- 
der dimensions there may be added a series defining certain other general rela- 
tions of value in fixing the cycle for given dimensions and initial and final 
pressures, and in predicting dimensions for specified total work to be done and 
its division between high- and low-pressure cylinders. 

Returning to the use of diagram points and translating into the general 
symbols as each expression is derived, there results, 



Receiver pressure = P d =P e = Pb^r- 



(rec.pr.) = (m.pr.)^-^- = (in.pr.)—— 



C^L 



r n Rl 

= (m.pr.)— -=■ 
KcKh 



High-pressure cylinder release pressure = P c = P b - 

/. (rel.pr.)#= (in.pr.)— 
Kh 

= (in.pr.)Zff 

Low-pressure cylinder release pressure = P/=P e y,- 

Vf 

1 



(rel.pr.)i= (in.pr.) 



= (in.pr.) 

(in.pr.) 
Rv 



RcRh 
Z h 



Rc 



(rel.pr.) g 
Rc 



(a) 
(b) 

(a) 
(*>). 

(a) 
(6) 
(c) 
(d) 



(308) 



(309) 



(310) 



WORK OF PISTON ENGINES 



261 



Division of work between cylinders may be made anything for a given load by 
suitably proportioning cylinders, and equations giving the necessary relations 
to be fulfilled can be set down. It is quite common for designers to fix on 
equal division of work for the most commonly recurring or average load or that 
corresponding to some high pressure cut-off or low-pressure terminal pressure, 
generally the latter. Therefore, a general expression for dimensional rela- 
tions to be fulfilled for equal division of work is useful. On the other hand, 
for an engine the dimensions of which are determined, it is often necessary to 
find the work division for the imposed conditions, so that the following equa- 
tions are of value. 



From Eqs. (296) and (297), noting that P d = P e = Pb^, 

v p. 



High-pressure cylinder work 
Low-pressure cylinder work 



ft r.[(i+k»*)-£] 



P>Vt 



1+log.pO- 



(■+<4:) 



l + l0g e R H -^ 

nc 

l+\o ge R L -(^)R c R H 

l+logeyr-jjry- 

, ,. 1 /bk.pr.\fic 



(a) 



(6) 



(311) 



This is a general expression for work division between the cylinders in 
terms of (a) ratio of expansion in each cylinder, initial and back-pressure 
ratio and cylinder ratio, or, in terms of (6) cut-off in each, associated with cylinder 
and pressure ratios. 

This expression Eq. (311) is less frequently used in its general form as above, 
than in special forms in which the work of the two cylinders in made equal, 
or the expression made equal to unity. The conditions thus found for equal 
division of work between cylinders may be expressed either (a) in terms of initial 
and back pressures, release pressure of low-pressure cylinder and ratio of L.P. 
admission volume to H.P. displacement, and cylinder ratio, or (b) cut-off in 
high- and low-pressure cylinders, initial and back pressures and cylinder ratio. 
Still more special conditions giving equality of work may be found (c) when the 



262 ENGINEERING THERMODYNAMICS 

cylinder ratio is made such that equality of work is obtained at all loads by 
equalizing high and low cut-offs. 

(a) To find the first set of conditions, equate Eqs. (296) and (297) from the 
first part of this section, and by simplification there results, 

, Vc v c . V f P 9 

loge v b -vr loge v e -p; 
log \v b v f ) = v e -p/ 



or 



I.Zs-A 7 '"'')- 



V 6 V t 



Introducing the usual symbols and putting in addition 

Low-pressure admission volume V e „ D (rel.pr.)ir 



High-pressure displacement volume V c (rec.pr.) ' 

Therefore, 

r( bk.pr.) 1 -I 

f Lcrei.prOi xA (in.pr.) "1* /010N 

Rc=\e , \ \ x \ (312) 

L (rel.pr.)i J 

This is of value when a given release pressure is to be reached in the low 
pressure cylinder and with a particular value of low-pressure cut-off volume 
as fixed by x in terms of high-pressure cylinder displacement. 

(6) Again for equal division of work, make Eq. (311) equal to unity, 
whence, 

,,, 1 1 ,,, 1 (bk.pr.)ik 



or 



flc 2 (bk.pr.) Z L 1 

Z H (in.pr.) ° Z H Z l 



which may be reduced to the following, solving for R c , 



Rc = 



^(bk^prO 
Z H (in.pr.) 



(313) 



Equal division of work for given initial and back pressure is to be obtained 
by satisfying these complex relations Eq. (313) between the two cut-offs, or their 
equivalent ratios of expansion in connection with a given cylinder ratio, or the 
relation between pressures and volumes in Eq. (312) equally complex. 



WOKK OF PISTON ENGINES 263 

(c) An assumption of equal cut-off in both cylinders gives results which are 
of interest and practical value, although it is a special case. Eq. (313) then 
becomes, when Zh = Zl or Rh = Rl, 



\bk.pr./ 



(314) 



As would be expected, this may also be derived from Eq. (312), since 

- = 7 — , V and x = -, . ' ' ^ under these conditions. 
x (rel.pr.)i, (bk.pr.) 

The receiver pressure under these conditions is constant, and is, from Eq. (308) 

(rec.pr.)=^^0 = ^^) = [ (in . pr0(bk . pr0 ]*. . . . (315 ) 
(.bk.pr.J 

The high-pressure release pressure is not affected by any change in the low- 
pressure cut-off, and hence Eq. (309) gives the value of high-pressure release, 
pressure for the case. Low-pressure release pressure Eq. (310) may be expressed 
for the case of equal cut-off, 



(rel.pr.)^ = ^^^ = Z^[(in.pr.)(bk.pr.)y (a) . 
\bk.pr v 



= ^-[(in.pr.)(bk.pr.)J 



(6) 



(316) 



The foregoing equations up to and including Eq. (311), are perfectly general, 
and take special forms for special conditions, the most important of which is 
that of complete expansion in both cylinders, the equations of condition for which 
are, referring to Fig. 82. 

Pc = Pd ', 

V d =V e ; 

which, when fulfilled, yield the diagram, Fig. 83. These equations of condi- 
tion are equivalent to fixing a relation between the cut-off in both the high- 
and low-pressure cylinders, and the volume of high-pressure cylinder with respect 
to the low-pressure volume, so that 

P 
V b = V f -5^, or symbolically, 

*b 

*-= D 4B: )-*<££■) <•» 



(317) 



264 



ENGINEERING THERMODYNAMICS 



Similarly the low-pressure cylinder cut-off volume must equal the high- 
pressure displacement volume or D H = Z L D L , 



7 D H 1 

Zl== W l = Rc {a) 



Rl — Rc 



(6) j 



(318) 



indicating that low-pressure cut-off is the reciprocal of the cylinder ratio. Making 
the necessary substitution there result the following equations for this cycle 
which, it must be noted, is that for most economical use of fluid in compound 

























A 




B 








(ir 
































<— z, 


.DhT" 


\ 






























































N 








c 




(r 


sc.pr.) 










D H = 






















Z L D L 




























M 




















E(rel 


































D L 
































H 



A 


B 


N \ ( 


N 


C 


M \| 



Fig. 83. — Special Case of Cycles V and VI, Complete Expansion in both Cylinders of 
Compound Engine with Infinite Receiver and Zero Clearance. 

cylinders without clearance and with infinite receiver, and in which the same 
work is done as in Cycle I, for simple engines at best cut-off. 
From Eq. (308) 

(rec.pr.) = (in.pr.)g| ; ^ ^ r -) % g^ -frQ-fc-pr.). . . (319) 

Rc (bk.pr.) 
From Eq. (309), 

, . N (in.pr.) ,. ,#c(bk.pr.) „ n . . ,. . , n „^ 

(rel.pr .)* = v -^- y = (in.pr.) - ^ J = # c (bk.pr.) = (in.pr.)x, . 320) 
Kh (m.pr.j 



WORK OF PISTON ENGINES 



265 



From Eq. (310), 



, . N (in.pr.) (in.pr.) /U1 N 

(rel - pr - ) ^fiir = p i (in.pr.) =(bk - pr - ) - 



Re (bk.pr.) 



From Eq. (311), 
High pressure cylinder work 



l+loge#„- 



Rl 
Re 



_l og e R H 

Low pressure cylinder work 1 ., D (bk.pr.) log e ^ 

(in.pr.) 

l0ge ^r- 



l0ge 



1 



(a) 



(b) 



(321) 



(322) 



For the case of most economical operation, that of complete and perfect 
expansion in both cylinders, there may be set down the four characteristic 
Eqs. (304), (305), (306), (307) with suitable modifications to meet the case. 
These become 



l0ge 



Work of cycle = 144(in.pr.)D I/ - 



m.pr. 

BE " p7/ -144(in.pr.)D £ l^. 

tiv 



/in.pr. \ 
Ibk.pr./ 



(323) 



•-CS.) 



(324) 



/ n / r x t -n \ W ,. N \bk.pr./ ,. AogeRv 

(m.e.p.) (ref. to L.P.) =77777 = (in.pr.)— 7^ v~ z -= (m.pr.)— %-A 

144D/, /in.pr. \ R v 

VbEprT/ 

Work per cu.ft. supplied = 144(in.pr.) log e (ri~-— ) = 144(in.pr.) log e R v . (325) 



Cu.ft. per hr. per IJLP.-, 13,750 fe: 

(m.e.p. ref. to L.P.) \in.pr. 



13,750 



1 



(m.e.p.ref. to L.P.) R v 



(a) 
(b) 



. (226) 



For equal division of work with complete expansion in both cylinders, the 
ratios of Eqs. (317) and (318) becomes 



*-*-©£)* (a) l 



(327) 



and this is evidently a case to which Eqs. (314) and (315) apply without change. 



266 ENGINEERING THERMODYNAMICS 

Example. 1. Method of calculating diagram, Fig. 82. 
Assumed data for Case A: 

P a =Pb = 100 lbs. per sq.in. abs. V a = Vn = V m = cu.ft. 

P n ^Pa =P e = 50 lbs. per sq.in. abs. V c = Vd = .6 cu.ft. 

P m =Pff= 10 lbs. per sq.in. abs. Yj = V g = 2 cu.ft. 

P c = 60 lbs. per sq.in. abs. V e = .8 cu.ft. 



To obtain point B: 



To obtain point F: 



V, = VcX~= .6 X^ n = .36 cu.ft. 



Ye .8 

P f =PeX— =50X-=201bs. per sq.in. 

V / 2, 



To construct the indicator cards: 

Lay off ND of the PV diagrams to equal the length of the card, and NA perpen- 
dicular to it at N to equal the height of the card. Cut off equals AB+ND. From A 
on card lay off this ratio times the length of the card. From D on the card lay off 
a perpendicular equal to CD of the PV diagram reduced by the same proportion 
as AN of the card is to AN of the diagram. Join the points B and C by a curve 
through points located from intermediate points on the PV diagram. The low- 
pressure card is constructed in same manner. 

Example. 2. A 12- and 18x24-in. steam engine without clearance runs on 150 
lbs. per square inch absolute initial pressure, 10 lbs. per square inch absolute back pressure, 
and has a speed of 125 R.P.M. What will be (a) the horse-power for | cut-off in H.P. 
cylinder, (b) pounds of steam per I.H.P. hour, (c) terminal pressures, (d) L.P. cut-off 
for continuous expansion, (e) work done in each cylinder. 

Note: § for 150 lbs. =.332. 

(a) From Eq. (305) 
(m.e.p.) referred to L.P. cylinder is 

(in.pr.)— 1 — (2 + log e R H +log e Rl-~) - (bk.pr.). 

tXHtiC \ tic } 

In this case 

R H =2, R c = (~\ 2 =2.25, R l =2.25, 

since vol. of L.P. cyl. at cut-off must be equal to the entire volume of the high for 
continuous expansion, hence 

(m.e.p.) =150Xr-4-^X(2+.69+.81-l)-10=73.3 lb. sq. inch, 



and 

33,000 



I.H.P. = (m ± P ;L Lan =282. 



(b) From Eq. (307) 



n * v, tud 13,750 Z„ 13,750^ .5 

Cu.ft. per hour per I.H.P. = - — ! — = ' ._ Xr-^r =41.7, 

(m.e.p.) R c 73.3 2.25 



WORK OF PISTON ENGINES 267 

(c) From Eq. (309) 

(rel.pr.)// = (in.pr.)^, 

= 150X^=75 lbs. sq.in, 

and from Eq. (310) we have 

(rel.pr.)/y 

(rel.pr.)z, = ^ , 

tic 

75 
= ——=33.3 lbs. sq.in. 
2.25 

^=-^=.444. 



(e) From Eq. (311) 



— — — 
H.P.work *° ge Z„ RcZ L 



L.P. work , , . 1 bk.pr. Re 

l + log e — — 

Z L (m.pr.) Z H 

1 



1+.69 



2.25 X. 44 .69 JM 
.456, 



, M 10 2.25 l.5l 

l+ - 8l -i50 X T5- 

or 

H.P.work =. 456 XL.P. work, 

also 

H.P. work+L.P. work =282 I.H.P. 

Hence 

H.P. work =88 I.H.P. 
and 

L.P. work = 194 I.H.P. 

Prob. 1. What must be the cylinder diameters of a cross compound engine to run 
on 100 lbs. per square inch absolute steam pressure, 18 ins. of mercury vacuum and to 
develop 150 H.P. at a speed of 200 R.P.M. with | cut-off in each cylinder, if cylinder 
ratio is 3 and stroke is 18 ins.? Engine is double-acting and assumed to have no 
clearance. 

Prob. 2. What will be the release pressure in each cylinder and the receiver 
pressure of the engine of Prob. 1? If cut-off were reduced to | in H.P. cylinder, 
how would these pressures be affected and to what extent? How would the horse- 
power change? 

Prob. 3. A 15- and 22 x30-in. infinite receiver engine has no clearance, a speed of 
150 R.P.M., initial pressure 125 lbs. per square inch gage. What will be the horse- 
power and steam consumption for a H.P. cut-off of J, j, f, f, and that value which 
will give complete expansion in high-pressure cylinder? Low-pressure cut-off to be 

}d at ^. 

Note: B for 150 lbs. gage = .363. 



268 ENGINEERING THERMODYNAMICS 

Prob. 4. What will be the release and receiver pressures, and the work done in each 
cylinder for Prob. 3? 

Prob. 5. An 18 and 24x30-in. infinite receiver engine is to be operated so as 
to give complete expansion in both cylinders. What will be the cut-off to accomplish 
this and what horse-power will result if the initial pressure is 100 lbs. and back pressure 
10 lbs. per square inch absolute? 

Prob. 6. Draw the PV diagram for following cases. Cylinder ratio 1 to 2.5, 
(in.pr.), 100 lbs. per square inch absolute, (bk.pr.), 20 lbs. per square inch absolute, 
H.P. cut-off (a)=h (b)=h (c)=f. L.P. cut-off (a) =4, (&)=-&, (c) =1 . 

Prob. 7. For the following conditions find the horse-power, steam used per hour, 
receiver pressure and release pressures. Engine, 10- and 15x24-in. 150 R.P.M., 
125 lbs. per square inch gage initial pressure, 2 lbs. per square inch absolute, back 
pressure, ^ cut-off in high-pressure cylinder, to cut-in low-pressure cylinder with 
infinite receiver. 

Note: 5 f or 125 lbs. =.311. 

Prob. 8. An infinite receiver engine is to develop 150 H.P. at 200 R.P.M. when 
initial pressure is 150 lbs. per square inch absolute. Cylinder ratio is 1 to 3 and 
back pressure is one atmosphere. What must be its size if the stroke is equal to 
the low-pressure cylinder diameter for | cut-off in the high-pressure cylinder and 
\ cut-off in the low-pressure cylinder? 

Prob. 9. Find by trial the cut-offs at which work division will be equal for an 
infinite receiver engine with a cylinder ratio of 2.5, an initial pressure of 100 lbs. per 
square inch absolute and a back pressure of 5 lbs. per square inch absolute? 

10. Compound Engine with Infinite Receiver, Exponential Law. No 
Clearance, Cycle VI. General Relations between Pressures, Dimensions, 
and Work. Again referring to Fig. 82, which may be used to represent this 
cycle also, the work of each cylinder may be expressed as follows, by the 
assistance of Eq. (254) derived in Section 4. 

W H = 1UD H \z H (m. W .) ( *~1*P ) ~ (recpr.)l . . . (328) 

W L = lUD L \z L (rec.pr.) ( *~f f "* ) ~ (bk.pr.) j , . . . (329) 

where Z H is the cut-off in the high pressure, = =■ and Z L} low-pressure cut-off 

Vd 

= ^. In combining these into a single equation for the total work, the term 
for receiver pressure (rec.pr.) should be eliminated. Referring to Fig. 82, 

(rec.pr.) = P d = P e = P b (pJ =(m.pr.)(^^)', • . . (330) 
hence 



^mm 1 



-(bk.pr.) , . (331) 



WORK OF PISTON ENGINES 269 

a rather complex expression which permits of little simplification, but offers 
no particular difficulty in solution. 

Mean effective pressure referred to the low-pressure cylinder is 



(m.e.p. ref. to L.P.) = (in.pr.) 



(ZjA ( s-Z H °-i \ _1_(_Zh_\ 
\Rc)\ s-l ) Rc\RcZ L ) 

+z <&) 8 C^f 1 )]- (bk - pr - ) - (332) 



Work per cubic feet fluid supplied may be found by dividing Eq. (331) by 
the supply volume, which in terms of low-pressure displacement is 

(Sup.Vol.)=Z)Jr (333) 

tic 

The consumption of fluid, cubic feet per hour per indicated horse-power is 

Consumption cu. ft. per hr. per I. H. P. = 7 L nn r ~TT\ rTt • • • (334) 

(m.e.p. ref. to L.P.) R c 

which is the same expression as for the logarithmic law. Multiplying this by 
Si, the initial density of the fluid, pounds per cubic feet, gives consumption, 
pounds per fluid hour per I.H.P. 

The receiver pressure has already been determined in Eq. (330). 

Release pressure of the high-pressure cylinder is 

(rel.pr.)^=(in.pr.)Z^ s , (335) 

and for the low-pressure cylinder, 



(rel.pr .) L = (in.pr.) (~) (a 



<^o {b) 



(336) 



where R v is the ratio of maximum volume in the low pressure, to, volume at 

r> 

cut-off in the high, and equals -==-. 

The distrubition of work between the high- and low-pressure cylinders may 
be found as follows, by means of Eqs. (328) and (329), eliminating (rec.pr.) by 
means of Eq. (330) 



W H = Zh \ s-l )~\Rct) 



(337) 



270 



ENGINEERING THERMODYNAMICS 



Equality of work in the two cylinders will be obtained if this expression 
is equal to unity, giving a complex relation between high- and low-pressure 
cut-offs, cylinder ratio and ratio of initial and back pressures, to be satisfied. 
It is found at once that the simple conditions for equality in the case of logarith- 
mic law will not give equality of work for the exponential law. There is, how- 
ever, a case under this law which yields itself to analysis, that of complete expan- 
sion in both cylinders, without over-expansion. The conditions for equality 
of work for this case will be treated after deriving work and mean effective 
pressure for it. 

Complete expansion, without over-expansion, in both cylinders may be 
represented by Fig. 83. 

7 _Jb _nc 

Zh 'W Zl ~w 



and since 



NC = Dh and ME = D L , 
D L ME 1 



Rc = -^r- 



d h nc z l . 



The true ratio of expansion =R V == =— — =— ^ but this is also equal to 
i 



/imprA 
\bk.pr./ 



due to the law of the curve, P b Vb=P e Ve s 



By means of Eq. (257) in Section (4) the work of the two cylinders may be 
evaluated, 



but since 



W H = 144(in.pr.)Z)tfZ„~- M - Z H * -A (a) 
= 144(^^01)^-^(1-^-1) (6) 
^ = 144(bk.pr.)Z)^-(^ zr - 1), 



(338) 



Re 



W L = 144(bk.pr.)D L — -{Rc s ~ l - 1) (a) 

s J. 

= 144{in.pr.)I> £r J 1 (g) S (flc s - 1 -l) (6) 



(339) 



WORK OF PISTON ENGINES 271 

The total work is evidently the same as that of a cylinder equal in size to 

the low-pressure cylinder with a cut-off equal to -^, working between the 

Kc 

given (in.pr.) and (bk.pr.) and may be stated by reference to Eq. (257), 

Section 4, or by taking the sum of W H and W L given above, 

Tf=144(in.pr.)/)7.g^ T j(l-^ s - l ) + (^) ! "W- 1 -l),[, 

which reduces to 

Tf = 144(in.p,)Z),|^jl-(|) S - 1 ! (340) 



For this case of complete expansion in both cylinders, the ratio of high- to 
low-pressure work is given by division and cancellation, 



©-'- 






(sr (ifc, - i - i) 



Equality of work, obtained by placing this expression equal to unity, pro- 
vides the condition that 



s-l 



or 



^-'-i-(tr-(te) • ■ 

f /in-prA 
\bk.pr.; 



«c=j^^ , (342) 

for equal work and complete expansion, and 




s-l 



i_ f/bk-PrA ' , jl-1 

, = /bk-PL-V R = \in.pr. / 

\in.pr./; c { 2 J 



(343) 



Since Z L =^- for complete expansion,' and (in Fig. 83) P c Vc=PfVf, the 
tic 

receiver pressure, P c , is 

(rec.pr.) = (bk.pr.) (yf = (bk.pr .)R C % (344) 

in which R c will have the value given above if work is equally distributed. 

Example 1. What will be (a) the horse-power, (b) consumption, (c) work ratio, 
(d) receiver and release pressures for the following conditions? Engine 12 and 
18x24 ins., running at 125 R.P.M. on initial air pressure of 150 lbs. per square 
inch absolute, and back pressure of 10 lbs. per square inch absolute, with | cut-off 
in high-pressure cylinder and continuous expansion in low-pressure cylinder. Exponent 
of expansion curve = 1.4 for compressed air, infinite receiver. 



272 



ENGINEERING THERMODYNAMICS 



(a) From Eq. (332) 



^-^m(^)-im 



«mm]-» 



which, on substituting values from above, gives for (m.e.p.) 63 lbs. per sq. inch. 
Hence, the indicated horse-power = 242. 
(6) From Eq. (334) 

Compressed air per hour per I.H.P. = _' _ -=- cu.ft., 



m.e.p. Re 



which, on substitution, gives 



f^«- » 



(c) From Eq. (337) 



Wh 
W l 



which gives 



Wh 
Wl 






2.25 



1.4 



JJ_.._A 

2.25 



1.4 



V2.25/ 



2.25 X 



A 



_10 
150 



and 

Hence 

and 



(d) From Eq. (330) 
(rec.pr.) = (in 



2.2 

W B + W L =242 I.H.P. 

17/7=56 I.H.P. 

W L = 184 I.H.P. 

.5 



-«*-"(=; 



1.4 



2.25, 



57 lbs. per sq.in. 



From Eq. (335) 



(rel.pr.)^ = (in.pr.)Z# s , 

= 150X(.5) 14 =57 lbs. per sq.in. 



WORK OF PISTON ENGINES 273 

From Eq. (336) 

(rel.pr.)x, = (in.pr.) +R S V , 

= 150-^21.85 =6.85 lbs. per sq.in. 

These values may be compared with those of Ex. 1, Section 9, which were for the 
same data with logarithmic expansion. 

Prob. 1. What will be the horse-power and steam used per hour by the follow- 
ing engine under the conditions given? Cylinders 18 and 30x48 ins., speed 100 
R.P.M., initial pressure 150 lbs. per square inch absolute, back-pressure 10 lbs. per 
square inch absolute, steam continually dry. Cut-off at first I in high-pressure and 



in low, and then £ in each infinite receiver. 



Prob. 2. The very large receiver of a compound pumping engine is fitted with safety 
valve which is to be set to blow at 25 per cent above ordinary pressure. The cylinder 
ratio is 1 to 3.5, and cut-offs are f in high and \ in low. If initial pressure is 125 
lbs. per square inch gage, for what must valve be set? What vacuum must be carried 
in the condenser to have complete expansion in low-pressure cylinder? Superheated 
steam. 

Prob. 3. A compound engine is to be designed to work on superheated steam of 
125 lbs. per square inch absolute, initial pressure and on an 18-inch vacuum. The 
load which it is to carry is 150 horse-power and piston speed is to be 500 ft. per 
minute at 200 R.P.M. Load is to be equally divided between cylinders and there is 
to be complete expansion in both cylinders. What must be cylinder sizes, and 
what cut-offs will be used for an infinite receiver? 

Prob. 4. How will the economy of the two following engines compare? Each is 
14 and 20x24 ins., runs at 200 R.P.M. , on compressed air of 100 lbs. per square 
inch gage pressure, and 15 lbs. per square inch absolute exhaust pressure. Low-pres- 
sure cut-off of each is \ and high pressure of one is i, the other, I. Infinite 
receivers. 

Prob. 5. A compound engine 12 and 18 X24 ins. is running at 200 R.P.M. on 
superheated steam of 100 lbs. per square inch absolute pressure and exhausting to a 
condenser in which pressure is 10 lbs. per square inch absolute. The cut-off is 3 in 
high-pressure cylinder and \ in low-pressure cylinder. Compare the power and steam 
consumption under this condition with corresponding values for wet steam under 
same conditions of pressure and cut-off and infinite receivers. 

Prob. 6. The initial pressure of an engine is 150 lbs. per square inch absolute, the 
back pressure one atmosphere, the cylinder ratio 3. As operated, both cut-offs are at \. 
What will be the receiver pressure, high-pressure release pressure, and low-pressure 
release pressure? What will be the new values of each if (a) high-pressure cut-off is 
made \, (b) I, without change of anything else, (c) low pressure cut-off is made I, 
(d) f, without change of anything else? Infinite receiver, s = 1.3. 

Prob. 7. In the above problem for | cut-off in each cylinder how will the release and 
receiver pressures change if (a) initial pressure is raised 25 per cent, (b) lowered 25 
per cent, (c) back pressure raised 25 per cent, (d) lowered 25 per cent? 

Prob. 8. How many pounds of initially dry steam per hour will be required to 
supply an 18-in. and 24 x30-in. engine running at \ cut-off in each cylinder if speed 



274 



ENGINEERING THERMODYNAMICS 



be 200 R.P.M., initial pressure 100 lbs. per square inch gage and back pressure 5 
lbs. per square inch absolute? Expansion to be adiabatic and receiver infinite. 
Note: d for 100 lbs. =.26. 

Sup. Vol. 

(in.pr.) 




OVER EXPANSION 

Fig. 84. — Work of Expansive Fluid in Compound Engines with Finite Receiver, Zero Clear- 
ance. Cycle VII, Logarithmic Expansion; Cycle VIII, Exponential Expansion. 

11. Compound Engine with Finite Receiver. Logarithmic Law. No 
Clearance, Cycle VII. General Relations between Dimensions and Work 
when H.P. Exhaust and L.P. Admission are not Coincident. The diagrams, 
Fig. 84, while showing only two degrees of expansion, that of over and under 
in both cylinders, suffice for the derivation of equations applicable to all 
degrees in either cylinder. Volumes measured from the axis AL are those 



WORK OF PISTON ENGINES 275 

occupied by the fluid in either cylinder alone, while fluid volumes entirely in the 
receiver, or partly in receiver, and in either cylinder at the same time are meas- 
ured from the axis A'V '. No confusion will result if all volumes represented by 
points be designated by the (V) with a subscript, and to these a constant rep- 
resenting the receiver volume be added when part of the fluid is in the receiver. 
Then, 

High-pressure cylinder work is 

W H = P h Y^l+\og e J^-P n O\og e (^^ (345) 

Low-pressure cylinder work is 

W L = P n O\og e (^^^+P e V e \og e ^-P V . . . . (346) 

Total work 

W = P b V b (l+\og e ^j +P n O lo & p±^ 

+P e V e \og e ^-P n O\og e (^^ ) J-P V . . (347) 

These expressions include some terms not known as initial data and may 
be reduced by the following relations, 

P b V b = P c V c = PeV e = PfV f , 

and 



P n O = P e (V e + 0) = P h V b (J^\ 



Hence 



H -=n,,[ 1+ ,4 :+ C^) 1( .(^) 






-P g V . (348) 



Dividing by the low-pressure cylinder displacement, V g , the result will be 
the mean effective pressure referred to the low-pressure cylinder, 

(M.E.P. ref. to L.P.) 

-p^i+io* y+{rvr) log « (^H+ l0 ^ v e 

'v e +o\ . /v c +o\~\ n 



276 



ENGINEERING THERMODYNAMICS 



A similar division but with the volume supplied, F& as the divisor, gives 
Work per cu.ft. supplied 

-(^Vm]-4: « 

Also as in previous cases 
Cu.ft. supplied per hr. per I.H.P. 






13,750 



X 



(m.e.p. ref. to L.P.) V g ' 



(351) 



Of course, the weight per hour per I.H.P. follows from Eq. (351) by introduc- 
ing the density as a multiplier. 

While the last four equations can be used for the solution of problems, it is 
much better to transform them by introducing dimensional relations as in the 
previous cases developed. 

Let (rec.pr.)i= maximum receiver pressure P n , which is also the initial admis- 
sion pressure for the low-pressure cylinder; 
(rec.pr.) 2 = minimum receiver pressure P e , which is the terminal admission 
pressure for the low-pressure cylinder and that at which 
expansion begins there; 



u _ receiver volume _ _0_ , _D H y _ y 

high-pressure cyl. displ. V c D H ' ' D L D L R c ' 

Other symbols necessary are unchanged from the meaning imposed in Section (9) . 
Substitution in Eqs. (348), (349), (350), and (351) gives the following set 
in a form for direct substitution of ordinary data: 

Work of cycle 



l+log e -^+log e -J- 



AlKc 



log, (l + ^)-log e (l+i)] \-Ui(bk.pr.)D L (a) 
144(in.p,)g { 1 +log, tf„+log, R L + (l +|f ) [log, (l+||) 



lo & (l+J)] } 



144(bk.pr.)2>i (6) 



(35) 



WORK OF PISTON ENGINES 



277 



(m.e.p. ref. to L.P.) 

= (in.pr.)~] l+log e ^-+log e -^-+ 



("*)M 



1 + 



ZlR 



L^C 



log, 1 + 



\)] 



(bk.pr.) (a) 



=( in -p r ^i{ 1+lo& ^ +logefli+ ( 1+ lf)h( 1+ S) 



1+^)]) -(bk.pr.) (6) 



Work per cu.ft. supplied 

= 144(in.pr.) J 1+log* ^-+log e ■= 



i+i*V^(i+l)] 



144(bk.pr.) 



Z H 



(a) 



+ ( 1+ i)h( 

= 144(in.pr.) { l+log.fo+lo & ^+(l+^) [log. (l+||) 

-log. (l+^)] | -144(bk.pr.)ifc#* (b) 



Cu.ft. supplied per hr. per I.H.P. 

13,750 



— X— 
(m.e.p. ref. to L.P.) R 



13,750 



X 



1 



(m.e.p. ref. to L.P.) R H R 



(a) 
(6) 



(353) 



(354) 



(355) 



It is desirable at this point to introduce a series of expressions fixing the 
relations between the dimensions, the cycle that may follow, and the fluctua- 
tions in the receiver pressure, and for the selection of cylinder and receiver 
dimensions for a required output of work and division of it between cylinders. 

In doing this it will be convenient to start with diagram points and finally 
substitute general symbols in each case. There will first be established the 
maximum and minimum receiver pressures and the fluctuations. 

Maximum receiver pressure 

-■--■^=mm-o v T> 



■■ (recpr.) 1 = (in.pr.)(-^ , ^ 

, Rl 



ZhD h , ZhDh 
( 



(in.pr.) 



y ^Z L R C 



*hV RhRc 



{a) 



Q>) 



(356) 



278 ENGINEERING THERMODYNAMICS 

Minimum receiver pressure 



P e = Pi 



TV 






Z D 7 

/. (rec.pr.)2 = (in.pr.) ~fj^= ( in *P r -)^| c (°) 



(in.pr.) 



Ri 



RhRg 



(W 



(357) 



Fluctuation in receiver] pressure = (P n —P e )=P b 







Z D 7 

:. (rec.pr.)i - (rec.pr.) 2 = (in.pr.) -^p = (in.pr.)-^ (a) 

: =(in - pr ' } i (6) 



(358) 



It is interesting to note that the minimum receiver pressure is exactly the same 
as the value of the constant-receiver pressure for infinite receiver, so that limit- 
ing the size of receiver does not affect the point E, but only raises point N higher, 
tending to throw more work on the L.P. cylinder for the same valve setting. 

The two release pressures P c and P/ can be evaluated as in the case of the 
infinite receiver, as both these points lie on the common expansion line, which 
is not at all affected by the receiver-pressure changes, and the values are the 
same as for the infinite receiver, and are here reproduced from Eqs. (309) and 
(310) with new numbers to make the set of equations complete: 



(rel.pr.)fl"= (in.pr.)-— (a 



= <in.pr.)Z* (6) 

1 



(359) 



(rel.pr.)i = (in.pr.) =-— (a) 

ttcJtlH 

= (in.pr.)§? (6) 
tic 

(in.pr.) 



Rv 

(rehprO^ 
Re 



(c) 
(d) 



(360) 



where R v is the ratio of maximum volume in the low- to the volume at cut-off 
in the high-pressure cylinder. 

Division of work between the cylinders cannot, as pointed out, be the same 
as for the infinite receiver, the tendency being to throw more work on the low 
as the receiver becomes smaller, assuming the cut-off to remain the same. As, 
therefore, equal division was obtainable in the case of infinite receiver with 



WORK OF PISTON ENGINES 



279 



equal cut-offs when the cylinder ratio was equal to the square root of initial 
over back pressure, it is evident that a finite receiver will require unequal cut- 
offs. As increase of low-pressure admission period or cut-off fraction lowers 
the receiver pressure and reduces the low-pressure work, it follows that with 
the finite receiver the low-pressure cut-off must be greater than the high for 
equal work division, and it is interesting to examine by analysis the ratio 
between them to determine if it should be constant or variable. 

For equal work division Eqs. (345) and (346) should be equal, hence by 
diagram points 



P b V b (l+\og e ^ -P n l0g e (-J") =PnO log e (~^~) + PeV e log e ~ f ~P g V g . 

( 1+te i- : )-p,,(-lto) ^(Y,±0) 

= P b Vbi—y—j l0g e (— ^— j-^PbVbXogey-PgVg. 



P b V 



1+1*^(1^1* (l4l)-(l+§)K*(l + 



RC 



)+log R L 

\m.pr. / 



hence for equal division of work, the following relations must be satisfied: 



r> 
(loge #tf-l0ge R L ) =l0g e -^ = 



h*M 



1 + 



Rc\ 



l+ : 



bk.pr. 
in.pr. 



RhRc — 1 . 



(361) 



It will be shown later that when expansion is complete in both cylinders 
and work equal that the high-pressure cylinder cut-off or the equivalent ratio 
of expansion bears a constant relation to that of the low, according to 

Rh 



Ri 



= a< 



(362) 



Rc=- 



(363) 



in which a is a constant depending only on the size of the receiver. It will 
also be shown that the cylinder ratio is a constant function of the initial and 
back pressures and the receiver volume for equal division of work according to 

1 / in.pr. \* 

,bk.prj ' 

in which (a) is the same constant as in Eq. (362). It is impo tant to know if 
these same values will also give equal division for this general case. Substi- 
tuting them in Eq. (361) 



2\og e a= 1 1 



mj)£i 
\ bk.pr./ 




2S0 



ENGINEERING THERMODYNAMICS 



Here there is only one variable, R L , the evaluation of which can be made by 
inspection, for if 

o \bk.pr.' 
the equation will become 

2\og e a=(l+y) log e (l+^J-2 = 2(l+y) l oge (\+I) -2, 



or 

which is a constant. 
1.00 



= 6 Ui+v)iog.(i + L)_ llf 



(364) 



1-40 



.20 









/ 


/ 


1 


/ 


1 


t 




r 


/ 
























/ 


/ / 


i 


1 






/ 




/ 


















n 




i 






lL 


■-1 

7 


i 


4 


J 

-4 


V 










rsVf 


&/ 




































\ 


// 


/ 




/ 
































1 


7 


/ 


1 
































i 




/. 


/ 


/ 














/ 


















/ 


/ 


7 


/ 




J 






























/ 


// 


/ 


/ 


/ 


/ 




























// 


A 


7 




7 






























7/ 


7 


/ 




/ 




























n 


/ 


// 


// 


^ , 


/ 






























J 


/, 


7/ 


/ 


// 


f 












RECEIVER VOLUME EOUALS 
yx H.P. DISPLACEMENT 






T 


// 


'// 


/ 


/j 
































T 


/// 


'h 


^ 


































/ 


% 


'// 


/ y 


































I / 


'///// 


/ 

























































































































































.20. 



.40 .60 

High Pressure Cut Off 



.100 



Fig. 85. 



-Diagram to Show Relation of High- and Low-Pressure Cut-offs for Equal Work in 
the Two Cylinders of a Finite-receiver Compound Engine with Zero Clearance 
and Logarithmic Law. 



As only one constant value of low-pressure ratio of expansion or cut-off 
satisfies the equation for equal division of work when there is a fixed ratio 
between the values for high and low, that necessary for equal division with 
complete expansion in both, it is evident that equal division of work between 
the two cylinders cannot be maintained at all values of cut-off by fixing the 
ratio between them. As the relation between these cut-offs is a matter of some 
interest and as it cannot be derived by a solution of the general equation it is 
given by the curve, Fig. 85, to scale, the points of which were calculated. 



WORK OF PISTON ENGINES 



281 



A special case of this cycle of sufficient importance to warrant derivation 
of equations because of the simplicity of their form and consequent value 
in estimating when exact solutions of a particular problem are impossible, 
is the case of complete and perfect expansion in both cylinders. For it the 
following equations of condition hold, referring to Fig. 84, 

Pc = Pd — Pe ', 





























A 












A 




B 








A B 












\ 














^V 


Q 










\ 


N 
























^^ c 


























E 




\ H 






N 




























NP 


M 
















' 




L 


























M ^^F 






















N, 




INDICATOR CARDS OF EQUAL 
BA8EAND HEIGHT 
















M 




















F 






* 






u 




J 










































L' 












L 




K 




1 












H 







Fig. 86. — Special Case of Cycles VII and VIII Complete Expansion, in both Cylinders of the 
Finite Receiver Compound Engine. Zero Clearance. 

which when fulfilled yield the diagram, Fig. 86. These equations of conditions 
are equivalent to fixing the cut-off in both high- and low-pressure cylinders, 
and the volume of the high- with respect to the low-pressure volume. Accordingly, 



P»' 

bk.pr.\ = Rc 
.pr. / R v 

7? 1 / in.pr. \ = R V 
H jKcVbk.pr./ R c 



V b =V t 

/bk.pTA 

\m.pr. / 



(a) 
(&) 



(365) 



Also for the low-pressure cylinder the cut-off volume must equal the whole 
high-pressure volume, or D H = Z L D L . Therefore, 



Z T = 




(366) 



282 



ENGINEERING THERMODYNAMICS 



Substituting these equations of condition in the characteristic set Eqs. (352) , 
(353), (354), and (355), there results the following for most economical operation: 

Work of cycle 

Tr=144(in.p,)« c (gg>, { l+lo g .[g^)i] +1* Re 

+(1+2/) log, (l+J-) -lo ge (l+i) J -lU(bk. W .)D L 
= 144(bk.pr.)Djog e (^) 



(a) 



144(in.pr.)D i 



/imprA 
Vbk.pr./ 



144(in.pr.)D, l ^ F (6) 



(367) 



W 



(m.e.p. ref. to L.P.) = 144:^j-p 



= (bk.pr.) loge (g^) = On.pr.; 



;.pr./ 



in 
bk 



= (m.pr.) ^ffe(368) 



IF 
Work per cu.ft. supplied = „ n 

= 144 



„ /bk.pr A ^ & Vbk.pr./ ^ & Vbk.pr./ 

tic \ '. ) L>H 

Vm.pr. / 



Cu.ft. supplied per hr. per I.H.P. 

13,750 x /bk-PrA 

(m.e.p. ref. to L. P.) \in-pr. / (m.e. 



= 144(in.pr.)logefly. .' (369) 
13,750 



p. ref. to L. P.) R v 



Xir • (370) 



For this special case of best economy the receiver and release pressures, 
of course, have special values obtained by substituting the equations of 
condition Eqs. (365) and (366), in Eqs. (356), (357), (358), (359), (360). 

(recpr.)i = (m.pr.) {^+ 1 ~) = (m.pr.) (g^+g;) 

-O-P'Ogg+i) - • • ■ • &D 

(rec.pr.) 2 = (in.pr.)^=(in.pr.)g=(bk.pr.) J Rc. . . . (372) 






WORK OF PISTON ENGINES 283 

Therefore 



(rec.pr.)i - (rec.pr.)2 = (m.pr.)-—-; 



(373) 



(rel.pr.)//=(in.pr.)B-=( bk -P r -)^c=(rec.pr.)2; .... (374) 

(rel.pr.) i . = (in.pr.)pV = (bk.pr.) (375) 

These last two expressions might have been set down at once, but are 
worked out as checks on the previous equations. 

For equal division of work in this special case the general Eq. (361) becomes 

1°* «.-■<*■&-■•* e-( 1+ 4£M( 1 '+h)( i+ ?)' 



or 



Therefore 



k»g-2[(l+v)tog.(nl)-l] 



|^ = e 2[(l+V)lo e ,(.+l-)- 1 ] = a2 I (376) 

Kl 

This term, a, has already been used in previous discussions of equality of 
work, while the derivation of its value has not been made up to this point. 

This indicates that ratio of cut-offs or individual ratios of expansion is a 
function of the receiver size for equal division of work. 

From Eq. (376) the cylinder ratio can be found in terms of a, and the ratio 
of expansion. Referring to Fig. 86, 

Ve 

Rb^Yi v?_ 2 

Rl Vr v b v f a > 

V e 

V e = aVVbV f , 

^"^i^r^aw^iv^"^^ • • * (377) 

whence the cylinder ratio is equal to a constant depending on the receiver size, 
multiplied by the value for the infinite receiver, i.e., the square root of the initial 
divided by back pressure. 



284 ENGINEERING THERMODYNAMICS 

The high-pressure cylinder ratio of expansion is 



^4:-V¥=Wf;-^-^ • • • <w> 



and the corresponding value for the low-pressure cylinder is 



Rl= Vj L.1&-I J§K=lVR r . . . . (379) 

V e aW b Vf aW b a\bk.pr. a ' 

For convenience in calculation Table, XII of values of a and a 2 is added for 
various size of receivers. 

Table XII. 



Receiver Vol. 


ra+2/)iog e (i+^-)-i] 


a 2 


H.P. Cyl. Disp. y 




.5 


1.915 


2.64 


.75 


1.624 


3.67 


1.0 


1.474 


2.17 


1.5 


1.322 


1.75 


2.0 


1.243 


1.55 


2.5 


1.198 


1.437 


3.0 


1.164 


1.359 


4.0 


1 . 1223 


1.262 


5.0 


1.0973 


1.204 


7.0 


1.0690 


1.143 


10.0 


1.0478 


1.098 


14.0 


1.0366 


1.068 


20.0 


1.0228 


1.046 


Infinite 


1.0 


1.0 



At the end of this chapter there is presented a chart which gives the relation 
between cylinder and receiver volumes, cylinder ratio, and high- and low-pressure 
cut-offs graphically. 

The corresponding values of maximum and minimum receiver pressure 
for equal division of work for this case of best economy are 



(recpr.), = (bk-pr.^g+l) = ^^* >fr-l) • (380) 

(reC .p r .) 2 = (b k .p r .)i^ ^^») m 

(382) 



/ \ / x V(in.pr.)(bk.pr.) 

(rec.pr.)i — (rec.pr.) 2 = — - — i1 -^- — ^ L - L 

ay 



WORK OF PISTON ENGINES 285 

Example 1. Method of calculating Diagram, Fig. 84. 
Assumed data for case A: 

P a =P b = 120 lbs. per sq.in. abs. V a = V n = V m =0 cu.ft. 
P m =Pg= 10 lbs. per sq.in. abs. Ft, = .4 cu.ft. 

F e = l cuit. 
= 1.2cu.ft. 

V c = .8 cuit. 



To find point C: 



To obtain point E: 



To obtain point D: 



D V b 120 X. 4 

P c =Pb-jr = « — =60 lbs. per sq. men. 



D F 6 120X.4 

P e =Pbrr = — ^ =48 lbs. per sq. inch. 



48x2 2 
P e (7,+0)=P d (F c +0) or P d = * * =53 lbs. per sq. inch. 



To obtain point iV: 



48 y 2 2 
P»(0 + F„) = P e (F e +0) or P n = • =88 lbs. per sq. inch. 



To obtain point F: 



_, P e Ve 48X1 „. „ . , 

P/=— -- = — - — =24 lbs. per sq. inch. 
Vf 2 



Example 2. Find (a) the horse-power, (b) steam used per hour, (c) the release 
and receiver pressures for a 12- and 18x24-in. engine with receiver twice as large as 
the low-pressure cylinder when the initial pressure is 150 lbs. per square inch absolute, 
back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-offs £ in 
high-pressure and such a value in the low pressure as to give complete expansion, and 
clearances zero. 

(a) From Eq. (353) 

(m.e.p.) = (in.pr.)— — I 1 +loge R H +log e R L 

^)M'+i)--K)]}-- 

which on substituting the above values gives 

hence I.H.P.=282. 

(6) From Eq. (355) we have 

n -w u . 13,750 1 13,750 1 - m 

Cu.ft, steam per hour per horse-power = ■ — : — - X D - = ' X:r— — — =41.7, 

(m.e.p.) R H R C 73.3 2x2.25 ' 



286 ENGINEERING THERMODYNAMICS 

(c) From Eqs. (356) and (357) for maximum and minimum receiver pressures 
respectively: 



(in.pr.)(^ 



Rl \ , ,. >, Rl 

hV RhRc/ RhRc 



(in.pr.) ( h -. p ) and (in.pr.) - 



maximum receiver 



iver pressure = 1 50 (-——+—-V—J =91.5 lbs. per sq. inch. 



2.25 
minimum receiver pressure = 150 X— ^,— = 75 lbs. per sq. inch. 



• From Eqs. (359) and (360) for release pressures 



(m.pr.)Z^ and , 



high pressure cylinder release pressure = 150 X. 5 =75 lbs., per sq. inch. 

150 
low pressure cylinder release pressure = ^777 =33.9 lbs. per sq. inch. 



These results may be compared with those of Example 1 of Sections 9 and 10, which 
are derived for same engine, with data to fit the special cycle described in the particular 
section. 

Note: In all the following problems clearance is to be neglected. 

Prob. 1. A 12- and 18x24-in. engine has a receiver equal to 5 times the volume of 
the high-pressure cylinder. It is running on an initial pressure of 150 lbs. per square 
inch, gage and exhausts to the atmosphere. It has a speed of 150 R.P.M. and the cut-offs 
are y 3 o and \ in high- and low-pressure cylinders respectively. What is the horse-power 
and the steam used in cubic feet per hour? 

Prob. 2. What will be the release pressures, and variation of receiver pressure for 
an engine in which the cylinder ratio is 3, cut-offs | and J, in high and low, initial pres- 
sure is 100 lbs. per square inch absolute, and receiver 2 times low-pressure cylinder 
volume? 

Prob. 3. Show whether or not the following engine will develope equal cylinder 
work for the conditions given. Cylinder diameters, 15 and 22 in., initial pressure 135 
lbs. per square inch gage, back pressure 10 lbs. per square inch absolute, cut-offs \ 
and f , receiver volume 4 times high-pressure cylinder, strokes equal. 

Prob. 4. For the same conditions as above, what low-pressure cut-off would give 
equal work? 

Prob. 6. What will be the most economical load for a 16- and 24x30-in. engine 
running at 125 R.P.M. on 150 lbs. per square inch absolute initial pressure and at- 
mospheric backpressure? What will be the economy at this load? 

Prob. 6. What will be the release and receiver pressures for the above engine if 
the receiver has a volume of 15 cu.ft.? 



WORK OF PISTON ENGINES 287 

Prob. 7. Find the cut-offs and cylinder ratio for equal work division and complete 
expansion when initial pressure is 150 lbs. per square inch absolute and back 
pressure is 10 lbs. per square inch absolute, receiver four H.P. volumes. 

Prob. 8. Will a 14- and 20x20-in. engine, with a receiver volume equal to 5 times 
the H.P: cylinder and running on 1 cut-off on the high-pressure cylinder and I cut-off 
on the low, with steam pressure of 100 lbs. per square inch gage and back pressure 
of 5 lbs. per square inch absolute, have complete expansion and equal work distri- 
bution? If not, what changes must be made in the cut-off or initial pressure? 

Prob. 9. What must be the size of an engine to give 200 1. H.P. at 150 R.P.M. on an 
initial steam pressure of 150 lbs. per square inch absolute, and 10 lbs. per square inch 
absolute back pressure, if the piston speed is limited to 450 ft. per minute and complete 
expansion and equal work distribution is required? Receiver is to be 6 times the volume 
of high-pressure cylinder and H.P. stroke equal to diameter. 

12. Compound Engine with Finite Receiver. Exponential Law, No 
Clearance. Cycle VIII. General Relations between Pressures, Dimensions, 
and Work, when High Pressure Exhaust and Low-pressure Admission are 
Independent. The diagram Fig. 84 may be used to represent this cycle, as well 
as cycle VII, by conceiving a slight change in the slope of the expansion and 
receiver lines. Using the same symbols as those of the preceding section, 
and the expression for work as found in Section 7, Chapter I, 

- 144,,, j f,,^^,^-,)-^, _ (-ij)"] j , 

but 

(rec.pr.)i=P n = P 6 (=J [—q-J >. 
or 

=M7")'fefc+')' 

and the last term in the equation for W H within the bracket may therefore be 
written 

(in.pr.)Z g ( y\ (Z H \ S ( y 
s 



or 






288 ENGINEERING THERMODYNAMICS 

and hence by simplifying the first two terms also, 

„ = 1 « (1 , P ,.,S.j,-Z,-(^)-( gfe;+1 )-[ 1 -fe)-']].(3S3, 

Work of the low-pressure cylinder may be expressed in terms of pressure 
and volumes at N, E, and G, but it is convenient to use instead of the pressure 
at N or at E, its equivalent in terms of the point B. The pressure at N is 

(rec.prO^Cin.p,)^)'^!)* 

and when multiplied by the receiver volume yD H , it becomes 

At E the product of pressure and volume is 

(rec.pr.) 2 xZ L D L =(m. V r.)Z H D H (^^~ K 

Using these quantities, the following equation gives the work of the low- 
pressure cylinder: 

+ (^) S ~ 1 [l-^ S "]}-144(bk.pr.)2) i , . (384) 

and the total work is, by adding (W H ) and (W L ) t 

-(jsfe-J'"'] + (&)" , ( I - z ' w )}- 144<bk - pr - )D " ' ' <385) 

This Eq. (385) is the general expression for work of the zero clearance com- 
pound engine with exponential expansion, no clearance, and finite receiver. 
From this the following expressions are derived : 

(m.e.p. ref. to L.P.) 

(in.pr.) Z H 



1 R c * 



-fc*ji+(ar(-«^)}-<»^--<» 



WORK OF PISTON ENGINES 
Work per cu.ft. supplied is 



289 



-(,-fe;)"X&)'"'<'- z '-'f'"<'"'-»')fe 

Cu.ft. supplied per hr. per I.H.P. 



13,750 



(m.e.p. ref. to L. P.) R c ' 
(rec.pr.)2 = (in.pr.)(^~j ; . • . . 

( ^ )pM y( i+ T/ ; 

(rel.pr.) i/ =(in.pr.)Z // s ; 



(rel.pr.)z, = (rec.pr.)2#i = (in.pr.) ( j^- 



(387) 

(388) 

(389) 

(390) 
(391) 
(392) 



If work is equally divided between the cylinders, W H , Eq. (383), and W L , 
Eq. (384), will become equal, hence 



s-Z^- 1 - 



II 



y ) \tUA 

y 

y+R c Z L 



'my*-*- 



s-l\_ 



(bk.pr.) ifc 
(in.pr.) Z* 



(s-1). . (393) 



This equation shows conditions to be fulfilled in order that an equal division 
of work may be obtained. It does not yield directly to a general solution. 
When expansion is complete in both cylinders, 



and 



bk.pr. 
in.pr. 



Zh 

Rc 



Introducing these values in the general expression Eq. (385) for work of 
this cycle, it may be reduced to the following: 



W 



^^(in.prOZ^^fl-d)^ 1 ] (394) 



From which are obtained 



(m.e.p. ref. to L. P.) = (in.pr.)-^ — - 1- 

Kc s — 1 L 

Mm 



Zh 

Rc 



Work per cu.ft. supplied = 144 (in.pr. 



(395) 
(396) 



290 ENGINEERING THERMODYNAMICS 

Cu.ft. supplied per hr. per LH.P. = 7 13?75 ° 

(m.e.p.ref. toL 



1M50 /b^prAl 

(m.e.p.ref. to L. P.) \in.pr. / * ldy/j 



If work is equally divided and complete expansion is maintained in both 
cylinders Eq. (381) becomes 

which may be simplified to the form, 

where R v is the ratio of maximum low-pressure volume, to the high-pressure 
volume at cut-off, 



hence 



r> Re 



rv Rc 

Zi H ^ 



Rv 
and the value of R v may be found from original data, 



fli 



/in.pr. V 
\bk.pr./ 



(399) 



Eq. (398) may easily be solved for Z H , from which the required cylinder 
ratio may be found by, 

Rc = Z H R v (400) 

This is the cylinder ratio which gives equal work in the two cylinders and 
complete expansion in both, when used with the value found for the high- 
pressure cut-off Z H , the assumed initial and back pressures, and the assumed 
ratio, y, of receiver volume to high-pressure displacement. 



WORK OF PISTON ENGINES 291 

Example. Find (a) the horse-power, (b) steam used per hour, (c) the release 
and receiver pressures of a 12- and 18x24-in. engine, with a receiver twice as large as 
the low-pressure cylinder when the initial pressure is 150 lbs. per square inch absolute, 
back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-offs \ in the 
high and such a value in the low as to give complete expansion. Exponent for ex- 
pansion curve = 1.4. 

(a) From Eq. (386) 

-kiEn+<&r<!>^>}-*- 

which, on substituting above values, gives 



.5 1 r4 /.5\- 4 / 4.5 A 1,4 I7 4 - 5 V 4 



.4 2.25 



2 - 25X 2.25 



4.5 



4 - 5+2 - 25X 2T2-5, 




[i-fern- 



or 



hence 

(b) From Eq. (388) 



(m.e.p.) =57.5 lbs. per sq.in., 
I.H.P.=221. 

13,750 Z H 



Cubic feet of steam per hour per horse-power 



m.e.p. Re ' 



13,750 .5 rort „ 
T7T X 2:25= 53 - 2cU - ft ' 



hence total pounds per hour will be 

53.2 X221X. 332 =3910. 
From Eqs. (389) to (392): 

(rec.p,) 1 = (in.p,)(||-J 8 (l + ? ff. 

(rec.pr.) 2 = (in.pr.) \^J , 

(rel.pr.)^ = (in.pr.)Z^ s , 
(rel.pr.) Z/ = (rec.pr.) 2 ^L S , 



292 ENGINEERING THERMODYNAMICS 

These, on substitution of the proper numerical values, become: 

/ g\ 1.4 / j\ 1.4 

(rec.pr.)i = 150x( ^J ( Xl+— J = 75 lbs. per sq. inch, 

(rec.pr.) 2 = 150x(.5) 1 - 4 =57 lbs., 

(rel.pr.)^ = 150 X (.5) L4 =57 lbs, 

(rel.pr.)z, =57 X (^V'' =32.1 lbs. " 

Note: In all the following problems clearance is assumed to be zero. 

Prob. 1. A 12x18x24 in. engine is running on superheated steam of 150 lbs. per 
square inch absolute pressure, and exhausts to the atmosphere. If the speed is 100 
R.P.M., high-pressure cut-off %, low pressure cut-off J, and receiver volume 10 
cu.ft., what horse-power will be developed and what steam used per hour? 

Prob. 2. What would be the effect on the power and the economy of (a) changing 
to wet steam in the above? (6) to compressed air? 

Prob. 3. What would be the receiver and the release pressures for each case? 

Prob. 4. Will there be equal work distribution between the two cylinders? 

Prob. 5. It is desired to obtain complete expansion in a 14x22x36-in. engine 
running on fluid which gives a value for s of 1.2. Initial pressure is 100 lbs. per 
square inch gage, and back pressure 5 lbs. per square inch absolute. What must be 
the cut-offs and what power will be developed at 500 ft. piston speed? Receiver = 3 
XH.P. volume. 

Prob. 6. How large must the receiver be for the above engine in order that the 
pressure in it shall not fluctuate more than 5 lbs. per sq. inch? 

Prob. 7. An engine is to run on steam which will give a value of 3 = 1.1, and to 
develope 500 horse-power at 100 R.P.M. Piston speed is not to exceed 500 ft. per 
minute. Steam pressure, 150 lbs. per square inch absolute, back pressure, 5 lbs. per 
square inch absolute. Complete expansion and equal work distribution, for this load are 
to be accomplished. What will be the cylinder sizes and the high-pressure cut-off if the 
receiver is to be 3 times the high-pressure cylinder volume? 

Prob. 8. What will be the steam used per hour by the engine of Prob. 7, and 
what will be the variation in the receiver pressure? 

Prob. 9. If the high-pressure cut-off were halved, how would the power and 
economy be affected? 

13. Compound Engine without Receiver, Logarithmic Law. No Clear- 
ance, Cycle IX. General Relations between Dimensions and Work when 
High-Pressure Exhaust and Low-Pressure Admission are Coincident. Such 
a peculiar case as this admits of but little modification of the cycle compared 
with the receiver cases, b; cause the low-pressure expansion is necessarily a direct 
continuation of the high pressure without any possible break. There can be no 
over-expansion in the high nor can expansion there be incomplete, as there is, 
properly speaking, no back pressure with which to compare the high-pressure 
cylinder terminal pressure. There may, however, be over and incomplete 
expansion in the low-pressure cylinder. It might appear that the high-pres- 
sure cylinder negative work was equal to the low-pressure admission work, as 
each is represented by the area below DC, Fig. 87A, but this is not the case, since 



WORK OF PISTON ENGINES 



293 



the diagram is drawn to two different scales of volumes, showing the pressure- 
stroke relation between high and low. This is apparent from the diagra m, F ig. 
87 C showing fluid volumes in each cylinder to a single scale on which A BCD 
is the work done in the high-pressure cylinder, ABD'EF the whole work, whence 
DCD'EF is the part done in the low-pressure cylinder. There is, of course, 
no low-pressure cut-off or even admission as ordinarily considered. The cycle, 
so far as the work to be done is concerned, is the same as for a simple engine, 





H.P.Cyl. Vols. 
1 2 




































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R 












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& 








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DICATOR CARDS OF EQUA 






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BASE AND HEIGHT 






i\ 3 
L.P.Cyl. Vols. 

GRAM ON"EQUA"L 
UNEQUAL VOLUM 






3A»E 






DIAGRAM OF FLUID WORK REGARDLESS 


























DIA 


STROKE 
E SCALE 










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DER 


DIS1 


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INDIVIDUAL CYLINDER WORK SHOWN TO SAME 
SCALE OF PRESSURE AND VOLUME 



N K L 

HIGH AND LOW PRESSURE CYLINDER DIAGRAMS PLOTTED TO SAME AXIS 



Fig. 87. — Work of Expansion in the No-receiver Compound Engine, Zero Clearance, Cycle 
IX, Logarithmic Expansion, Cycle X, Exponential Coincident Piston Movement. 

and the only reason for introducing formulas for overall work, work per cubic 
feet supplied, {m.e.p. referred to low), and fluid consumption, is to put them into 
form for immediate substitution of dimensional relations. Because of the absence 
of cut-off in the low, the distribution of work between high and low will 
depend solely on the cylinder ratio and high-pressure cut-off, for, the earlier the 
high-pressure cut-off, and the larger the high-pressure cylinder, the greater 
the fraction of the total work that will be done there, as there is only a fixed 
amount available, and the less there will be left to be done in the low 



294 ENGINEERING THERMODYNAMICS 

The diagrams of the two cylinders are plotted to combined axes in 
Fig. 87D. The points Q and R sd> equal heights. KN is the L. P. displace- 
ment, and KG that of the H.P. It has been shown in Section 8, that 
the expansion lines CD and CD' may be plotted to the axes LN and LXM, 
the point X being the intersection of NQ and KR extended, and that the distance 



and 



also 



^^k-^=B^D^^V ■ ■ ■ W 



Hence the work area under CD is 

W cd = GLXPc log e p-144(rel.pr.) i , n Z)g2 n log, ^ 

but 

(rel.pr.)^=(in.pr.)Z^, 

hence 

IF^=144(in.pr.)Z^Z)^|l+log e ^-^-log e i?c|. . (404) 

Again the work area under CD' is 

W, d , =KLXP C Io & f| = 144(rel.pr.)« 7 ^%- log, g*. 
hence 

TF i = 144(in.pr.)Z^Z)^^^ I ^og e fl c -144(bk.pr.)Z>z., . (405) 
and the total work, 



1 1 

Zu Re — 1 

. Re 



W = 144(in.pr.)Z„D„ l+log e ±— ^— \og b R c 



Re 



i log B ^}-144(bk.pr.)Z)L 



= 144(in.pr.)Z // D„ j l+log e ^+(^ 1 -^r 1 )log e ^} -144(bk.pr.)i). 



WOEK OF PISTON ENGINES 295 

But 



and 
so that 



\Rc-l Rc-l) ' 



1 r> 

log e ~-+log e R c = log* ^ = log e Rv, 



W = lU(m.pv.)Z H D H (l+\og e ^)-lU(bk.pr.)D L} . . (406) 



which shows by its similarity to the work of the simple engine that, as before 
stated, the total work is the same for this cycle as if the entire expansion were 
made to take place in a single cylinder. 

This same result could have been attained in another way sufficiently 
interesting to warrant setting it down. Since the low-pressure work is repre- 
sented truly to scale by C'D'EF, Fig. 87C, the mean effective pressure of the 
low-pressure cylinder is given by the area divided by V e . By contracting all 
volumes proportionately, CD' takes the position CD' and CF the position 
CF', hence 

area CD'EF 



V e -V f 



~Pe 



represents the mean effective pressure in the low-pressure cylinder just as 
truly. Therefore, 

L.P. cylinder work= ( _ P e J V e 

= p ^(^) io 4:- p ^- 

As the high-pressure work is (total— low), 
H.P. cylinder work = P 6 7^1+lo & ^ -PeVe-PcV^^-^—^ log e ^ +P e V e 



= P b V l 



1+loge Fr^ iy ) loge F c 

V c 



Introducing symbols 
L.P. cylinder work = 144(iii.pr.)^Z) ff ^ s ^ i Vo&5c-144(bk.pr.)i>i. (407) 

H.P. cylinder work = 144(in.pr.)Z^ \i + \ oge ^-^- \og c Rc\ 

= 144(m.pr.)Z^Jl+log e J-- — ~log,&l. . (408) 
which check with Eqs. (404) and (405). 



296 



ENGINEEKING THERMODYNAMICS 



Dividing the total work by the low-pressure cylinder volume and the high- 
pressure admission volume in turn, 



(a) 



(m.e.p. ref. to L.P.) = (in.pr.)|^(l+Io & |?) - (bk.pr.) 

= (in.pr.) ~[1 +lo & (BhRc)] - (bk.pr.) (6) 
Untie J 

Work per cu.ft. supplied = 144(in.pr.)Z#( 1 +log e -~\ — (bk.pr .)R C (a) 



. (409) 



• (4io; 



= 144(in.pr.)^-[l+lo&(&i2c)]- (bk.pr.)22c (&) 



Cu.ft. supplied per hr. per I.H.P. 



13,750 Zh , s 

(m.e.p. ref. to L.P.) * R c W 



13,750 



(m.e.p. ref. to L.P.) R H Rc 



(6) 



• (411) 



For equal division of work there can obviously be only one setting of the 
high-pressure cut-off for a given cylinder ratio and any change of load to be met 
by a change of initial pressure or of high-pressure cut-off will necessarily unbalance 
the work. Equating the high-pressure and low-pressure work expressions, 
Eqs. (404) and (405), 

1+loge ] 



1 , D Rc , D /bk.prA R c 
Z H R c —1 Rc—l \m.pr. / Z H 



or 



,,. 1 , (bk.pr.) R c Rc+l, r> n 
Zi H (m.pr.J L H He— 1 

Another relation exists between Z H and i£ c , namely, that 

J_ = Rv 

Zh Rc 
where Rv is the ratio of volumetric expansion. Then 
1 ., R v Rc+1 * D , /bk.prA 

1+loge ^ - -5 ^ l0g e #c+ ( — £~ ) # F = 0, 



Rc Rc — 1 



in.pr. 



but 



hence 



. R v Rc+l } D , \(Rv\( 1X1, £ F 

loge 5TE=i log « ^ =logc [\rc) [ "5+i ]] = log « — ^ ' 



#c*c 



2ie, 



Iog,*^- 1 , 



1+ v 



^)Rv. 
.pr.y 



/bk.pr 
in 



(412) 



WORK OF PISTON ENGINES 



297 



With this formula it is possible to find the necessary ratio of cylinder 
displacements for given initial and back pressures and for given ratio of. 
expansion R v - 

For convenience in solving this, a curve is given in Fig. 88 to find value 



2R 



of R c when R C R C 1 has been found. 



35 
> 
3 

1 



/ 



25 



50 



75 



100 2Rc 125 



150 



Values of (Rc) &** 



2R 



Fig. 88. — Curve to Show Relation between Values of Re and (ifo) s c _1 for Use in Solving 
Eq. (412), Giving Cylinder Ratio in Terms of Ratio of Expansion for the No-receiver 
Compound Engine without Clearance. 

The complete expansion case of this cycle results from the condition 

Pa = Pe or (rel.pr.^Kbk.pr.) or R v =(^l\ 

Vbk.pr./ ' 

which when applied to Fig. 87, transforms the diagrams to the form Fig. 89. 
It also follows that 



(bk.vr.)D L =(m.pr.)Z H D H 



and 



Rc_ / in.pr. \ 
Z^ Vbk.pr./ 



These conditions will, of course, reduce the total work Eq. (406) to the 
common value for all cycles with logarithmic expansion and likewise those 
for mean effective pressure, work per cubic foot supplied, and consumption. 
For the equal division of work under this condition, Eq. (412), becomes 



r> R - 

Kc C 



since 



1 =7.395, (413) 

/bk.prA 

\m~Dr~/^ F = 1 and R may re P resent ratio of expansion or ratio of 



298 



ENGINEERING THERMODYNAMICS 



initial to back pressures, these being equal. Fig. 90 gives a curve showing the 
relation between cylinder ratio and ratio of expansion established by the above 
condition. 



p 




H.P. Cyl. Vols. 
1 2 


/\ 






B 
























\ 












\ 










® 


\ 














\ 












\ 


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D 










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L 


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A 






B 






























































































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C 






















































































F 






















D 























5 4 3 2 1 
L.P. Cyl. Vols. 



p 






















fs 






3 












































































































© 


















C 










C 




















\ 
\ 
\ 
















































X 


-k^ 






D 
















■^^ 


»„^ 


















1 





Fig. 89. — Special Case of Cycles IX and X. Complete Expansion in both Cylinders of the 
No-Receiver Compound Engine, Zero Clearnace. 



Example 1. Method of calculating Diagram, Fig. 87. 

A. As described in the text this diagram is drawn to two-volume scales, so that 
there may be two volumes for one point. 



WORK OF PISTON ENGINES 



299 



Assumed data: 

P a =P b = 120 lbs. per sq.in. abs. V a = V c = V d = V g = V f = cu.ft. 

P e =P f = 10 lbs. per sq.in. abs. V b = 1 cu.ft. 

V c =2 cu.ft. 



To locate point C: 



To locate point D: 



_ P b V b 120X1 „„ 

P c =— - — = — - — =60 lbs. per sq.in. 

Vc ' " 



_ PcV c 60X2 

P d =— r-— = — -—=24 lbs. per sq.m. 
Vd 5 



To locate intermediate points from C to D. The volume at any intermediate 
point is (the volume of low-pressure cylinder up to that point) +( volume of high- 

















































































































































































































































































s 








































s 


y 








































d. 


/ 











































10 15 

Values of R 



20 



Fig. 90. — Curve to Show Relation between Values of Re, the Cylinder Ratio, and R the Ratio 
of Initial to Back Pressure for Complete Expansion in the No-receiver Compound 
Engine without Clearance (Eq. (413).) 



pressure cylinder from that point to end of stroke), e.g., at I stroke the volume in 
low, is .75x5, and the volume in the high is .25x2, or total 4.25, and the pressure 
at that point is found by the PV relation as above. 



B. Assumed data: 



Pa =Pb = 120 lbs. per sq.in. abs. 
P e =Pj= 10 lbs. per sq.in. abs. 



7 o = V>=0cu.ft. 

V d = V e =5 cuit. 
V = l cuit. 



To locate point D: 



P d = 



P b V b 120X1 



=24 lbs. per sq.in. 



Intermediate points from B to D found by assuming volumes and computing 
pressures from the PV relation as above. 

C. Figure ABCD constructed as in A. Figure C'D'EF is figure CDEF of A to the 
same pressure scale but to a volume scale 2.5 times as large. 

D. Figures constructed as in C. 



300 ENGINEERING THERMODYNAMICS 

To draw indicator cards. The volume and pressure scales are chosen and from 
diagram A, a distance AB is laid off to the volume scale, AD is then laid off equal to 
AD of diagram A to the pressure scale. Point C is located to these scales and joined 
to B and D by drawing curves through the intermediate points plotted from the PV 
diagram to the scales of the card. For the low-pressure card EF is laid off to the 
volume scale, and FC and ED' to pressure scale. C and D' are then joined in same 
manner as C and D for high-pressure card. 

Example 2. Find (a) the horse-power, and (b) steam used per hour for a 12 X 18 X 24 
in. engine with no clearance when initial pressure is 150 lbs. per square inch absolute, 
back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., and cut-off in 
the high-pressure cylinder is |, there being no receiver. 

(a) From Eq. (409) we have 

(m.e.p.) = (in.pr.) [1 +log e (R H Rc)] -(bk.pr.), 



= 150X o i oc X(l+.8)-10=50 lbs. sq.in. 
Z XZ.Zo 



hence 



I.H.P.=192. 

(b) From Eq. (411) we have 

Cubic feet of steam per I.H.P. per hour=-r— rX— tt> 

(m.e.p.) RhRc 

_13 1 750 v ^ fil9 

" 50 X 2 X 2.25" bl ' A 

hence the weight of steam used per hour will be 

61.2 X.332X 192 =3890 pounds. 

Example 3. What will be the cylinder ratio and the high-pressure cut-off to give 
equal work distribution for a ratio of expansion of 6, an initial pressure of 150 lbs. per 
square inch absolute and back pressure of 10 lbs. per square inch absolute? 

Ratio of back to initial pressures is .067 and 



hence from Eq. (412) 



or 



and from Fig. 88 



Rv=6, 

2R C 

log e ^ =1.40, 

6 

2R C 

Rc Rc-i=24M, 
Re =2.8. 



7? o o 

From the relation Z# = — = high-pressure cut off =- J — =.446. 
Kv o 



WORK OF PISTON ENGINES 



301 



Prob. 1. A compound locomotive has no receiver and runs on an initial pressure of 
175 lbs. per square inch gage and atmospheric exhaust. The cylinders are 18 and 
30x42 in. The steam pressure may be varied, as may also the cut-off to a limited 
degree. For a speed of 200 R.P.M. and a constant cut-off of f, find how the power 
will vary with initial pressure and for constant initial pressure equal to boiler pres- 
sure show how the power at the same speed will vary from \ cut-off to full stroke. 

Prob. 2. Show how the steam used per horse-power hour will vary in above 
problem. 

Note: d for 175 lbs. =.416. 

Prob. 3. With the cut-off at f , what should the initial pressure be to give equal 
work distribution? 

Prob. 4. With full boiler pressure and f cut-off what would be terminal pressure 
in the low-pressure cylinder? 

Prob. 5. What must be size of cylinders for a tandem compound engine with 
negligable receiver volume to run at 125 R.P.M. with complete expansion and equal 
work distribution on an initial pressure of 125 lbs. per square inch gage and a back 
pressure of 5 lbs. per square inch absolute, when carrying a load of 500 horse-power, 
the piston speed to be less than 500 ft. per minute? 

Prob. 6. What will be the steam used by the above engine in pounds per hour? 

Note: 8 for 125 lbs. =.311. 

Prob. 7. A builder gives following data for a tandem compound steam engine. 
Check the horse-power and see if the work is equally divided at the rated load. 
Cylinders 10 ins. and 17^X15 ins., initial pressure 125 lbs., speed 250 R.P.M., horse- 
power 155. Neglect the receiver volume. 

Prob. 8. Another manufacturer gives for his engine the following, check this: 
Cylinders 20 and 32x18 ins., initial pressure 100 lbs., atmospheric exhaust, speed 
200 R.P.M., horse-power 400. Neglect the receiver volume. 



14. Compound Engine without Receiver, Exponential Law, Cycle X. 
General Relations between Dimensions and Work when High-pressure 
Exhaust and Low-pressure Admission are Coincident. Referring to Fig. 
S7D it is desirable first to evaluate the work areas CDKG and C'D'NK. As 
before, 



T7T ~ 1 



GL = D H 



Rc-l 



and KL = D H 



Re 
Rc-V 



hence 

but 
so that 



(rel.pr.)jy = (in.pr.)Z^, 



']. 



TF^^in.pr.) 



DhZh 
s-l 



1- 



8-Z B ^- 1 -Z B » 



Re 



Rc-l 



• (414) 



_ P C XKL \ (KLV-n _ (rel. P r.)„ D H R C \ ( 1 y-H 



ENGINEERING THERMODYNAMICS 



whence 



o J. 



i- L - r 

Re 
'Rc-l 



144(bk.pr.)Z>£. • • (415) 



It is to be expected that the sum of high- and low-pressure work will be 
of a form similar to that which would be obtained if all work were performed 
in a single cylinder of a displacement equal to that of the low pressure, adding, 



Tr = Tr H +TF£ = 144(in.pr.) 



DjrZ, 



8-1 



,_'*.-i-^-i(izy_) 



+Z H '- 1 R ( 






Rc-l 



144(bk.pr.)2)i 



= 144(in.pr.) 



D H Z H 

8-1 



s -Z^-i+Z^-^l-(iy 11 -144(bk.pr.)i>i> 



whence, substituting b^=^- 



TF 



= 144(in.pr.)^J[s-(1 ; ) S " l ]-144(bk.pr.)i) i ; (416) 



(m.e.p.refaoLPO^^I^-^^" 1 ]-^,) ..... (417) 
Work per cu.ft. supplied = 144^2^) L_ /_*V" 1 -I44(bk.pr.)|^ . (418) 



Cu.ft. supplied per hr. per I.H.P.= 



13,750 



(m.e.p. ref. to L.P.) R c 
13,750 



(m.e.p. ref. to L.P.) R v ' 



5". - (419) 



Conditions for equal division of work between high- and low-pressure 
cylinders may be obtained by equating Eqs. (414) and (415), 



8-Z h -i-Zh 



.8-1 



w 



IV- 

c, 



Rc-l 



= Z H °-iR c 



\Rc) 
Rc-l ' 



/bk.pr.\fl c 



WORK OF PISTON ENGINES 303 

Rearranging 

is^-['-&n««-'-(!^)i<-»=»- 

The last term in the first member of this equation may be expressed as 

\m.pr. / 



and the relation 






Rv 

exists between Z H and Rc, hence, making these substitutions, 

|±J [ ^- 1 -l H -^- 1 =B F .-x[ g+ (gg-) iSF(s - 1 )]. . (420) 

which is not a simple relation, but can be solved by trial. 

The assumption of complete expansion in the low-pressure cylinder (it is 
always complete in high, for this cycle), leads to this following relations: 

(my*-- 



hence 



144(bk.pr.)Z>L = 144(in.pr.)|~D^ 



and from Eq. (414), 

W 1 A AC sD H Z H [ / 1 \ RcS—\,~\ 



but 



1 

Rc t> ■, Rvs — l s— 1 , 1 /bk.prA* 

-—=R V , and w--^r=w~^i^ and > ^~ = l 7— 
L H L H Rc Rv s l Rv \in.pr. / 

r = 144(in.p,)^ fi ^ I [l-(g^) V ]. . . . (421) 
(m .e.p. re f.toL.P.)=in.p,| s 4 I [l-(£f)" 1 ]. • • (422) 



304 



ENGINEERING THERMODYNAMICS 



The expression for equality of work Eq. (420) becomes, for this case of complete 
expansion, 



txc — 1 



. (423) 



by which it is not difficult to find the ratio of expansion R v , which gives equality 
of work for given values of s, and R c , the cylinder ratio. Values for R v for 
various values of R c and s are given by the curves of Fig. 91. 



10 



£ 5 
> 



























1.5 


1.4 


Values of 
1.3 1.2 


S 


1.1 


























y> 


s 


































/, 




/ 


































^ 


'<? 


^ 
































& 


g 


^ 
































>t 


4 


'# 


>* 
































A 





<y 





























































































































































































10 
Values of Rv 



15 



Fig. 91.— Curves to Show Relation between R c the Cylinder Ratio, and R v the Ratio of 
Expansion, for Various Values of (s), Applied to the No-receiver Compound Engine 
without Clearance, when the Expansion is not Logarithmic. 



Example 1. Find (a) the horse-power, and (b) the steam used per hour for a 
12- and 18x24-in. engine with no receiver when the initial pressure is 150 lbs. per 
square inch absolute, back pressure 10 lbs. per square inch absolute, speed 125 
R.P.M., cut-off in high-pressure cylinder is J, there being no receiver and steam 
having expansion, such that s = 1.3. 

From Eq. (417) 

<»->-«a-(i;rw>. 

which, on substituting the above values, becomes 
150 .5 r / .1 \ 1 

TW 1 - 3 - (il) J ~ 10=C3 ' 3 lbs ' per sq ' in - 

hence the indicated horse-power =243. 



WOEK OF PISTON ENGINES 305 

(6) From Eq. (419) the steam used per hour in cu.ft. per horse-power is 

13/750 Zn 
m.e.p. Re 

which, for the data given above, becomes 

13,750 .5 ian 

T 3 :r x ^r 4S - 2cu - ft - 

or pounds per hour total, is, 48.2 X243X. 332 =3880. 

Example 2. What will be the high-pressure cut-off and cylinder ratio to give 
equal work distribution and complete expansion for an initial pressure of 150 lbs. per 
square inch absolute, and back pressure of 10 lbs. per square inch absolute? 

From relation R V S = \7T-— ), #f=6.9 and from this, by the curve of Fig. 91, 
\bk.pr./ 

Rc=5A. 



For complete expansion 



7 R ° M -S 



Prob. 1. A tandem compound engine without receiver has cylinders 18- and 
30x42-ins. and runs at 200 R.P.M. What will be the horse-power developed at 
this speed if the initial pressure is 175 lbs. per square inch gage, back pressure 
atmosphere, high-pressure cut-off J, and s has a value of (a) 1.1, (6) 1.3? Compare 
the results with Prob. 1 of Sec. 13. 

Prob. 2. What will be the weight of steam used per horse-power per hour for 
the two cases of the above problem? Compare these results with those of Prob. 2, 
Sec. 13. 

Note: S=.416. 

Prob. 3. What must be the cut-off in a 10- and 15x20-in. compressed air engine 
running on 100 lbs. per square inch gage initial pressure and atmospheric back pres- 
sure, to give complete expansion, and what will be the horse-power per 100 ft. per 
minute piston speed, s being 1.4? 

Prob. 4. It is desired to run the following engine at its most economical load. 
What will this load be and how much steam will be needed per hour? 

Cylinders 20- and 32x18 ins., speed 150 R.P.M., steam pressure 100 per square 
inch gage, atmospheric exhaust, dry saturated steam. 

Prob. 5. Should the load increase 50 per cent in Prob. 4, how would the cut-off 
change and what would be th effect on the amount of steam used? 

Prob. 6. What would be gain in power and the economy of the engine of Prob. 4 
were superheated steam used, for which s = 1.3? 

Prob. 7. In a 14- and 20x24-in. engine will the work be equally divided 
between the cylinders for the following conditions? If not, what per cent will be done 
in each? Steam pressure 100 lbs. per square inch absolute, back pressure 10 lbs. per 
square inch absolute, s = 1.2, cut-off = |. 

Prob. 8. What would be the work and steam used by the above engine if there 
were complete expansion and equal distribution? 



30( 



ENGINEERING THERMODYNAMICS 



15. Compound Engine with Infinite Receiver. Logarithmic Law. With 
Clearance and Compression, Cycle XI. General Relations between Pressures, 
Dimensions and Work. In terms of point pressures and volumes, Fig. 92, the 



QA 
r 




B 






(im 


ar)- 












Z„Df 


\ 
















\ 






















- D 


\ 




















H \ 






F 




\ 




















4 






V 


















it 


p xh] 


3h 


\ 


CO 


*el.p] 


r-)H. 












^ 


E 




D 


1 

\ M (rec.pr. 


> 










i 






H\ 
















\ 
\ 


-f — 


ZlI 


>ir^ 




















[ 


















l__ 




4Cl~ 


K 


K 


















tbk.pr.) 


4^|x,r 


, 




Di 










J 


incomplete expansion and: compression. ^ 
F. 


/ 


■£■ 


1 






















A< 


ZhDh 


B 






(in. 


«.) 












5 
























X„D„\ 




















lij 


r 






















H \ 


> 




G J 


u 


h\ 


M 


D 

-(-r< 


xipr 














3fc 


A 


w 


)h 


















\\ 




eL.pi 




















s 


"'h 














k\ 






\ 


V 










J 


( bk.pr.) 




L 














SB 








)l 






u l 














•( rel.pr.) 






















1 






























INDICATOR CARDS OF EQUAL BASE 
AND HEIGHT FOR CASE OF INCOM- 
PLETE EXPANSION AND COMPRESSION. 



H.P. 



^ 




INDICATOR CARDS OF EQUAL BASE 
AND HEIGHT FOR CASE OF OVER 
EXPANSION AND COMPRESSION. 



OVER EXPANSION AND COMPRESSION. v 

Fig. 92. — Work of Expansive Fluid in Compound Engines with Infinite Receiver, with Clearance, 
Cycle X Logarithmic, and Cycle IX Exponential Expansion, and Compression. 

work of the two cylinders may be written down at once as if each were inde- 
pendent of the other, the connection between them being fixed first by making 
the back pressure of the high equal to the initial pressure of the low, or to 



WORK OF PISTON ENGINES 



307 



the receiver pressure, and second by making the volume admitted to the low 
equal to that discharged from the high reduced to the same pressure. This 
last condition may be introduced in either of two ways, 

(a) EM = NH, 

(b) [(PV) on H.P. expansion line— (PV) on H.P. comp. line], 

= [(PV) on L.P. expansion line — (PV) on L.P. comp. line]. 

Without introducing the last relation 

TFH = PftF 6 (l+lo & Q-P / 7 / (l+lo & ^)-(P B -P / )Fa-P d (7 Jl -7.); (424) 

W L = PnVn(^+\og e ~^-PiV l (\+\og c ^-(P g -P l )V g -P j {V j -V^ . (425) 

W=W S +W L ; 

W = P b V b log, ^ + P h V h l0g c P - P e V e log, ~~P k V k log, £ 

Vb Vh Vf Vi 

+ Pt>V b -P f V f -P a V a + P f Vj-P d V d + P e Ve 

+P*Vn-PiVi-P g V g +PiVi-P j V j +P t V t . 

The second condition is 

P b V b -P e V e = P fl V h -P k V k ; or P b V b +P k V k = P e V e +P h V h , 



or 



P b V b + P k V k +P e Ve + P h V h = 2(P b V b +P k V t ). 



Substituting 



W = P b V b log, ~+P h V h log, ^-PeVe l0g e ~PtV t l0g e ^ ] 

Vb Vh Vf Vi 

+2(P b V b +P k V k )-P a V a -P g V -P d V d -PjVj 



(426) 



(417; 



This expression, Eq. (427) contains, however, the receiver pressure which is 
related to the release pressure by 

t \r»nr>r>7i"^i'/i \ L.P. max.vol. 

(rec.pr.)=P^P e = P, = A = P i? -=(rel.p r .), Lpcut _ offvol . 
Introducing this 

W = P t V t l0g.^+P t 7, l0 & ^--Plf-Ve log, J?-P*7* loge ^ ] 

V b Vh Vh Vf Vi 



Vi 



v t 



+2(P & F 6 +P*y*)-P a 7 a -P^F, -P^Va-PjVj 

* h V h 



(428) 



308 



ENGINEEEING THERMODYNAMICS 



Introducing the usual symbols in Eqs. (427) and (428) and in addition 
the following: 

Z — cut-off as fraction of stroke, so that Z H D H is the displacement volume 

up to cut-off. 
c = clearance volume divided by displacement, so that c H D H is the clearance 

volume and (Z h -\-ch)Dh is the volume in the high-pressure cylinder 

at cut-off. 
X = that fraction of the stroke during which compression is taking place so 

that {X H -\rc H )D H is the volume in the high-pressure cylinder when 

compression begins. 



Applying the general symbols to Eq. (427), 

(m.pr.)(Z/7+c#)Z>/,loge (^r--— ) 
\Z H +c H / 



JF = 144 



•+cj 



imc.j)r.)(Z L +c L )D L \oge (^ 
(rec.pr.) (X H +c H )D H log e (— — - ) 
(bk.pr.)(X £ +c i )D i lo& (— c — ) 



+2(in.pr.) (Z H +c B )D H - (in.pr.)c H D H - (rec.pr.) (1 +c H )D H 
+2(bk.pr.) (X L +c L )D L - (vec.pr.)c L D L - (bk.pr.) (1 +c L )D L J 



. (429) 



This expression gives the work in terms of initial, receiver and back pressures, 
the valve periods, cut-off and compression, the clearances and cylinder dis- 
placements. 

Substitution of the symbols in Eq. (428) will give another equivalent expres- 
sion in terms of the same quantities except that low-pressure cylinder release 
pressure will take the place of receiver pressure. This is 



W = 144 



(m.vr.)(Z H +c H )D H log e 



1+Ch 
Zh+ch 



+ (rel.pr.) L (l+c L )Z) / ,log e 



\Zl+cJ 



A~~^)(X H +c H )D H log/- 



h+Dh \ 

-(bk.pv.)(X L +c L )D L 



loge(^) h ,3,n 



+2(m.-pT.)(Z H +CH)DH-(m.pr.)c H DH- (rel.pr.)z,( ~^- j (1 +c H )D H 

\Zl+Cl/ 

+2(bk.pr.)(X z .+ CZ/ )^-(re].pr.) L ('i^J CL Z) i -(bk.pr.)(l+ Cz ,)Z) i 



WORK OF PISTON ENGINES 



309 



It is sometimes more convenient to involve the cylinder ratio and low- 
pressure displacement than the two displacements as involved in Eq. (430) 
and the ratios of expansion instead of cut-offs. This may be done by 



^ C = A 






and 



Rh = ^ = 



\+c„ 



V* Z„+c H ' 



KL V h Z L +C L ' 



(431) 



and it should be noted here that the ratio of expansion in each cylinder is no 
longer the reciprocal of its cut-off, as was the case when clearance was zero, nor 
is the whole ratio of expansion equal to the product of the two separate ones 
because the low-pressure cylinder expansion line is not a continuation of that 
in the high. Making these substitutions for cylinder and expansion ratios, 
Eq. (430) becomes, 



W = 
144Dz, 



(in.pr.)(l +c#)— — loge Rn + (rel.pr.) Z/ (l +c L ) log e R L 

liHtlC 

- (rel.pr.M^i/ + c *)f* lo &(^^) ~ ( bk 'P r ') &* + c ^ lo &( 

+2(in.pr.)(l +c^)-i- -(in.pr.)^- (rel.pr.)^(l +c H ) 
KhKc tic He 

+2(bk.pr.)(X L +c Zi )-(rel.pr.)£^if£-(bk.pr.)(l+cz # ) 



Xl+Cl 



(432) 



It is interesting to note that this reduces to Eq. 304 of Section 9, by making- 
clearance and compression zero. 

From any of the expressions for work, but more particularly (430) and 
(432), the usual expressions for (m.e.p.) referred to low-pressure cylinder, work 
per cubic foot supplied, and consumption per hour per I.H.P. can be found, 
but as these are long they are not set down, but merely indicated as follows: 



Work per cu.ft. supplied 



W 
(m.e.p. ref . to L.P.) = j^- . . . 

= W 

D H [(Z H +c H )-(X H +c H )(^^] 



(433) 
(434) 



Cu.ft. sup. per hr. per I.H.P. 
13,750 



(m.e.p. ref. to L.P.) 



[(Zh 



+c H )-(X H +c H ) 



rec.prAI 1 
in.pr. J ]Rc 



13,750 



(m.e.p. ref. to L 



Poh^)-^+-)(S£)](5r)- (435) 



310 



ENGINEERING THERMODYNAMICS 



As the receiver pressure is related to the initial and back pressures and to 
the relation between the amount taken out of the receiver to that put in, which 
is a function of the compression as well as the cut-off and cylinder ratio, it is 
expressed only by a complicated function which may be derived from the 
equivalence of volumes in the high and low, reduced to equal pressure. 

P b V b rPeV e = PnV h -P t Vt J Or PnVn+PeVe = P b V b + P & V k =P h (V h +V e ). 

Therefore, 



V* 



Vn + Ve 



-f-pi. 



V h + Re- 



introducing symbols 

(rec.pr.) = (in.pr.) 
Hence 



(Zh+Ch)D h 



(rec.pr.) = (in.pr.) 



(Z l +c l )D l +(X h +ch)Dh 

(Z h +Ch) 



-Kbk.pr.) 



(X L +c L )D L 



(Z L +c L )Rc-\-(X H +c H ) 



Kbk.pr.) 



{Z L +c L )D L +{X H +c k )D H ' 

(X L +c L )R c 



(Z L +c L )R c +(X H +c H y 



(436) 



This Eq. (436) gives the receiver pressure in terms of initial and back pressures, 
the two clearances and compressions, the cylinder ratio and the cut-off in each 
cylinder. 

Proceeding in a similar way, the release pressures can be found in terms of 
initial data, 



or 



Vr 

C — -Lbj-r J 

Vb 



(rel.pr.)tf= (in.pr.) 



(in.pr.) p-. 
tin 



Zh±Ch\ 

l+c B .) 

1 



(a) 
(b) 



(437) 



And 



or 



(rel.pr.)i = (in.pr.) 



= (in.pr.) 



Pi — PffFT — Pb 

V i 



( (Z H + c H ) 
\(l + c L )Rc 




1 + 



(X H +c n ) 
(Z L +c L )R c 



+ (bk.pr.) 



1 + 



1 



1 + 



1±Ch\ 
1 + cJ'RhRc 
(X h +c h )R l 
{l+c L )Rc 



+ (bk.pr.) 



.1 + 



(l+czj__ 
(Xh±c h ) 
(Z L +c L )R c 
(X L + c L ) 

(X H + C H )Rl 



(1+c l )Rv J 



[fl) 



(6) 



(438) 



WORK OF PISTON ENGINES 



311 



These three pressures all reduce to those of Eqs. (308), (309), (310), Section 
9, when clearance and compression are zero. 

Equal work in both cylinders is, of course, possible, but it may be secured 
by an almost infinite variety of combinations of clearance, compression and 
cut-off in the two cylinders for various ratios of expansion; it is, therefore, not 
worth while setting down the equation of condition to be satisfied, but reference 
may be had to Eqs. (424) and (425), which must be made equal to each other, the 
result of which must be combined with the equation of cylinder relations. 























A 




B 
















1 








































\ 






















\ 


















G 
\ 


V 




h\ 


c 


















\ 


v 




















\ 






















SK 














-^ 1 









































INDICATOR CARDS OF EQUAL 
BASE AND HEIGHT 



V 

Fig. 93. — Special Case of Cycles XI and XII Complete Expansion and Compression in both 
Cylinders, of Compound Engine with Clearance and Infinite Receiver. 

There are certain special cases of this cycle for which equations expressing 
important relations are simpler, and they are for that reason worth investigat- 
ing. Those that will be examined are 

(a) Complete expansion and compression in both cylinders, Fig. 93. 

(6" Complete expansion in both cylinders with no compression, any clearance, 
Fig. 94. 

(c) Any amount of expansion and compression but equal in both cylinders, 
equal clearance percentages and a cylinder ratio equal to the square root of 
the ratio of initial to back pressures, Fig. 95. 

Case (a) When both expansion and compression are complete in both cylin- 
ders, Fig. 93, 

(in.pr.) 



W H = 144(in.pr.)Z*Z>* log. (gg) , 
TF i = 144(rec.pr.)Z i Z) i log e (§^), 



(439) 



(440) 



312 ENGINEERING THERMODYNAMICS 

but 

Z H D H (m.j)T) = ZzjDz, (rec.pr.) 
and 

. / (in.pr .)\ M /(rec.pr.)\ , /(in.pr. ) (rec.pr.)\ , (in.pr. ) 

log « fccTpTo) +logc ( (bkTpFo) =loge fc^) (bEfo) =l0& (bkfa' 

hence 

IF=144(in.pr.)Z^log e gg^ (441) 

(m.e.p.ref.toL.P.) = (in.pr.)jlog, ({jg^) (442) 

Work per cu.ft. supplied = 144(in.pr.) log e (tut^t) (443) 

Consumption, cuit. per hr. per LH.P. = -. ' ■=£-. . . . (444) 

(m.e.p. ref. to L.P.) R c 



Equality of work in high- and low-pressure cylinders is obtained by making 

/ (in.pr. ) \ /(re^prOX = /(in.pr. )\ i 

\(rec.pr.)7 \(bk.pr.)/ \(bk.pr.)/ ' 

or 

(rec.pr.)= [(in.pr.) (bk.pr.)]* (445) 

It is desirable to know what clearances and displacements will permit of 
equal work and complete expansion and compression. 

(rec.pr.) = (m.pr.)^- = (m.pr.) 



V c -~— \ i+c H 



hence 



or calling 



=(bk, P ,)£ =(bk .p,)(^), 

/( in.pr. )\ = ±+Ch_ = /(in .pr. ) *\ 
\(rec.pr.)/ Z H +c H \(bk.pr.) / ' 

/ (rec.pr.) \ l+c L = / (in.pr. ) \ * 
\(bk.pr.)/ Z L +c L \(bk.pr.)/ ' 



/(in.pr. )\ _ 



&— -g-j ■, and, Z L = --^ . . . . (446) 

Equating discharge of high and intake of low-pressure cylinders, 
ZhD h R p $ = Z l D l or 1 ?r = Rc = T L Rp*. 



WORK OF PISTON ENGINES 
Inserting in this the values just found for Z H and Z L , 



He 



1+Cl — ClR, 






313 



(447) 



which is the required relation between cylinder sizes, clearances and ratio of 
pressures, which, together with cut-offs given in Eq. (446), will give equal work 
and complete expansion and compression. The compression in the high- 
pressure cylinder is such that 

Xb=cb(Rp*-1) I 
and for L.P. cylinder. v -._/»_*_ -n 

P 



X L = c L (R P *-l). 





























L A 




B 
























\ 
























\ 
























\ 


























\ 


















EM 






\ 


C 














r» 






H\ 
































K 








































- 1 

















































A 


B 


E X( 


G 


H 


K 


V^ 



INDICATOR CARDS OF EQUAL 
BASE AND HEIGHT 



V 
Fig. 94. — Special Case of Cycles XI and XII. Complete Expansion and Zero Compression 
in both Cylinders of Compounds Engine with Clearance and Infinite Receiver. 

Case (6) With complete expansion and no compression, both cylinders, any 
clearance, Fig. 94, 

TF* = 144D„[(in.pr.)(Z*+c*)loge (^~ } ) -^[(in.pr.)-(rec.pr.)]] (449) 

W L = lUD L [(rec. W .)(Z L +c L )\og e ({^—j) -cj(rec.pr.)-(bk.pr.)]] (450) 

with the added requirement that the high-pressure discharge volume, EC = low 
pressure admission volume FH, or 



£>* = £>J 



Z L +c L -c L 



(bk.pr. ) 
(rec 



•pr.)/J' 



(451) 



314 ENGINEERING THERMODYNAMICS 

and 

(rec.pr.) = (bk.pr.) ^ = (bk.pr.)^-~S • 

hence 

r (bk.pr .)l = Z L +c L 

L(rec.pr.)J l+c L ' 
which substituted in Eq. (451) and rearranging gives 



(452) 



Rc = * , , , (453) 



„ , ( Z L +c L \ 



Eq. (453) indicates that for this special case of complete expansion and no com- 
pression the cylinder ratio required to give this case, is determined entirely by 
the L.P. cut-off and clearance. If the cylinder ratio and clearance are fixed, the 
required cut-off in the L.P. cylinder can be found by solving Eq. (453) for Z L , 

Z L = l -±^-c L , (454) 

lie 



and from Eq. (452), 

(rec.pr.) = (bk.pr.) 



D —C L -\-C L 
He 



= (bk.pr.)#c. . . (455) 



Cut-off in the high-pressure cylinder is determined by clearance, initial 
pressure and receiver pressure, which in turn depends on low-pressure cut-off 
and clearance Eq.(452), or may be reduced to cylinder ratio and low-pressure 
clearance by Eq.(454), as follows 

V c= l+c H = / ( in.pr . )\ = /(in .pr. ) \ Z L +c L 
V b Zh+Ch \(rec.pr.)/ \(bk.pr.)/ l+c L ' 
hence 



Z H = t> ,rl\ _ n c H . 



( l+Cir)(l+(fc ) 

R P {Z L +c L ) 

Eliminate Z L by Eq. (454), 



(l+to)(l+Ci) M . .R c , At . aS 

ii c H =(l+to)o — c H . . . . (456) 

i-j-Cl ,. \ ftp 



(R P ^-c L +t L \ 



Since the high- and low-pressure cut-offs are functions of cylinder and clearance 
dimensions, and of Rp, the rato of initial and back pressures, the work of highl- 
and low-pressure cylinders may be expressed entirely in terms of these quantities. 

J^ = 144D*(in.pr.){(l+c^ . . (457) 



Wi 



= 144D«i2c(bk.pr.)[(H-cJlogefic-Cz.(/2c-l)]. • • . L (458) 



WORK OF PISTON ENGINES 315 

Hence, total work by addition is 



W=lUD H (m.pv.)^ 

rip 



(l + C*)lo & (^)-C*(^-l 



R 



+ (l+c L )\og c Rc-c L {R c -l) 



(459) 



Expressions might be easily written for mean effective pressure referred 
to the L.P. cylinder, work per cubic foot fluid supplied, and consumption, but 
will be omitted for brevity. It is important to note, however, the volume of 
fluid used per cycle is not AB, but is LB, Fig. 94, and is, 



(Sup.Vol.) =D h \(Z h +Ch) -Ch^-°— PJ 4] =Da\&s+te)-cJr I ■ ■ 



(460) 



W 
(m.e.p.ref.toL.P.)=j ii5 ^ c (461) 

W 
(Work per cu. ft. supplied) =-^ vTT (462) 

Consumption cu.ft per hr. per I.H.P. = -^ , A (Z H +c H ) 

Equality of work, secured by equating Eqs. (457) and (458) gives 

(l+^)log e ^-^^-l) = (l+cJlog CJ Rc-^(^-l). . (464) 

This equation may be satisfied in an infinite number of ways. One case 
worth noting is that of equal clearances, when it is evident that if 

7? 

Ch = c l , and -^ = Rc, or R C = VR P 
Kc 

the Eq. (464) is satisfied. This last condition is the same as that which satisfied 
Case (a) with complete compression. 
Case (c), Fig. 95, assumes that 

ch = cl = c, Zh = Zl = Z, Xh = Xl = X, 
and 



316 



ENGINEERING THERMODYNAMICS 



and corresponds to the first special case considered in Section 9, which lead 
in the no-clearance case to equality of high- and low-pressure work. 



^A 




3 















1 








1 

A 




ZD„— i 
















cl 


S 


\ 


> 














\ 




F 


H 


\ 
















\ H.P. >v 








\ 
















\ 












\ 














1 








1 






\ 


C 
















X 


XDh 






\ 
\ 
\ 




'H 








\ 


f\ 


G 






s 

D 


\H 


— (r 


jc.pr.) 






\ \ 












\ \^ 


<-e-D I 


(l 








' 










\ 






















sl.pr.) L 










j'XI 


J K| 




















K 


















>"■"" 












d: 












H 





Fig. 95. — Special Case of Cycles XI and XII, Equal Per Cent Clearance in Each Cylinder of 
Compound Engine with Infinite Receiver and Cylinder Ratio Equal to the Square Root 
of Initial Divided by Back Pressure. 

The assumptions already made are sufficient to determine the receiver pres- 
sure. By Eq. (436) 

rv , , /in.pr. \* 

/ \ r \ ^+ c . /ui \ Vbk.pr. 

(rec.pr.) = (m.pr.) T . r^ h(bk.pr.) 



^m i+x+c 



)ife. ) • 



"S:) -^ + - 



(Z+c) 



m.pr. 



[(in.pr.) (bk.pr.)]* 



+X+c 



bk.pr v 

= [(in.pr.) (bk.pr.)]* (465) 

The work of the high-pressure cylinder may now be evaluated. 



W H = 14±D H (in.pT.) Jz|\+loge . 



- 144ZM (in.pr.) (bk.pr.)]*j (X+c) log e ^t^)+l-x| . . 



(466) 



WORK OF PISTON ENGINES 317 

The low-pressure cylinder work may be similarly stated, 



W L =-= 144Z)J(in.pr.) (bk.pr .)]* | Z [l +log e (j^ 



(467) 



-144Z) L (bk.pr.) l(X+c) loge (— J^+l-X 1, 

but 

D4(in.pr.)(bk.pr.)]* = 2)^c[(in.pr.)(bk.pr.)]* 

-^(SS)Vm.pr.)(bk.pr.)l* 

= D/,(in.pr.),1 

and similarly, 

D^bk.pr.) = £*[ (in.pr.)(bk.pr.)]* ) 

With these substitutions the value of low-pressure work,W L ,~Eq. (467), becomes 
equal to high pressure work, Eq. (466), hence the total work 

W = 2W H = 2W L (468) 

Example 1. Method of calculating Diagrams, Fig. 92. 
Assumed data: 

Pq=P a =Pb = 120 lbs. per square inch abs. V a = V/= .12 cu.ft. 

P n =Pg=Pe=Pd=Ph= 50 lbs. per square inch abs. Vb = A cu.ft. 

Pi =Pj = 10 lbs. per square inch abs. V c = V* = .8 cu ft. 

V g = Vi = . 16 cu.ft. 
F,= F/= 2 cu.ft. 

V e = .2 CU.ft. 

(F ft -7n)=(F„-7e). 7*= .4cu.ft. 

The above may be expressed in initial pressure, etc., and in terms of cut-off, etc., 
but as the relation of the lettered points to these terms is shown on the diagram values 
for cut-off, etc., they will not be given here, as they may readily be found from values of 
the lettered points. 



To locate point C: 



To locate point F: 



To locate point Q: 



p p F & 120 X. 4 

P C =P 6 — = — =60 lbs. per sq.m. 

V c «o 



p PeVe 50X.2 

i J j = ——= — — — =83.3 lbs. per sq.m. 



r.-^-SB-*"-* 



318 ENGINEERING THERMODYNAMICS 

To locate point L: 



_ PjcVjc 10 X. 4 

Pi = —— = = 25 lbs. per sq.m. 

Vl .10 



.To locate point N: 



PuVk 4 



To locate point H: 

(V fl -Vn)=(V m -V e ), or F ft = F m + F ra -7 e = .96 + .08-.2 = .84cu.ft., 
since 



To locate point I 



48 
PmV m = P 6 7 6 , = 7 W =- = .96 cu.ft. 
50 



_ P.F/, 50X.84 

Pi = — — = — - — =21 lbs. per sq.m. 

V i & 



Example 2. Find (a) the horse-power, (b) steam used per hour, and (c) receiver 
and release pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per 
cent clearance in high-pressure cylinder, and 4 per cent clearance in low-pressure cylinder, 
when initial pressure is 150 lbs, per square inch absolute, back pressure 10 lbs. per 
square inch absolute, speed 125 R.P.M., cut-off in high-pressure cylinder is J, low- 
pressure cut-off is such as to give complete H.P. expansion, and compression is 15 per 
cent in high and complete in low. 

(a) For complete high-pressure expansion the receiver pressure must be equal to the 
high-pressure release, and to maintain the receiver pressure constant the low-pressure 
cylinder must take as Lauch steam per stroke as the high-pressure discharges. With 
initial pressure and cut-off as given, the release pre3sure for the high-pressure cylinder 
may be found from the relation (in.pr.) (c H +Z H ) = (rel.pr.)# (c H +D H ) or 150 X (.56) 
= (rel.pr.)#(1.04), or (rel.pr.)# = 79.3 lbs. Since there is 15 per cent compression in 
high-pressure cylinder there is exhausted each stroke 85 per cent of its volume. Also 
since compression in low-pressure cylinder is complete, the low-pressure clearance is 
full of steam at the receiver pressure at the beginning of the stroke. Hence the 
low-pressure displacement up to cut-off must equal .85D H or L.P. cut-off = .852)^, 
divided by cylinder ratio, or .85-7-2.25 = .378. As compression is complete, the 
per cent compression may be found from the relation cz, X(rec.pr.) =(cz,+Xi)(bk.pr), 
or .04X79.3 = (.04 +X L ) 10, or X L = .28. 

From Eq. (432), (m.e.p.) referred to low-pressure cylinder is obtained by dividing 
by 144 Dl, and on substituting the above values it becomes, 

150(.1-K06)(— |--) log, 2 +30(1 +.04) log e 2.64-30(.15 + .06) (|~) log e (^~) 

-10(.28+.04)log e (J*$L) +2Xl50(l+.06)(-y -150X^ 

^2 64 
-30X~(l+.06)+2xl0(.28 + .04)-30x2.64x.04-10(l+.04)=60.5 lbs. per. sq.in., 

Z.Zo 

hence 

I.H.P.=235. 



WOEK OF PISTON ENGINES 

(b) From Eq. (435) by substituting the above values 

r 

13,750 



319 



Cu.ft. steam per hour per horse-power = 



60.5 



( .3 8+ .04)-(.28 + .04)~-j^f]=45.5, 



_10\79. 



or pounds per hour will be 3550. 

(c) Release pressure for high-pressure cylinder has been shown to be 79.3 lbs. 
and may be checked by Eq. (437), as follows: 

(rel.pr.k = 150(^^t^') 



Receiver pressure has already been shown to be equal to this quantity and may 
be checked by Eq. (436) 



(rec.pr.)tf = 



150 X (.5 + .06) 



+ 



10x(.28 + .04)2.25 



(.378 + .04)2.25 + (.28 + .04) (.378 + .04)2.25 + (.28 + .04) 



79.3 lbs. 



Low-pressure release pressure is found from Eq. (438) to be 



(rel.pr.)z,= 150 



1+.06 



X 



1 



1 +.04 2x2.25 



1 + 



(.15 + .06)2.64 



L, (1 +.04)2.25 J 



+ 10 



.28 + .0 4 
1+.04 



(.15 + .06)2.64 
L + (1 +.04)2.25 J 



= 30 lbs. 



Prob. 1. What will be the horse-power and steam used by the following engine 
for the data as given? 

Engine 20 and 28x36 ins., running at 100 R.P.M., clearance 5 per cent in high 
pressure. 3 per cent in low. From cards H.P. cut-off = .3, L.P. = .4, H.P. compression, 
.1, L.P., .2. Gages show (in.pr.) to be 150 lbs., (r c.pr.) 60 lbs., (bk.pr.) 26 ins. 
Hg. (barometer = 30 ins.). 

Prob. 2. What must be the cut-offs and the cylinder ratio of an engine to give 
equal work and complete expansion and ompression for 200 lbs. per square inch 
absolute initial pressure and atmosphe+c exhaust, if clea ance is 5 per cent in the 
high and 3 per cent in the low-pressure cylinder? What will the horse-power for an 
engine with a low-pressure cylinder 24x36 ins., running at 100 R.P.M. for this case? 

Prob. 3. Should there be no compression, how would the results of Prob. 2 be 
altered? 

Prob. 4. What will be the total steam used by engines of Probs. 2 and 3? 

Prob. 5. For an 11- and 19x24-in. engine with 5 per cent clearance in each 
cylinder, \ cut-off in each cylinder, and 20 per cent compression in each cylinder, what 
will be the horse-power and the steam consumption when the speed is 125 R.P.M. , the 
initial pressure 150 lbs. per square inch gage, and back pressure at atmosphere? 



16. Compound Engine with Infinite Receiver. Exponential Law, with 
Clearance and Compression, Cycle XII. General Relation between Pressures, 
Dimensions and Work. Referring to Fig. 92, of the preceding section, which 



320 ENGINEERING THERMODYNAMICS 

will represent this cycle by a slight change of slope of the expansion and com- 
pression lines, the high-pressure work may be expressed in terms of dimensions, 
ratios and pressures. Since this must contain receiver pressure as a factor, 
and since that is not an item of original data, it is convenient first to state 
receiver pressure in terms of fundamental data: 



V m -V e = Vn-V n . 

Vm = V J^-Y and V n =V k (™^'Y s 
\rec.pr./ \rec.pr./ 

v JjmeJ +v VWtprA- = Vh +V e , 

\ rec.pr./ \rec.pr./ 

, \ r A \in.pr. / / 
(rec.pr.) = (m.pr.)y V h +V e / ' 

or in terms of dimensions and pressures, 

(z B +c B +Rc(X L +c L ) (£!E*EA 
/ \ r \\ \in.pr. / 

(rec.pr.) = (in.pr.) I - 



But 



Whence, 



whence 



Rc(Z l +c l )+X h +ch 

The high-pressure work may be stated as follows: 
(Z b +ch) 



W H =lteD H \ (in.pr.) 



1 



.^^) S " 1 ]-(rec.pr.)(l-X.) 



-(rec.pr.)--^^^— ^— j 1- J -(m.pr.)c* J 
= 144D,(in.pr).(-- I -^-^ TqF ^j j - c H 

Z H +c H +R c (X L +c L ) I^L) 
\m.pr. / 

Rc(Z l +c l )+Xh+Ch 

"^^^^y- l 1 _j +1 _ x 1 (471) 



144D.(in.pr.)^ - Bc{Zl+ c£)+X b +c b ' J X 



Xi 

s- 



WORK OF PISTON ENGINES 



321 



W L = 144D L \ (rec.pr.) ( ^^[ S - (-^g)'" 1 ] -(bk.pr.)(l-X.) 

'PS*)"-]-*— M- 



(bk.pr.)— — r 



(472) 



Zh+ch+R< 



= 144D^c(m.pr.)' 



\in.pr. / 



Rc(Z l +c l )+X h +ch 

z l + Cl \ / z L +c L y-i-] 



-144D^c(bk.pr0{^p[(^^) S -l]+l-X, 1. . (473) 

The expression for total work need not be written here, as it is simply the sum 
of Eqs. (470) and (472) or of (471) and (473), the former containing receiver pres- 
sure and the latter containing only dimensions, initial and back pressures and 
both, the exponent of the expansion, s, 

The volume of high-pressure fluid supplied per cycle is QB, Fig. 92, which 
may be expressed either in terms of high pressure or of low pressure points, 
thus; 



(Sup.Vo\.)=D H [z H +c H - (X H +c H )(j^j S (a) 



(474) 



The following quantities will be indicated, and may be evaluated by sub- 
stitution from the preceding: 



W W 

(m.e.p.ref.toL.P.) = I3i5 - = ni ^ 



HJ^c 



Work per cu.ft. fluid supplied = 



W 



(Sup.Vol.)' 
Consumption cu.ft. per hr. per I.H.P. 

13,750 



(475) 
(476) 



(Sup.Vol.) 



(m.e.p. ref. to L.P.) D L 



. . . (477) 



Equal division of work between high- and low-pressure cylinders requires 
that Eqs. (470) and (472), or (471) and (473) be made equal. The latter will 



322 ENGINEERING THERMODYNAMICS 

give an expression showing the required relation between dimensions and initial 
and final pressures, cut-off and compression in high- and low-pressure cylinders. 
In this expression there are so many variables that an infinite number of com- 
binations may be made to give equality of work. 

It is desirable to examine the results of assuming special conditions such as 
those of the previous section, the most important of which is that of complete 
expansion and compression in both cylinders, which is represented by Fig. 93. 

TF„=144(in.pr.)ZHD. s 4 I [l-(^) ' ]. ... (478) 

W*-144^)Z^[l-(g&)-], . . . (479) 
but 

\rec.pr./ 
hence 

,= lM(,p r )^j[l-(gf] 

1_ S-_l 

/rec.prA / in.pr. y r /bk.pr.X • 1 
\ in.pr. /\rec.pr./ [ \rec.pr./ J J 

= 144(in.pr.)D^Z^^— - I _ / rec.prA * 
1 \ in.pr. / 

8—1 S—l S—l . 

/re c.pr. \ s _ A^P£A s /bk.pr. \ T" ! 
\ in.pr. / \ in.pr. / \rec.pr./ J 

= 144(in.p,)^^[l-(^) M ] (480) 

The receiver pressure may be found as follows. In Fig. 93, EC = GH: 

Equating 

l+CH .j}^K Rc \i+ CL -j^^Y](^^Y . (48D 

Vrec.pr./ \bk.pr./ J \rec.pr./ v ' 






WORK OF PISTON ENGINES 



323 



When this is solved for receiver pressure it results in an equation of the 
second degree, which is somewhat cumbersome, and will not be stated here. 
Eq. (481) is, however, used later to find R c . 

If work is to be equally distributed between high- and low-pressure cylinders, 
from Eqs. (478) and (479), 



s-l 

rec.pr. \ s 
b in.pr. / 



/rec.pr 
\ in.pr. 



!-l 



8-1 I 

rec.prA s /bk.pr. V 
in.pr. / \rec.pr./ 



-l 

s 



or 



8-1 

^ rec.prA • =1+ /bk^r 
\in.pr. 



in.pr. 



s 



1 
































































































































































































































































































\ 


\ 
















































\\ 
















































\ 


^ 














































\ 


^ 


^N 














































^ 








































































































































b 

S 


■Lit 

k.4 














































s 


=1.3 

-1t2 
=1.1 














































s 














































s 


=1.0 



































































































25 



50 



75 



^lues of M) 



100 



Fig. 96. — Curves to Show Receiver Pressure to Give Equal Work Distribution when Expansion 
and Compression are Complete in both Cylinders of the Compound Engine with Infinite 
Receiver, with Clearance when Expansion and Compression are not Logarithmic. 



hence, for equal division of work, 



(rec.pr.) = (in.pr.) 



^/b^pr. 
\ in.pr. 



(482) 



which, if satisfied, will give equality of work in the two cylinders, for this case 
of perfect compression and expansion. 

In Fig. 98, is given a set of curves for use in determining the value of the 
ratio of (rec.pr.) to (in.pr.) as expressed by Eq. (482). 



324 



ENGINEERING THERMODYNAMICS 



When (rec.pr.) has been found by Eq. (482) it is possible by means of (481) 
and the clearances to find Re- The events of the stroke must have the follow- 
ing values to maintain complete and perfect compression and expansion. 



Z H =(i+ CB )(¥^)' -c H ; (483) 

\ m.pr. / 

(484) 



^=(H-c,)( bk ^) s "-c,; 
\rec.pr./ ' 



LVrec.pr./ J' 



X L = c L 



Km-'] 



(485) 
(486) 



Example. Find (a) the horse-power, (b) compressed air used per hour, and (c) receiver 
and releaie pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per cent 
clearance in the high-pressure cylinder, and 4 per cent in the low-pressure cylinder, when 
initial pressure is 150 lbs. per square inch absolute, back pressure 10 lbs. per square 
inch absolute, speed 125 R.P.M., cut-off in high-pressure cylinder J, low-pressure 
cut-off such as to give complete expansion in high-pressure cylinder, compression in 
high-pressure cylinder 15 per cent, and complete in low. Expansion such that s — 1.4. 

(6) As in example of Se:tion 15, receiver pressure equal high-pressure release 
pressure, and low-pressure volume at cut-off must equal volume of steam exhausted 
from high pressure. Release pressure may be found from relation (in. pr. )(<?#+£#)* 
= (relpr.) -(c h +DhY : or 150(.06+.5) 14 = (rel.pr.) -(.06+1) 14 , or(rel.pr.) =60 lbs. As 
in the previous example, the low-pressure cut-off is .38, and the low-pressure compression 
may be found from the relation c L s (rec.pr.) =(ci+Xi) s (bk.pr.),or(.04) 1,4 X 60 = 
(M+X L ) 1A (10), or X L = .09. 

From the sum of equations (471) and (473) divided by 144D^i?c, and with the 
proper values substituted the following expression for (m.e.p.) results: 



150 .5 + .06 



2.25 .4 



'.15 + .06 
.4 


.38 + .04 - 
.4 



1.4 



15 



• 5 + - 06 V 4 l OR 

i+m) J- 06 



.5 + .06+2.25(.15 + .06) 



150 



2.25 { 2.25(.38 + .04)+.15 + .06 



/10\- 71 I 1 - 4 
W 



m'-'i 
"-my}-" 



+ 1-.15 +150 



1.4 



.10 



.09 + .04 



.4 



.09 + .04Y 4 
.04 



.5 + .06+2.25(.15+.04)(^ 
2.25(.38 + .04)+.15 + .06 



1 +1 -.09 =55 lbs. per sq.in. 



hence the horse-power is 214. 






WORK OF PISTON ENGINES 325 



From Eq. (477) with proper values substitute, 

Cu.ft.perI.H.P. hr.=i|Px[(.38+.04)gy 7 '-(.09+.04)( ] | )j ?1 =50, 

or total steam per hour will be 

50X214 = 10700 cu.ft. 

(c) Release for the high-pressure cylinder has already been given as 60 lbs. and the 

receiver pressure the same. The latter quantity may be checked by equation (469) 

and will be found to be the same. The low-pressure release pressure may be found 

from the relation (rec.pr.) {Z L +c L ) 1A = (rel.pr.) Z/ (l -\-c L ) lA , which on proper substitution 

gives 

i 



(rel.pr.)z,=60f ' ' — j =27 lbs. per sq. inch 



Prob. 1. What will be the horse-power and steam used per hour by an 18- and 
24x30-in. engine with 5 per cent clearance in each cylinder and with infinite receiver 
running on 100 lbs. per square inch gage initial pressure, and 5 lbs. per square inch 
absolute back pressure, when the speed is 100 R.P.M. and the cut-off in high-pressure 
cylinder is \ and in low ^? 

Note: 8*1.3 and S = .2. 

Prob. 2. What must be the receiver pressure for equal work distribution when the 
initial pressure has the following values for a fixed back pressure of 10 lbs. per 
square inch absolute? 200, 175, 150, 125, 100, and 75 lbs. per square inch gage? 

Prob. 3. For the case of 150 lbs. per square inch gage initial pressure and 14 lbs. 
per square inch absolute back pressure, what will be the required high-pressure cylinder 
size for an air engine with a low-pressure cjdinder 18x24 ins., to give equality of work, 
clearance in both cylinders being 5 per cent? 

Prob. 4. What will be the horse-power and air consumption of the above engine 
when running at a speed of 150 R.P.M. , and under the conditions of perfect expansion 
and compression? 

17. Compound Engine with Finite Receiver. Logarithmic Law, with 
Clearance and Compression, Cycle XIII. General Relations between Pressures, 
Dimensions, and Work when H.P. Exhaust and L.P. Admission are Inde- 
pendent. As this cycle, Fig. 97, is made up of expansion and compression lines 
referred to the different origins together with constant pressure, and constant 
volume lines, the work for high- and low-pressure cylinders and for the cycle can 
be set down at once. These should be combined, however, with the relation 
noted for the case of infinite clearance which might be termed the condition 
for a steady state 

[(PV) on H.P. expansion line]— [(PV) on H.P compression line] 

= [(PV) on L.P. expansion line— PV on L.P. compression line,] 

or, P b V b -P e Ve = PnVn-P,cV k ...... (487) 



326 



ENGINEERING THERMODYNAMICS 




OVER EXPANSION AND COMPRESSION 



Fig. 97.— Work of Expansive Fluid in Compound Engine with Finite Receiver and with 
Clearance. Cycle XIII, Logarithmic Cycle XIV, Exponential Expansion, and Com- 
pression. 



WORK OF PISTON ENGINES 327 

Besides this there is a relation between H.P. exhaust and L.P. admission 
pressures, corresponding to the equality that existed for the infinite receiver, 
that may be set down as follows: 

PmV m = P b V b , and P m (V m +0)=P d (V a +0); 
/. P m V m = P b V b = P d (V d +0)-P m O, and P m = P h ; 

_ p b v b +p h o u „„, 

Pd ~ v d +o (488) 



Also 



PnV H = PtV k , and P n (V n +0)=P g (V g +0); 
PnV n = P t V k = P g (V +O)-P n O, and P n = P e , 

n P t V t + P e O 



v g +o 



(489) 



These two expressions for the pressure at D and at G are not available in 
their present form, since they involve two unknown pressures — those at H and 
E, but two other equations of relation can be set down from which four equa- 
tions, the four unknown pressures P e , P d) Pg and P/,, can be found. These 
other equations are 



and 



P d (V d +0)=P e (V e +0), or Pa = Pe(j^ 



(490) 



P,(V,+0)=P h {V>+0), or P, = P n (^±&). . . (491) 
Equating (488) to (490), 



P b V b +P h O 



= P *(va+$' ° r p * v *+ p *0= p *W'+0). 



v d +o 

and 

p _ P b V b +P h O 

6 ~ Ve + O ' 

Equating (489) to (491) 

^f^(£g)> or P t r l+ P^=P„(V h +0) 

and 

n _ Pn(Vn+Q)-P*Vt 

Therefore = P b V b +P h O Pn(Vn+Q)-P k V k 

Ve + O 

P b V b O+P h 2 =P h (Vn+0)(Ve+0)-P k V k (V e +0); 
P»[(V h +0)(V e +0)-0 2 ]=P b V b O+P IC V b (Ve+0), 

^ p b v b o+p k v k (Ve+o ) = p b v b o+p k Vic(Ve+o) 

" (V h + OKV e + 0)-0 2 V h V e + VeO+V h O ' 



328 ENGINEERING THERMODYNAMICS 

Therefore 

= P b V b O + P* V k (Ve + Q) 
* V h (V e + 0) + V e O ' 

Substitution will give 



Pe = 



P b V b (Vn + 0)+PjcVicO 
{V e +0)(V h +0)-0 2 

_ \P b V b O+P & V k (V c +0)'] \Vn+OT\ 

r ° L v Jl (Ve-hO)+v e o \[v g +o\ 

p _ \ p b v b (Vn+o)+PicV jcO] \ Ve+o ~\ 
d L(v.+o)(yM+o)-o*\lv a +o\ 



(a) 

(&) 

(c) 

(d) 



(492) 



It will be found that the use of these pressures is equivalent to the applica- 
tion of the equation of condition given in Eq. (487), for substitution of them 
reduces to an identity, therefore the work of the two cylinders can be set down 
by inspection in terms of point pressures and volumes and the above pressures 
substituted. The result will be the work in terms of the pressures and cylinder 
dimensions. 



TTir = P»7 B (l+lo & ^-P / 7,(l+lo & Q-P.7.1o & ^^-(P.-P J 



fWa 



= P 6 7 & (l+l0ge 1£) -P e Ve-PeVe log, y-P.V e Og e ^~-P a V a +P e Ve 



™ l+1 ^)- w <*(©(|3§) 



-PaVa- 



Therefore 



W H = P,vJl+\0ge y) -PaVa 

_ \ P i V i (V„+0)+P l FsO l /FA (V d +Q\ 

I (F+oxf,+o)-o 2 J VeloSe \vJ\v<+o) 



(4' 



(F+0)(F,+0) 

Tf i =p ! ,y^i+ioge y-^S) + p " F » lo & y- p ' v > ( 1+lo & 

-PAVj-v t ) 

=p g v„ log, (f^) + p » v » i°& y- PtV * lo & i^-PiW- v *)- 



p t -p, \v 



Therefore 



Wl= [ v„(v e +o)+v e o \T g +o) J F ' l0& Vtv 

, [ P>V b O+P*V*( Ve+0)~] 1! .\ „ V, 

+ [ v h (y.+o)+vjo~\ Yh l0& f, 



+0 



■p*v t ioe.w-Pj<y i -v t y 



(494) 



WORK OF PISTON ENGINES 
Adding W H and W L 

P b V b (l+\0ge yj ~P k V k l0g e ^~ P a V a ~ Pj(Vj-V k ) 



w= 



_ \ P b V b (V h +0)+P*V kO'] v ! /£A ( Va+Q \ 

L (Ve+o)(v h +o)-o 2 J Vc ioge \vj \v.+o) 
, rp»7 i o+fty*(7a +o) (v*+o\ ] v lo _ 

+ L ^(7e+0)+7 e 0-J n l0ge 7/ 



JVhO 



329 



(495) 



While this Eq. (495) for the cyclic work is in terms of initial data, it is not 
of very much value by reason of its complex form. To show more clearly 
that only primary terms are included in it, the substitution of the usual symbols 
will be made. 



TF=144X 



(in.pr.) (Z H +c h )D h [l H-log* ^^] 



-(bk.pr.)(X L +c L )D L \cge 



Xl±cl 



-(in.pr.)ctfDtf -(bk.pr.)(l -X L )D L 

(in.pr.) (Z H +ch)D h [{Z l +c l )D l + 0] + (bk.pr.) (X L +c L ) D L 



[{X H +c H )D H +0}[(Z l +c l )D l +0] -0* 



X 



(X h +ch)D 



H l0g e f 



X K +C H \ 



Ch 



{l+c H )D 



(X h +c h )D. 



H+OJ 



(in.pr.) (Z H +Ch)DhO + (bk.pr.) (X L +c l )Dl [{X h +c h )Dh +0] 
(Z l +cl)DlKXh+ch)Dh+0]+(X h +ch)DhO 



(Z l +c l )Dl+0 
ClD l +0 



C L Dl l0g e 



(Z l +cl)Dl+0 
clD l +0 



(in.pr.) {Z H +ch)DhO + (bk.pr.) [{X L +c l )Dl][X h +c h )Dh +0] 
{Z l +cl)D l [(Xh+c h )Dh+0]+{Xh+ch)DhO 



(Z L +C L )D L l0g e ( 



1+cl 
Zl+cl 



(496) 



Such equations as this are almost, if not quite, useless in the solution of 
problems requiring numerical answers in engine design, or in estimation of engine 
performance, and this fact justifies the conclusion that in cases of finite receivers 
graphic methods are to be used rather than the analytic for all design work. When 
estimates of power of a given engine are needed, this graphic work is itself seldom 
justifiable, as results of sufficient accuracy for all practical engine operation 
problems can be obtained by using the formulas derived for infinite receiver when 
reasonably, large and zero receivers when small and the pistons move together. 



330 



ENGINEEEING THERMODYNAMICS 



It might also be possible to derive an expression for work with an equivalent 
constant-receiver pressure, that would give the same total work and approxi- 
mately the same work division as for this case, but this case so seldom arises that 
it is omitted here. 

Inspection of the work equations makes it clear that any attempt to find 
equations of condition for equal division of work for the general case must be 
hopeless. It is, however, worth while to do this for one special case, that of 
complete expansion and compression in both stages, yielding the diagram Fig. 
98. This is of value in drawing general conclusions on the influence of receiver 
size by comparing with the similar case for the infinite receiver. 

By referring to Fig. 98, it will be seen by inspection that cylinder sizes, 
clearances and events of the stroke must have particular relative values in order 













P 












n nr) A B 














A 




B 














\ 
\ 


\ 










— \i 


n.pr.Jn i \ 










(c H r 


h)\ 


4r 




















XfiVA 





--(rec.pr.)! 














>J 


^v 


s ^l v 


H 


E 










(CL 


D,V 










^jC 


(rec 


•^ 


< 




-0= 


V»Y 




> 






-v 


> — 


! . 




\2 


\^.i 




















K 












J 
















-x, 


r>r* 






Dl 










(bk.r 





































R Q V 

Fig. 98. — Special Case of Cycles XIII and XIV, Complete Expansion and Compression in 
both Cylinders of Compound Engine with Clearance and Finite Receiver. 



to give the condition assumed, i.e., complete expansion and compression. It 
is, therefore, desirable to state the expressions for work in terms which may 
be regarded as fundamental. For this purpose are chosen, initial pressure 
(in.pr.), back pressure (bk.pr.); high-pressure displacement, D H ; cylinder 
ratio, R c ; high-pressure clearance, c H ; and ratio of receiver volume to high- 
pressure displacement, y. Call 



/imprA 
Vbk.pr./ 



Ei 



It will be convenient first to find values of maximum receiver pressure 
(rec.pr.)i, and minimum (rec.pr.)2; high-pressure cut-off Z H , and compression 
X H ) low-pressure clearance c L , cut-off Z L , and compression, Xl, in terms of 
these quantities. Nearly all of these are dependent upon the value of cl and 
it will, therefore, be evaluated first. 



WORK OF PISTON ENGINES 331 

From the points C and «/, Fig. 98, 



From A and E, 



(rec.pr.)2 = (bk.pr.)^ ( ^ ) . . . . . . . (497) 



(rec.pr.)i = (in.pr.)-^, (498) 



and from E and C, 

(vec^hl+c^+y 

(rec.pr.)2 R c c L +y 
Dividing Eq. (498) by Eq. (497) and equating to Eq. (499), 

RpCh(1+c h ) = l +c H +y 
R<?Cl{X-\-c l ) RcCl+V ' 

Multiplying out and arranging with respect to c L , the relation to be 
fulfilled in order that complete expansion and compression may be possible is, 

CL 2 [Rc?(l+CH+y)]+CL[Rc 2 (l+c H +y)-RcRpCH(l+c H )] 

-[yRpc H (l+c H )] = 0- • (500) 

This is equivalent to 

c L 2 l+c L m-n = 0, (501) 

and the value of c L is 

te= (m*+y-m (502) 

It is much simpler in numerical calculation to evaluate I, m, and n and 
insert their values in Eq. (502) than to make substitutions in Eq. (500), which 
would make a very cumbersome formula. (rec.pr.)2 and (rec.pr.)i may 
now be evaluated from Eqs. (497) and (498) by use of the now known value of c L . 

High-pressure cut-off Zh may be found from the relation of points B and 
J, Fig. 98, 

Rp(Z h +Ch)=Rc(1+Cl) 
or 

Z h = ^(1+Cl)-c h (503) 

rip 

Low-pressure cut-off, Z L from, 

Rc(Z l +cl) = 1+c h , 



or 



Z L =^^-c L (504) 

High-pressure compression, X H , 

X h = c l Rc-Ch . . . (505) 



332 ENGINEERING THERMODYNAMICS 

Low-pressure compression, X L , by the use of points A and K t 

Rc(X l +c l )=RpCh, 
or 

X l = ^ch-c l (506) 

If c L is regarded as being part of the original data, though it is related to 
R c , R P , c H and y as indicated in Eq. (488), the expressions for high- and low-pres- 
sure work and may be stated as follows: 

F,=144(in.p,)2>,{|(l +Ci )[l+lo & |g±g] ; j 



+ (l+ te )lo 6 « i±|] - [g(l+c,)-c„] -c H log.fg 



(508) 



Adding these two equations gives the total work of the cycle as follows: 



W = lU(in.pr.)D H R c (l +c L ) {i-Jl+lo & |? ( ' 



iJ c (l + c 7j ) 



. 1 , D 1+Ci 



1+ch+v t 1+cg+y . 1+ch+v } l + c H + y 
Rp(1+c h ) i0ge Rcc L +y ^Rpil+cn) 10ge ite+?/ 



144(in.pr.)^ 



c H 



l + lo & ^] + c,lo & |g + |(l + c i )- to 



This, however, may be greatly simplified. 



and 



2^(1 -he*) 1+cl . 

l0§c ^(T+^) +l0ge ^T+^ =l0ge ^ 



logc^-^ + loge ^^ = l0g e R P . 
Ch KcCl 



Hence 



W = 144(in.pr.)Z)J |^(1 +Ci)-c Jlog, R P . 

= 144(in.pr.)^^log^ P (509) 



WORK OF PISTON ENGINES 333 

From this may be obtained mean effective pressure referred to the low- 
pressure cylinder, work per cubic foot supplied, and consumption per hour 
per indicated horse-power, all leading to the same results as were found for the 
case of complete expansion and compression with infinite receiver (Section 15,) 
and will not be repeated here. 

To find the conditions of equal division of work between cylinders, equate 
Eqs. (507) and (508). 

2_ ( i +CL )-2^+-(l+c,)^log e ^^^-log e - TI ^ y J 

Rp l+c H RcCl+v L RcCl c h J 

which may be simplified to the form, 

, l+CH+y ^„ l+c„+y , ,_ R* p (1+Ch) , RpCh \,R h pC H ..1 n 
1 "T+^ 1 ^^^iT +l0& "Sc(l^ Reel' 1 ] = °' 

(510) 

This equation reduces to Eq. (376) of Section 11, when ch and cl are put 
equal to zero. In its present form, however, Eq. (510) it is not capable of 
solution, and it again becomes apparent that for such cases the graphical solution 
of the problem is most satisfactory. 

Example 1. Method of calculating Diagram, Fig. 97. 
Assumed data: 

Pa =Pb = 120 lbs. per square inch abs. Vj = Vt= 2 cu.ft. 

P m =Ph= 30 lbs. per square inch abs. V c = Vd = Vic= .8 cu.ft. 

Pj= 10 lbs. per square inch abs. V g = Vi= .24 cu.ft. 

V e = .2 CU.ft. 

Va = V f = .12 cu.ft. 
F = 1.2 cu.ft. 
F & = .4 cu.ft, 



To locate point C: 






p P b V b 120 X. 4 

Pc- y - 8 =60 lbs. pe 


To locate point M: 






T/ P b V b 120 X. 4 tfl ^ 
Vm= P m = 30 - L6cU ' ft - 


To locate point D: 





Pa(V d +0)=P m (V m +0), or P d = 30^^=42 lbs. per sq.in. 



334 ENGINEERING THERMODYNAMICS 

To locate point E: 

Pe = (Ve+0)=P m (V m +0), or P e =^^ X30 =60 lbs. per sq.in. 



To locate point F: 
To locate point L: 

To locate point N: 

since P n =Pe 

To locate pcint G: 



p P e Ve 60X.2 

Pf=—— = — -7— =100 lbs. per sq.in. 
V f AZ 



Pi=— Tf— = — --—33.3 lbs. per sq.in. 
Vi .24 



.. P k V k 10 X. 8 

Fra= "p7 = "6^ = - 13cu ' ft - 



^(^+0)=P„(7„+0) or, P, = 60^i|^| =55.5 lbs. per sq.in. 



To locate point H: 

Pn(Vn+O)=P g {V +O) Vn = 



P g V g +P g O-P h O 



or 



T7 .24X55.5+55.5X1.2-30X1.2 . Ae . 
V h =■ ^ = 1.46 cu.ft. 

To locate point I: 

_ P*F„ 1.46X30 

P<=— — = =21.7 lbs. per sq.in. 

K* 2 

Example 2. Find the horse-power of a 12- and 18x24-in. engine, running at 125 
R.P.M., with a receiver volume twice as large as the low-pressure cylinder, 6 per cent 
clearance in the high-pressure cylinder, 4 per cent in the low, when the initial pressure 
is 150 lbs. per square inch absolute, back pressure 10 lbs. per square inch absolute, high- 
pressure cut-off ^, low-pressure f , high-pressure compression 10 per cent, low 3 per cent. 
From Eq. (496) divided by 144Di, and with the values as given above, the (m.e.p.) 
is equal to following expression: 

150x.56x^(l+log e ^°|)-10x.34 1og e |i-150X.06x^--10X.7 

150 X .56[.79 X2.25 +4.5] +10(.34)2.25 X4.5 _1 J.6 1.06x2.25+4.5 

[.16+4.5][.79x2.25+4.5] -(4.5) 2 X ' X 2.25 ° ge .06 X .16 X2.25+4.5 



1 50x.56x4.5 + 10x.34x2.25[.16+2.25] [". 79x2.25+4.5 1 .79x2.25+4.5 

.79 X2.25[.16+2.25] + (.16X2.25) [.04x2.25+4.5_|' X ' ° ge .04x2.25+4.5 



150x.56x4.5 + 10(.34x2.25)[.16+2 
+ .79x2.25[.16+4.5] + (.16x2.25) 

hence the horse-power will be 191. 



— I .79 log e ™r- =49.2 lbs., per sq.in. 



WORK OF PISTON ENGINES 335 

Prob. 1. Find the work done in the high-pressure cylinder and in the low-pressure 
cylinder of the following engine under the conditions given. 

Engine 14 and 30x28 ins., 100 R.P.M., 5 per cent clearance in each cylinder, 
high-pressure cut-off i 3 o, low-pressure cut-off to, high-pressure compression i 2 o, low- 
pressure compression tV, initial pressure 100 lbs. per square inch gage, back pressure 
5 lbs. per square inch absolute, and receiver volume 3 times the high-pressure 
displacement. Logarithmic expansion. 

Prob. 2. The following data are available: initial pressure 200 lbs. per square inch 
absolute, back pressure 10 lbs. per square inch absolute, engine 10x15x22 ins., with 
5 per cent clearance in the high- and low-pressure cylinders, speed 100 R.P.M. What will 
be the cut-offs, and compression percentages to give complete expansion and compression. 
Logarithmic expansion? 

Prob. 3. What will be the work done by the above engine working under these 
conditions? 

Prob. 4. What must be the low-pressure clearance, cut-offs, and compression 
percentages, to give complete expansion and compression for a similar engine work- 
ing under the same conditions as those of Prob. 2, but equipped with a receiver twice 
as large as the high-pressure cylinder? 

18. Compound Engine with Finite Receiver, Exponential Law, with 
Clearance and Compression, Cycle XIV. General Relations between 
Pressures, Dimensions, and Work when H.P. Exhaust and L.P. Admission 
are Independent. It cannot be expected that the treatment of this cycle by 
formulas will give satisfactory results, since even with the logarithmic expansion 
law, Cycle XIII gave formulas of unmanageable form. For the computation of 
work done during the cycle, however, and for the purpose of checking pressures and 
work determined by graphical means, it is desirable to have set down the relations 
of dimensional proportions, initial and final pressures, and valve adjustments, to 
the receiver pressures, release pressures and work of the individual cylinders. 

The conditions of a steady state, explained previously, require that (Fig. 97) 

v >- v °MW -HW > (Hi) 

which is the same as to say, that the quantity of fluid passing per cycle in the 
high-pressure cylinder must equal that passing in the low. Expressed in terms 
of dimensions, 

Mch+z h ) -d h ( Cb +x h ) \(J^k]° 

L (m.pr.) J 

= D H R c{ c L+ Z L ) \ ^g^kl ' ~D H R c (c L +X L ) [tell ' 
I (m.pr.) J | (m.pr.) J 

or, rearranging, and using R P = [™' pT '\ . 

(bk.pr.) 

c«+Z„+R c (c L +X L ) i-! =( CH+ xJ^^hy 
Rp; 'I (m.pr.) J 

+Rc(c,+Z L )\ {cU ^° e ^ Y , -(512) 
L (m.pr.) J 



336 



ENGINEERING THERMODYNAMICS 



an equation which contains two unknown pressures (rec.pr.)i and (cut offpr.)z,. 
To evaluate either, another equation must be found : 



V n = V* 



where P n = P c , so that 



Hence 



or 



Pn 



i 
V. = RcI>h(ol+X l ) f, (bk VV \ ]' (513) 



- p /Z-+2Y 



(cut-off pr.)i= (rec.pr.)i 



Ph ~ Pn \V h +Ol 

y+Rc{cL+XL) \(^SL)] 
'[(rec.pr.jij 



y+Rc(c L +Z L ) 



2/(rec.pr.)^i + ffcfe+Xi)(bk.pr.)" 



y+Rc(c L +Z L ) 



(514) 



which constitutes a second equation between (cut-off pr.) L and (rec.pr.)i, which, 
used with Eq. (512) makes it possible to solve for the unknown. By substitu- 
tion in Eq. (512) and rearranging, 



(rec.pr.)i = (bk.pr.) 



RpT(cH+Z H )[y+Rc(cL+Z L )]+R c y(c L +X L )~\' 



(c H +X H )[y+Rc(c L +Z L )]+R c y(c L +Z L 



?]'■ 



(515) 



This expression is of great assistance even in the graphical construction 
of the diagram, as otherwise, with all events known a long process of trial and 
error must be gone through with. It should also be noted that when s = 1 this 
expression does not become indeterminate and can, therefore, be used to solve 
for maximum receiver pressure for Cycle XIII, as well as Cycle XIV. 

Cut-off pressure of the low-pressure cylinder, which is same as the pressure at 
H or at M, Fig. 97, is now found most easily by inserting the value found by 
Eq. (515) for (rec.pr.)i in Eq. (514). 

Enough information has been gathered now to set down the expressions for 
work. 

Wh = IUDh I (m.pr.) ~jz:y~ [ S ~ \T+^"/ J ~ c *( m -P r ') 



{c H +X H ) 
s-1 

(y+Cn+X n) 

5-1 



<—).[(^)""'-'] 



(rec.pr.)i 1 



( y+c B +X. 
\ y+Cn+1 



r] 



(516) 



WORK OF PISTON ENGINES 337 

_(bk.pr.)^±^[(^^)'" l -l]-(bk.pr.)(l-X i ^cl. . (517) 

Addition of these two Eqs. (516) and (517) gives an expression for the total 
work W, and equating them gives conditions which must be fulfilled to give 
equality of work in the high- and low-pressure cylinders. Since these equations 
so obtained cannot be simplified or put into more useful form, there is no object in 
inserting them here, but if needed for any purpose they may be easily written. 
In finding the conditions of equal work, the volumes of (rec.pr.)i and (cut-off pr.)z, 
must be inserted from Eqs. (514) and (515) in (516) and (517), in order to have 
terms in the two equations consist of fundamental data. This, however, increases 
greatly the complication of the formula. 

After finding the total work of the cycle, the mean effective pressure referred 
to the low pressure is obtained by dividing by 144 X Dl. 

To assist in finding the work per cubic foot supplied and consumption, 
and the cubic feet or pounds per hour per I.H.P. it is important to know the 
volume of fluid supplied per cycle, 

r i 

— /(rec.pr.)i\*l „-.^ 

(8up.VoL)=QB = D H \(c H +Z H )-(c H +X H ) ((E^)) J* * (518) 

Example. Find the horse-power of and compressed air steam used by a 12- and 
18x24-in- engine running at 125 R.P.M., with a receiver volume twice as large as the 
low-pressure cylinder, 6 per cent clearance in the high-pressure cylinder, 4 per cent in 
the low, when the initial pressure is 150 lbs. per square inch absolute, back pressure 
10 lbs. per square inch absolute, high-pressure cut-off J, low-pressure cut-off f, 
high-pressure compression 10 per cent, low-pressure compression 30 per cent, and 
expansion and compression follow the law PV 1A =c. 

From Eq. (515) (rec.pr.)i is found to be as follows when values for this problem 
are substituted: 

. , n [~ 15- 7 (.56)[4.5 +2.25(.79)] +4.5 X2.25(.34) "| • . 

(recpr.h-loL . 1< $ [4 .5 + 2.25X.79]+4.5X2.25X.79 J " 8L7 lh *' ^ abS ° lute ' 

and by using this in Eq. (514) 

,-■'-. / 4.5X81.7' +2.25X.34X10 \ . ? 

(cut-oft pr.)i= I 4 5 +2 25 X 79 / Sq,m * absolute * 



338 ENGINEERING THERMODYNAMICS 

It is now possible by use of Eqs. (516) and (517) by addition and division by 
144Di, to obtain (m.e.p.). Substituting the values found above and carrying out 
the process just mentioned. 

( „.,,., = i( 1M xfx[,,-g)']-.«X >S ,-fxS 2 [( : »)'-,] 

wHts)>* 4i± ^[(i£tsr-'] 



4.5 +.16 



.4 

.34 



ro 4.5 +2.25 X. 79 r, / 4.5 + 2.25X-79 V] 2.25 X.: 

+53X j L 1 -(, 4 .5 + 2.25XlW J _10X_ ^ 



|V 4 - 



-10X2.25X.7 j =51.5 lbs. sq.in. 

hence the horse-power will be 200. 

By means of Eq. (518) the supply volume may be found. This gives upon 
substituting of the proper values: 



(Sup.Vol.) =D J.56-16x(||Y 71 l =.462), 



T ^ 13,750 Sup.Vol. 
Cubic feet per hour per I.H.P. = ;— — ■ r X — — , 

13,750 .46 
51.5 X 2.25" 54 ' 5 ' 

hence the total volume of air per hour will be 

54.5X200 = 10900 cu.ft. 

Prob. 1. What will be the receiver pressure and L.P. cut-off pressure for a 
cross-compound compressed air engine with 5 per cent clearance in each cylinder, run- 
ning on 100 lbs. per square inch gage initial pressure and atmospheric exhaust, when 
the high-pressure cut-off is I, low-pressure f , high-pressure compression 15 per cent, 
low 25 per cent, and s = 1.4. Receiver volume is twice the high-pressure cylinder 
volume. 

Prob. 2. Find the superheated steam per hour necessary to supply a 14- and 21 X28- 
in. engine with 5 per cent clearance in each cylinder and a receiver twice" the size 
of the high-pressure cylinder when the initial pressure is 125 lbs. per square inch gage, 
back pressure 7 lbs. per square inch absolute, speed 100 R.P.M., high-pressure 
cut-off £, low-pressure A, high-pressure compression 15 per cent, low pressure 40 per 
cent and s = 1.3. 

Note: § = .3. 

Prob. 3. If the high-pressure cut-off is changed to £ without change of any other 
factor in the engine of Prob. 2, how will the horse-power, total steam per hour, 
and steam per horse-power per hour be affected? If it is changed to I ? 

Prob. 4. A boiler capable of supplying 5000 lbs, of steam per hour at rated load fur- 
nishes steam for a 12- and 18 x24-in. engine with 5 per cent clearance in each cylinder and 
running at 125 R.P.M. The receiver is three times as large as the high-pressure cylinder, 



WORK OF PISTON ENGINES 



339 



the initial pressure 150 lbs. per square inch gage,, back pressure 5 lbs per square inch 
absolute, the low-pressure cut-off fixed at \ and low-pressure compression fixed at 
30 per cent. At what per cent of its capacity will boiler be working for these follow- 
ing cases, when = 1.2 for all and 20% of the steam condenses during admission? 

(a) high-pressure cut-off I, high-pressure compression 80 i er cent, 

(b) high-pressure cut-off §, high-pressure compression 20 per cent, 

(c) high-pressure cut-off f, high-pressure compression 10 per cent. 
Note: §=.33. 




Fig. 99. — Work of Expansion in Compound Engine without Receiver and with Clearance. 
Cycle XV, Logarithmic Expansion; Cycle XVI, Exponential, High-pressure Exhaust 
and Low-pressure Admission Coincident. 

19. Compound Engine without Receiver. Logarithmic Law, with Clear- 
ance and Compression, Cycle XV. General Relations between Pressures, 
Dimensions, and Work when H.P. Exhaust and L.P. Admission are Coinci- 
dent. The graphical construction for this cycle has been described to some 
extent in connection with the first description of the cycle, given in Section 8, of 
this chapter, and is represented here by Fig. 99 in more detail. 

To show that the expansion from D to E is the same as if volumes were 
measured from the axis ML, consider a point Y on BE. If the hypothesis 
is correct 



P d XKM = P y X(KM+Y'K) (519) 



340 ENGINEERING THERMODYNAMICS 

The true volume when the piston is at the end D of the stroke, is 
D H (l-Jrc H -\-R c c L ), and at Y, the true volume is 

D H (l+c H +R c c L )-D H y+D L y, 

where y is the fraction of the return stroke that has been completed in both 
cylinders when the point Y has been reached. Then 

F a DH(l+C H + RcCL)=PyD H (l+CH + RcCL)+PyD H (Rc+l)y. 

Dividing though by (R c —l), 

P d j- ^+CH+flcej =p ^ D J + c+R c c L +DB ^ (520) 

This equation may be observed to be similar in form to Eq. (519). More- 
over, the last term within the bracket, D H y, is equal to the corresponding term 
Y'K, in Eq. (519), hence, 



KM = D, 



1 ±f±f-^] , (521) 

tic— 1 J 



Similarly, the distance QM, or equivalent volume at D f is 

m = DL V + R?-\ -] (522) 

The following quantities will be evaluated preparatory to writing the expres- 
sions for work: 

(rel.pr.) s =(m.pr.)^±g?; (523) 

. ^=^^™Sl= WCi+Xi) ^i£- • (524) 

SC 



(in.pr.)z, = P d ' = (rel.pr.)#==; 



(in.pr.) 



Zh-\-c h 1 
1+Ch J 



D H \ 



1+Ch+Rc(ci ;+ Xl) (^') 1+** 



(in.pr.) Zh+ch 



D H [l+c H +R c c L } 



Z H +c H +R c (X L +c L )t U 



= (inpr)- (in ' pr - ) (525) 

, . n, x r v KM \ 1+Ch +RcCl 1 

(cut-on pr.)z,= (m.pr.)z, =--= = (m.pr.)z, r~i r^5 ftf, t^ti — ^\ 



= (in.pr.) 



Z a +CH+Rc{X L +c L )^^ 



l+c H +RcC L +(Rc-l)(l-X H ) 



(526) 



WORK OF PISTON ENGINES 
The ratio of expansion from E' to G is equal to 



(cut-off pr.)z, _ 1+cl 
(rel.pr.)z, ~(l-X H )+c L 



341 



(527) 



Hence 

W H = 144Z>,/(in.pr.) I (Z H +c H ) | 1 +log e 

Z H +CH+RciX L +c L ) 



Zh+c h \ 

( bk.pr.) 

(in.pr. ) 



£/J+Ctf+#c(Xz,+Cz,) 



(bk. pr.)" 
(in.pr. ) 



log e [ - -^ 



rCz, + (ifo-l)( l-Xg) " 






Ch-\-Xh 

{Ch+Xh) lOge - -C/f 



CH 

(bk.pr.) 
(in.pr. ) 



(528) 



+ 



fic-1 



(in.pr. ) 



X 



loge 



1+c h +RcClHRc-1)Q . -X 

1+Ch+RcCl 



3] 



(1+CL-X^)l0gc 



l+c*+#cC^+(#c-l)(l-X„) 

-144Z^(bk.pr.) { l-X i +(X i +c i ) log e 



l+c £ 



l+Cz.-Xtf/J 

X^+c^ | 

* r 



(529) 



The total work found by adding W H and W L as given above, leads to the 
following : 



= 144IMin.pr.) { (Z H +c H ) log e Q^) 



+ ^ + k + c, + ^(X, +C ,)^llog e r i+^+^ + (^-l)(l-Z g ) -| 

(in.pr.) J L 1+Ch+RcCl J 

(bk.pr.)' 



+ 



Zh+Ch+Rc(Xl+c l ) 



(in.pr. ) 



1+Ch+RcCl+(Rc-1)(1-X h ) 



Rc(1+c l -X h ) X 



/_ 1+Cl 



Hi^-x;)-^^'-^)] 



-144Z)i(bk.pr.) 1 -Zi + (X £ + Ci ) 



Xi+cz, 



Ci 



(530) 



342 



ENGINEERING THERMODYNAMICS 



This is the general expression for the work of the compound engine without 
receiver, with clearance and compression, when high-pressure exhaust and low- 
pressure admission are simultaneous and expansion and compression logarithmic, 
in terms' of fundamental data regarding dimensions and valve periods. 

From this the usual expressions for mean effective pressure, work per 
cubic foot supplied, and consumption per hour per I.H.P., may be easily 
written, provided the supply volume is known. This is given by 



(Sup.Vol.) =A'B = DhUz h +ch) - (ch+X h ) 



(cut-off pr.)z, 
(in.pr.) _ 



Z H +c H +Rc(X L +c L ) 



= D h \Z h +c h -(ch+X h ) 



(b^prO 
(in.pr. ) 



1+Ch+RcCl+(Rc-1)(1-X h ). 



(531) 



To find the conditions which must be fulfilled to give equal work in the two 
cylinders, equate Eqs. (528) and (529). 



■*d£S) 



Re +1 
Re -I 



Z h +Ch+Rc(X l +c l ) 
Zh+c h +Rc(X l +cl) 



(bk.pr.) 
(in.pr.) 

(bk.pr .) 
( in.pr.) 



lo n — i+c H+ r c c l — J 



1+Ch+RcCl + (Rc-1)Q.-Xh). 



Rc(l+c L -X H ) lo& 



l+c L -X H ) 



+ {C H X H ) log e 



Ch+Xh 
Ch 



+R J^)\l-X L +(X L +c L )loJ^±^) 1 =0. (532) 

( in.Dr.) \ c L / \ 



These expressions are perfectly general for this cycle, and expressed in 
terms of fundamental data, but are so complicated that their use is very 
limited, as in the case of some of the general expressions previously derived 
for other cycles. 

As in other cycles, it is desirable to investigate a special case, that of 
complete expansion and compression in both cylinders, Fig. 100. First it is 
necessary to determine what are regarded as fundamental data in this case, and 
then to evaluate secondary quantities in terms of these quantities. The following 
items are assumed to be known: (in.pr.), (bk.pr.), which is equal to (rel.pr.)i, 
R c , c H , and c L , and D H , which are dimensions, and it is known that the 
pressure at the end of compression in L.P. is equal to (rel.pr.)^. 

Referring to the diagram, displacements, clearances, and the axis for the 
common expansion, ML, can all be laid out, and the location of the points 
A and G determined. 

The points E and E' are at the end of the common expansion within the 
two cylinders, and at beginning of high-pressure compression and of separate 
low-pressure expansion, hence p e = pL 



WOEK OF PISTON ENGINES 
From the points A and E: 

From the points G and E', 



Pe' = p e = (bk.pr.) 



1+Cl 



l+c L -X H ' 



343 

























A 


B 






jL 
























^iC 


iZ M D 


A 


























c h Dh 






































































^*" 


-7 


\ 






























/ 
























1 .^ 

4^- 


i^T 


D,, 


\ !> 




















^* 




/ 


f" 


-+ — i 


j 


./Jj 


) 
















'E'^ 







~^- 


! 

i 
i 

.4-__. 


-4— 












G . 






















k^ 


^ 


7^— 






D, 


1 


*—. Xl Dl- 

1 


l 


V 


1 

i 

1 









V F K M V 

Fig. 100. — Special Case of Cycles XV and XVI. Complete Expansion and Compression in 
both Cylinders of Compound Engine without Receiver and with Clearance High- 
Pressure Exhaust and Low-Pressure Admission Coincident 



and equating, 

whence 
where 



<^>xfe-<^>H^ 



X H = C H 



(1 +c l )(Rp-1) 
1+Cl+c h R p : 



(533) 



Rp = 



m.pr. 
bk.pr. 



Substituting the value of X H in either of the expressions for p e , which is 
the low-pressure cut-off pressure, 

(cut-off W .) L = p e = (bk.pr.) p |{^^ g ] = ?V (534) 



344 ENGINEERING THERMODYNAMICS 

It may be noted here that the cylinder ratio does not enter into this, but 
only clearances and pressures. In the no-clearance case, it may be remembered 
that the point E or E' was not present, as it coincided with G. 

Next, to find the high-pressure release pressure, p d} by means of points E 
and D, and their relation to the axis ML, Fig. 100. 

1+Ch+RcCl- 



( , v EM Jl+c L +c H Rc 

(rel.pr.).=^= Pe = = (bk.pr.)|_ 1+Cl+Ch ~_ 



1-X, 



Rc-1 



1+Ch+RcCl 

Rc-1 



= (bk.pr.) 1+Ch+RoC -.. ■ (535) 

Knowing the release pressure of the high-pressure cylinder, it is possible to 
find the high-»pressure cut-off and compression necessary to give the required 
performance. 

Z^tl+c^-^-^ Pf+^ l-c, ■ (536) 
(m.pr.) R P I 1+Ch+RcCl J 

(rel.pr.)g \ Rc(1+c l )+RpC H J ,™. 

Xl = Cl -W^)- Cl = Cl [ 1+c h +R c c l _1 J- * * ' (537) 

The work of the two cylinders is as follows: 

w -iAAri r \ [0-+Ch)[Rc(1 + Cl)+RpCh~] [ 1 M R p (1+Ch+RcCl)1 

^ = 144D,(m.pr.) j-g- ^—^-- J [l+loge^^y^-J -* 

_ \ 1+Ch+RcCl ] \ Rc(1+C l )+RpCh ] _1_ ] [ ig c (l+c £ )+fipc g ] [ 1+c^+to 1 
|_ Bc-1 JL 1+Ch+RcCl ]Rp 0ge l 1+Ch+RcCl RI+Cl+RpCh] 

= 144D.(m.pr.)|-^- • 1+fty+Jfefe [ 1 + to fr &y (i +te )+ gp ^ j^ 
_ \ R c (1+Cl)+RpCh ] 1 pfc(l+fe)+fo>c g ] [" 1+Cl+Ch 1 

■ WtSSjSt]) «»> 



WORK OF PISTON ENGINES 345 

These expressions, when added and simplified, give the following for total 
work per cycle, 

Wtf w**'+-'[ i S*- 1 ]H- • (540) 

in which of course D H Rc may be used instead of D L and — .y- 1 - instead of 

lip 

(bk.pr.) and then 

W=lteD s Z H (m.pr.)\og e Rp, . ... (541) 

Z H having the value of Eq. (53(3). 

Equality of work the in high- and low-pressure cylinders results, if W 
Eq. (538) equals W L , Eq. (539), or df 

2W H = W, or 2W L = W, 

all of which lead to equivalent expressions. Simplification of these expres- 
sions, however, does not lead to any direct solution, and hence the equations 
will not be given here. 

Example. Find (a) the horse-power and (b) steam used per hour for a 12- and 18 x24-in. 
tandem compound engine with no receiver, 6 per cent clearance in the high-pressure 
cylinder, and 4 per cent in the low, when the initial pressure is 150 lbs. per square inch 
absolute, back pressure 10 lbs. per square inch absolute, speed 125 R.P.M., high- 
pressure cut-off |, high-pressure compression 15 per cent and low-pressure compression 
is complete. 

(a) Since the low-pressure compression is complete, the pressure at end of compression 
must be equal to the release pressure of the high. This latter quantity may be found 
rom the relation (m.pr. )(Z H +c H ) = (rel.pr.)jy(l+cjr), or 

(rel.pr.)# = 150y---=79.3 lbs. persq.in. absolute. 

Low-pressure compression may be found from the relation (rel.pr.)^(cz,) = (bk.pr.) 
(cl+X l ), or Xl = .28. (m.e.p.) may be found from Eq. (540) divided by 144Dz,, which 
on substitution gives 

Si""*'*! 



+ 



.5 + .06+2.25(.28+.04)-^ 
150 

1 + .06+2.25 X.04+1.75+.85 



2.25(1.04-.15)log ei: i± 4 ^ 5 



-(.06+.15) loge —~ I -10 1 1.28 + (.28 + .04) log e '^~ \ 
and the horse-power will be 271, 



346 ENGINEERING THERMODYNAMICS 

(b) Since the consumption in cubic feet per hour per horse-power is equal to 

13,750 Sup.Vol. 
X- 



(m.e.p.) D L ' 

and supply volume is given by Eq. (531), this becomes 

13,750 J_l / ■56 + 2.25(.32)^ \l 

69.7 X 2.25 I ' \1.06+.09+1.25x.85/J ' 

hence the consumption per hour will be 

44X271X.332=4000 pounds. 

Prob. 1. A Vauclain compound locomotive has cylinders 18 and 30x42 ins., with 5 
per cent clearance in each and runs on a boiler pressure of 175 lbs. per square inch gage 
and atmospheric exhaust. The steam pressure may be varied as may also the cut-off to a 
limited degree. For a speed of 200 R.P.M., a cut-off f and 10 per cent compression 
in each cylinder, find how the horse-power will vary with the initial pressures of 175, 
150, 125, and 100 lbs. gage. 

Prob. 2. When the cut-off is reduced to § in the above engine compression 
increases in the high-pressure cylinder to 20 per cent- For the case of 175 lbs. gage 
initial pressure find the change in horse-power. 

Prob. 3. Find the steam used by the engine per hour for the first case of 
Prob. 1 and for Prob. 2. 

Prob. 4. It is desired to run a 12- and 18x24-in. no-receiver engine with 5 per cent 
clearance in each cylinder, under the best possible hypothetical economy conditions for an 
initial pressure of 200 lbs. per square inch absolute and atmospheric exhaust. To 
give what cut-off and compression must the valves be set and what horse-power 
will result for 100 R.P.M.? 

20. Compound Engine Without Receiver. Exponential Law, with Clear- 
ance and Compression, Cycle XVI. General Relations between Pressures, 
Dimensions, and Work, when H.P. Exhaust and L.P. Admission are Coin- 
cident. Again referring to Fig. 99, it may be observed that reasoning 
similar to that in Section 19 but using the exponential law, would show that 
the same formulas and graphical constructions will serve to locate the 
axes of the diagram, hence, as before, 



1+Ch+RcCl 
JS.1V1 = U H 

and 



KM=D H ^ % \ --, ...... (542) 



QM = P* 1+ p g+ f — (543) 

tic— 1 



Release pressure in the high-pressure cylinder is 



,( T+fr 1 



WORK OF PISTON ENGINES 



347 



Immediately after release the pressure is equalized in the high-pressure 
cylinder and the low-pressure clearance. The pressure after equalization, 
termed (in.pr.) L , is found by the relation of the volume at S and that at D', 
Fig. 99, measured from the axis KT. 



1+c h +R c (cl+X l ) 



(in.pr.)z, = (rel.pr.)tf 
which, by means of Eq. (544) becomes 
(in.pr.)z, = (bk.pr.) 



(bk.pr .) 
(rel.pr.) 



7 



l+C H + R c C L 



R~*p(ch+Z h ) + R c ( Cl +X l ) 
1+c h +RcC L 



(545) 



The expansion of the fluid goes on as it passes from the high-pressure 
cylinder to the greater volume in the low-pressure, as indicated by D'E r 
and DE, and when the communicating valve closes, the pressure has become 



(cut-off pr.)^ = (in.pr.)z, 



1+C h+RcCl 

Rc-l 



1+Ch + RcCl 

Rc-l 



+ (l-X H )\ 



which, by means of Eq. (545) reduces to 



(cut-off pr 



.) i = (bk.pr.)[ r 



R~>p(ch+Z h )+Rc(cl+X l ) 



(c L +X L ) j 
h)(Rc-1)\ 



(546) 



+c h +RcCl+(1-X 1 

After cut-off in the low pressure, expansion goes on in that cylinder alone 
to the end of the stroke, when release occurs at a pressure 

l+c L -X E 



(rel.pr.)z, = (cut-off pr.) L 
or by substitution from equation (546), 



1+cl 



(rel.pr.)z, = (bk.pr.) - 



R s p(c h +Z h )+Rc(c l +X l ) 
+CH+RCCL+Q.-XHKR, 



*l) l+c L -X H 



. (547) 



In terms of these quantities the work of the high- and low-pressure 
cylinders can be written out as follows: 



^ = 144i),j (in.pr.) C -^[ S -(|±J) S - 1 ] - Ci , (in .pr. 



(bk.pr.) [Rs p (c h +Z h )+R c (cl+Xl) 






] ] s \i-( i+ch+Rccl y-n 

s-l L Rc-l J L \ l+c H +RcC L +(l-X H )(R c -l) ) J 



348 ENGINEEKING THERMODYNAMICS 

and 



1 



L \ 8-1 L Rc-l J 

r / i+ch+Rccl y-n 

^ L \l+c H +Rcc L +a-X) (Rc-l)J I 



(bk.pr.),. , « J B-i-feH-Z^+fic^+Xi) ~| S L /1+c^-Xtf 






l+Ci 



(bk.prOfe+Xjr/Ci+X^- 



s-1 



V 



Ci 



-,]-a- 



Xz,)(bk.pr.) . (549) 



These are general expressions for work of high- and low-pressure C3dinders 
for this cycle, and from them may be obtained the total work of the cycle, 
mean effective pressure referred to the low-pressure cylinder, and by equating 
them may be obtained the relation which must exist between dimensions, 
events, and pressures to give equal division of work. It would, however, 
be of no advantage to state these in full here, as they can be obtained from 
the above when needed. 

The supply volume, cubic feet per cycle, is represented by A'B, Fig. 99, 
and its value is found by referring to points B and E as follows: 

(Sup.Vol.) - D H [^+Z H ) - (c a +X H ) ( (c "gy * ) ' ] ' 

-D H \c H +Z H "I— [i +CH+RcCL+ (1 _ Xh) {R ~T) \- (550) 

LIS p 

Work per cubic foot supplied is found from Eqs. (548), (549), and (550). 
Work per cu.ft. supplied = , q g v H ' ' ' ' ' (^51) 

Consumption, cubic feet per hour per I.H.P., is found from mean effective 
pressure referred to L.P. cyl. and supply volume as follows: 



Consumption, cu.ft. per hr. per I.H.P. 

13,750 (Sup.Vol.) 

(m.e.p. ref. to L.P.) D L 



(552) 



This will give pounds consumption by introducing the factor of density. 
Further than . this, it will be found more practicable to use graphical 
methods instead of computations with this cycle. 



WORK OF PISTON ENGINES 349 

Example. Find (a) the horse-power and (b) consumption of a 12- and 18x24-in. 
no-receiver engine having 6 per cent clearance in the high pressure cylinder and 4 per 
cent in the low when the initial pressure is 150 lbs. per square inch absolute, back 
pressure 10 lbs. per square inch abolute, speed 125 R.P.M., high-pressure cut-off J, 
high-pressure compression 15 per cent, and low-pressure compression is complete. 

(a) The per cent of low-pressure compression may be found as in the Example of 
Section 19, using the value of s in this case of 1.4. Then 

(in.pr.)(Ztf+c*) 1 - 4 = (rel.pr.)„(l +ch) 1a , 
or 

(rel.pr.)tf =61.5 lbs. sq. inch absolute, 
and 

(rel.pr.)^Xcx, 1 - 4 = (bk.pr.)(c i +Zz ( ) 1 - 4 , 
or 

X L = M 



From the sum of Eqs. (548) and (549) divided by 144Dz, and with proper 
values substituted, (m.e.p.) =48.5 lbs.; hence the horse-power is 189. 

i q *7cn 
(b) From Eqs. (550) the value for (Sup. Vol.), which when multiplied by — - — , 

and divided by D L , gives cubic feet air per hour per I.H.P. 



v J_rnfi+S— ^ 15*(.56) +2.25(15) ] = .63 cu.ft. per 

2.25L (15) 7 1 +.06+2.25 X.04 + (1-.15)(2.25-1)J hour per I.H 



Prob. 1. If the locomotive of Prob. 1, Section 19, should be equipped with superheater 
so that the steam expanded in such a way that 8=1.3, what would be the effect upon 
the horsepower for conditions of that problem and on the cylinder event pressures? 

Prob. 2. A 30- and 42x54-in.no receiver steam pumping engine runs at 30 R.P.M. 
and has 3 per cent clearance in the high-pressure cylinder and 2 per cent in low. There 
is no compression in either cylinder. Initial pres ure is 120 lbs. per square inch gage, 
and back pressure 28 ins. of mercury (barometer reading 30 ins.). The steam is such 
that the expansion exponent is 1.25. What will be the horse-power of, and the steam 
used by the engine when the cut-off in the high is §? 

Prob. 3. By how much would the power change if the cut-off were shortened to f 
and then to \, and what would be the effect of these changes on the economy? 

21. Triple-Expansion Engine with Infinite Receiver. Logarithmic Law. 
No Clearance, Cycle XVII. General Relations between Pressures, Dimen- 
sions and Work. Fig. 101 represents the cycle of the triple-expansion engine 
with infinite receiver, no clearance, showing one case of incomplete expansion 
in all cylinders, and another where overexpansion takes place in all cylinders. 



350 



ENGINEEEING THERMODYNAMICS 



The reasoning which follows applies equally well to either case, and to any 
combination of under or overexpansion in the respective cylinders. 

It is desired to express the work of the respective cylinders and the 
total work in terms of dimensions, initial and back pressures, and the cut-offs 
of the respective cylinders. To do this, it is convenient first to express the 




INCOMPLETE EXPANSION 



CYCLE XVII 



OVER EXPANSION 



Fig. 101. — Work of Expansive Fluid in Triple-Expansion Engine with Infinite Receiver and 
Zero Clearance. Cycle XVII, Logarithmic Expansion. 

first receiver pressure (1st rec.pr.) and second receiver pressure (2d rec.pr.) in 
terms of these quantities. The subscript I refers to the intermediate cylinder. 



P f =Pi 



V/ 



or 



and 



or 



Z D 

(1st rec.pr.) = (in.pr.)-^-. 



■Pi — Pvy> 



(2d rec.pr.) = (in.pr.)f^. 



(553) 



(554) 






(556) 



WORK OF PISTON ENGINES 351 

Work of high-pressure cylinder is 

W H = P b V b (l+\og e ^-P d V d , 

= 144(in.pr.)Di/jz i ,(l+lo & ^)-^^|. . . (555) 

Work of intermediate cylinder is 

= 144(in.p,)^{^(l+lo & i ; )-^|g 

- 144(in.pr.)i>iJz„(l+log, gj)-§*§ " ' 

Work of low-pressure cylinder is 

IF i -P«7 i (l+lo&|?)--P«7* 

= 144(in.pr.)Z)H&^l+lo gB ^-144(bk.pr.)Z) i . . . (557) 

The total work by addition is 
W = 144(in.pr.)2>*Z* j (l +log e ^j + (l +log c ~) + (l +log e jr) 

= 144(in.prO^^J3+log e ^|^--^g--^-| - 144(bk.pr.)A> . (558) 

Mean effective pressure referred to the low-pressure cylinder is found by 
dividing W by 144Z) 7 ,, and is therefore 

(m.e.p. ref. to L.P.) 

= (in.pr.)Z^J3+log c ^--|^--^-}-(bk.p,). . (559) 



352 ENGINEERING THERMODYNAMICS 

Work done per cubic foot supplied is equal to W divided by the supply 
volume AB or Z H D H , 



Work per cu.ft. supplied 



1 D H Dj 1 , Jjn _, ___ x D L 



= 144(in. P rO^ (560) 

The volume of fluid supplied per hour per indicated horse -power is 



Consumption, cu.ft. per hr. per I.H.P. 

13,750 Z H D H 

(m.e.p. ref. to L.P.) D L 



(561) 



The weight of fluid used per hour hour per indicated horse-power is of 
course found by multiplying this volume Eq. (561) by the density of the fluid 
used. 

The conditions which will provide for equal division of work between the 
three cylinders may be expressed in the following ways: 

W H = Wj = W L , 

which is equivalent to, first : 



or 



1/D H 



'o&v ^\tV-)= 1o & 



h z l \dJ' 



*z H zADjj * e z 

hence 

^l=i(t)-it) W) 

Similarly from W H = W L , 

. & Z H ZjyDjJ (rn.pr.) \D H / Z 7 v ; 

These two equations, (562) and (563), show the necessary relations between 

*~ *• h (a), (a). - (gg). 

in order that work shall be equally divided. Since there are six independent 
quantities entering (as above) and only two equations, there must be four 
of these quantities fixed by conditions of the 'problem, in order that the other 



WORK Off PISTON ENGINES 353 

two may be found. For instance, if the cylinder ratios, the pressure ratio, 
and one cut-off are known, the other two cut-offs may be found, though the 
solution is difficult. 

Again, if cut-offs are equal, and the ratio of initial to back pressure is known, 
it is possible to find the cylinder ratios. This forms a special case which is of 
sufficient importance to require investigation. 

If Zh = Zi = Z l , Eq. (562) becomes 



D h IV 



and Eq. (563) reduces to 







DjD L (in.pr.) 
D H 2 (bk.pr.)' 


but from Eq. (564), 








D L Dj D L (DA* 
Dj D H D h \Dh) ' 


and therefore 






• 




DjD L (DjV 
Dh 2 \Dh) 


and 




Dj D l /in.pr. \* 
D H Dj Vbk.pr./ 


which, along ^ 


rith the condition assumed that 






Zh — Zj — Zl 



(564) 



(565) 



(566) 



(567) 



constitute one set of conditions that will make work equal in the three cylinders 
This is not an uncommon method of design, since by merely maintaining 
equal cut-offs, the work division may be kept equal. 

The work done in any one cylinder under these conditions Eqs. (566) 
and (567) is then 

1\ _ /bk.prA 3 
,in.pr 

and the total work 



WH = W J = W L = 14A(jR.pT.)D H \Zn+log fi ^)-\^^) . (568) 



1\ /bk.pr .\* 
m.pr. 



IF = 432(in.pr.)IV Z( 1+log^l -( j^) (569) 



in which Z represents the cut-off in each cylinder, all being equal. 



354 



ENGINEERING THERMODYNAMICS 



A special case of the triple-expansion engine with infinite receiver and no 
clearance which demands attention is that of complete expansion in all 
cylinders, represented by Fig. 102. Here 



_Dj_ (bk.pr.) 
L D L (2d rec.pr.)' 

y _D H= (2d rec.pr.) ^ 
Dj (1st rec.pr.)' 




(570) 



(571) 



Fig. 102. — Special Case of Cycle XVIII Complete Expansion in Triple-expansion Engine 
with Infinite Receiver, Zero Clearance, Logarithmic Expansion. 

and 



7 _ (1st rec.pr.) _ /bk.prA D L 
(in.pr.) ~ \in.pr. ) D H ' 



hence the receiver pressures are as follows: 

(1st rec.pr.) = (bk.pr. )-=— . 



(572) 



(573) 



WORK OF PISTON ENGINES 355 

and 

(2d rec.pr.) = (bk.pr.)^ "... (574) 

The work of the respective cylinders, expressed in terms of initial and 
back pressures and displacements is then, 

,r.= 1 «, i ,p,, D ,,(;±j)g[ 1+1 „^^)^)i_, 44l> , P 4 ;i) , 



■ i fas w) 



Similarly, 



and 



(^), 



T^z, = 144(bk.pr.)D L log e ^ (577) 



Total work, by addition, is 

*=144(b k .p,)4o & (gg> t)+Io & g-; + lo ge f ] 

= 144(>k.prOB i log.jjg^ (578) 

If for this special case of complete expansion equality of work is to be 
obtained, then from Eqs. (575), (576), and (577), 

■ ' (in.pr.) D H = Di = Dl • {r , 7Q , 

(bk.prO.Di D H IV ■ l ; 

which is readily seen to be the same result as was obtained when all cut-offs 
were equalized, Eqs. (564) and (565). This case of complete expansion and 
equal work in all cylinders is a special case of that previously discussed where 
cut-offs are made equal. Hence for this case cut-offs are equal, 

D H _D I _D L (bk.pr.) , . 

Dj D L D h (m.pr.) 

Example. A triple-expansion engine 12- and 18- and 27x24-ins., with infinite 
receiver and no clearance, runs at 125 R.P.M. on an initial pressure of 150 lbs. 
per square inch absolute, and a back pressure 10 lbs. per square inch absolute. 



356 ENGINEEEING THEEMODYNAMICS 

If the cut-offs in the different cylinders, beginning with the high, are \, f, and f, 
what will be (a) the horse-power, (b) steam consumed per hour, (c) release and 
receiver pressures? 
(a) From Eq, (559) 

( m .e.p.)=(in.p,)^{3 + lo ge ^-^-^}-(bk.p,) ) 

= 150X.5X^{3 + loga4.2---|-- 3>< 8 2 _ 25 } -10=39 lb, per sq.in, 



hence 



39X2X573X250 
UL *'~ 33,000 ^- 338 * 



(6) From Eq. (561). 

n , . - , . , - 13,750 Z H D H , 

Cubic feet per horse-power per hour= > x — - — 

(m.e.p.) D L 

13,750 r 1 - n 

= -i9- X - 5X 5^ =34 ^ 

hence total pounds per hour will be, 

34.9 X.338X. 332 =3920. 



(c) From Eq. (553) 



1st (rec.pr.)=(m.pr.)-^-£p 



.5 
= 150 X — =89 lbs. per sq.in absolute. 

.o7o y^Jj.Zo 



From Eq. (554) 



2d (rec.pr.) =(m.pr.)-— — , 



.5 
= 150 X ' =3.75. per sq.in absolute. 

.o7o Xo.uo 

High-pressure release pressure may be found from relation (m.pr.) ZhDh 
= (rel.pr.)//!)// •'• (rel.pr.)# =75 lbs. Similarly 1st (ree.pr.)Z/D/ = (rel.pr.)/Dj, or 
(rel.pr.)/=33.4. Similarly 2d (rec.pr.)Z i Z>£ = (rel.pr.)i t Di, or (rel.pr.)z, = 14.8. 

Prob. 1. What would be the horse-power and steam used per hour by a 10- and 

16- and 25x20-in. infinite receiver, no-clearance engine, running at 185 R.P.M. 

on an initial pressure of 180 lbs. per square inch gage and atmospheric exhaust. 

Cut-offs .4, .35, and .3. 

Prob. 2. The following data are reported for a test of a triple engine: 

Size 20x33x52x42 ins., speed 93 R.P.M., initial pessure 200 lbs. per square inch 

gage, back pressure one atmosphere, H.P. cut-off .5, horse-power 16Q0, steam per hsore- 



WOEK OF PISTON ENGINES 357 

power per hour 17 lbs. Check these results, using cut-offs in other cylinders to give 
approximately even work distribution. 

Prob. 3. What change in cylinder sizes would have to be made in the above engine 
to have equal work with a cut-off of \ in each cylinder, keeping the high pressure the 
same size as before? 

Prob. 4. What would be the horse-power of a triple-expansion engine whose low- 
pressure cylinder was 36x3 ins., when running on 150 lbs. per square inch absolute 
initial pressure and 10 lbs. per square inch absolute back pressure, with a cut-off in 
each cylinder of .4 and equal work distribution? Make necessary assumptions. 

Prob. 5. A triple engine 18x24x36x30 ins., running at 100 R. P.M. on an initial 
pressure of 200 lbs. per square inch absolute and back pressure of 20 lbs. per 
square inch absolute, is to be run at such cut-offs as will give complete expansion in 
all cylinders. What will these be, what receiver pressures will result, what horse- 
power can be produced under these conditions, and how much steam will be needed 
per hour? 

Note: § for 200 lbs. =.437. 

22. Multiple-Expansion Engine. General Case. Any Relation between 
Cylinder and Receiver. Determination of Pressure Volume-Diagram and 
Work, by Graphic Methods. It is possible to arrange multiple-expansion 
engines in an almost infinite variety of ways with respect to the pressure-volume 
changes of the fluid that take place in their cylinders and receivers. There 
may be two or three cylinder compounds of equal or unequal strokes, pistons 
moving together by connection to one piston rod, or separate piston rods 
with a common cross-head or even with completely independent main parts 
and cranks at 0°, 180°, displaced with either one leading, or the pistons may 
not move together, being connected to separate cranks at any angle apart, 
and any order of lead. Moreover, there may be receivers of large or small 
size, and there may be as a consequence almost any relation between H.P. 
discharge to receiver and low-pressure receipt from it, any amount of fluid 
passing to correspond to engine load demands and consequently any relation 
of cut-offs, compressions, and receiver-pressure fluctuations. Triple and 
quadruple engines offer even greater varieties of combination of related factors, 
so that problems of practical value cannot be solved by analytical methods 
with anything like the same ease as is possible by graphic means, and in 
some cases not at all. 

These problems that demand solution are of two classes. 

1. To find the work distribution and total work for cylinders of given 
dimensions, clearances, receiver volumes and mechanical connection or move- 
ment relation, with given initial and back pressures, and given valve gear at 
any setting of that valve gear or at a variety of settings. 

2. To find the cylinder relations to give any proportion of the total work 
in any cylinder at any given valve setting or any fraction of initial pressure 
or any value of release pressure or total number of expansions. 

The essential differences between these two classes of problems is that in the 
first the cylinder dimensions are given, while in the second they are to be found. 



358 ENGINEERING THERMODYNAMICS 

In general, however, the same methods will do for both with merely a 
change in the order, and in what follows the dimensions of cylinders, 
valve periods, receiver volume, initial and back pressures will be assumed 
and the diagrams found. By working to scale these diagrams will give the 
work by evaluation of their area, by means either of cross-section paper directly, 
on which strips can be measured and added, or by the planimeter. Thus 
will high- and low-pressure work be evaluated through the foot-pound equiva- 
lent per square inch of diagram, and the total work or the equivalent mean 
effective pressure found by the methods of mean ordinates referred to the 
pressure scale of ordinates. 

In the finding of the pressure-volume diagram point by point there is but one 
common principle to be applied, and that is that for a given mass the product 
of pressure and volume is to be taken as constant (for nearly all steam prob- 
lems, which is the almost sole application of this work) and when two masses 
come together at originally different pressures and mix, the product of the result- 
ing pressure and the new volume, is equal to the sum of the PV products of the 
two parts before mixture. At the beginning of operations in the high-pressure 
cylinder, a known volume of steam is admitted at a given pressure and its 
pressure and volume are easily traced up to the time when it communicates 
with the receiver in which the pressure is unknown, and there difficulty is 
encountered, but this can be met by working from the other end of the 
series of processes. The low-pressure cylinder, having a known compression 
volume at the back pressure, there will be in it at the time of opening to 
receiver a known volume, its clearance, at a known compression pressure. 
The resulting receiver pressure will then be that for the mixture. These two 
receiver pressures are not equal ordinarily, but are related by various compres- 
sions and expansions, involving high- and low-pressure cylinder partial 
displacements, grouped with receiver volumes in various ways. 

Take for an illustrative example the case of a two-cylinder, single-acting, 
cross-compound engine with slide valves, cylinder diameters 12f and 20 ins. 
with 24 ins. stroke for both. High-pressure clearance is 10 per cent, low- 
pressure clearance 8 per cent. Receiver volume 4000 cu.ins. High-pressure 
crank following by 90°. Find the mean effective pressure for the high- and low- 
pressure cylinders, for a cut-off of 50 per cent in the high, and 60 per cent 
in the low, a compression of 10 per cent in the high and 20 per cent in the low, 
initial pressure 105 lbs. per square inch gage, back pressure 5 lbs. per 
square inch absolute, expansion according to logarithmic law. 

On a horizontal line SZ, Fig. 103, lay off the distances 

TU = low-pressure cylinder displacement volume in cubic inches to scale. 
UV = low-pressure cylinder clearance volume in cubic inches to scale. 
VW = receiver volume in cubic inches to scale. 

WX = high-pressure cylinder clearance volume in cubic inches to scale. 
XY = high-pressure cylinder displacement volume in cubic inches to scale. 



WORK OF PISTON ENGINES 



359 



Through these points draw verticals produced above and below, T'T", 
U'U", V'V", WW", X'X", Y'Y". Then will WW and WZ be PV 
coordinates for the high pressure diagram in the quadrant W'WZ, and V'V 



























W 


X' 






Y' 






























V 




Lbs. 
Sq. 


per 

nch 


A 




i 








































H 




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-1 










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f\ 


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r 1 
















i 

i 








<\ J 






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C 


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-4-- 






r 








-30~ - 












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(6) 


K 


































./ 


















V 




























L 












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1 






Z 

ches 


°T 


8 7 
Thousands of 


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Cubic Inches i 


2 1 


U 


V— 


1000 c 


u. In 


_w 


i.L 


2f> 3 

r 


Y 


Thou 
CRA 


sand Cu. Ii 




CRA 


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-y 


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cp 
o 


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a 
o 

W 

o 


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o 
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D 3 


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Ex 


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essur 


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330 


. 












> 






















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O 


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i 


50 60 70 















































T" 



U"V 



W"X' 



Fig. 103. — Graphical Solution of Compound Engine with Finite Receiver and with Clearance 
Illustrating General Method of Procedure for any Multiple Expansion Engine. 

and VS the PV coordinates for the low pressure diagram in the reversed 
quadrant V'VS. 

Lay off AB to represent the high-pressure admission at a height XA rep- 



360 ENGINEERING THERMODYNAMICS 

resenting absolute initial pressure; lay off LM at a height TL representing 
low-pressure exhaust at a constant absolute back pressure to the same scale. 

Locate point B at the cut-off point AB = .50XY on the initial pressure 
line, and. drop a vertical BB 2 and draw similar verticals J J 2 , GG 2 , MM 2 , at 
suitable fractional displacements to represent L.P. cut-off, H.P. and L.P. 
compression volumes resprectively. 

This operation will fix two other points besides the points A and L, B the 
H.P. cut-off at the initial pressure and M the low-pressure compression at the 
back pressure. Through the former draw an expansion line BC and through 
the latter a compression line MN, locating two more points, C and N, at the 
end of the outstroke of the high and instroke of the low. 

At point C the H.P. cylinder steam releases to the receiver of unknown 
pressure, and at N, the L.P. cylinder steam is opened to both the receiver and 
high-pressure cylinder at unknown pressure and volume. 

To properly locate these pressures and volumes from the previously known 
pressures and volumes in a simple manner, the construction below the line 
SZ is used. 

Lay off on W'W" the high-pressure crank angles 0-360°, and to the right 
of each lay off from the clearance line XX" the displacement of the piston at the 
various crank angles for the proper rod to crank ratio, locating the curve 
A 2 B 2 C 2 E 2 F 2 G 2 H 2 . This is facilitated by Table XIII at the end of the 
chapter, but may be laid out graphically by drawing the crank circle and 
sweeping arcs with the connecting rod as radius. 

Opposite H.P. crank angle 270° locate L.P. crank angle 0° = 360° and 
draw to left of the low-pressure clearance line UU" the crank angle dis- 
placement curve for that piston. 

It will be noted that steam volumes are given in the lower diagram by the 
distances from either of these curves toward the other as far as circum- 
stances call for open valves. Thus H.P. cylinder volumes are distances from 
the H.P. displacement curve to WW", but when H.P. cylinder is in communi- 
cation with receiver, the volume of fluid is the distance from H.P. displacement 
curve to VV", and when H.P. cylinder, receiver and L.P. cylinder are all 
three in communication the volume is given by the distance from H.P. 
displacement curve to L.P. displacement curve. This pair of displacement 
curves located one with respect to the other as called for by the crank angle 
relations, which may be made to correspond to any other angular relation, 
by sliding the low up or down with respect to high-pressure curve, serve 
as an easy means of finding and indicating the volumes of fluid occupying 
any of the spaces that it may fill at any point of either stroke. 

On each curve locate the points corresponding to valve periods by the 
intersection of the curve with verticals to the upper diagram, such as BB 2 . 
These points being located, the whole operation can be easily traced. 

At H.P. cut-off (B) the volume of steam is B S B 2 . During H.P. expansion 
(B to C) steam in the high increases in volume from B 3 B 2 to C 3 C 2 . 

During H.P. release (C to D) the volume of steam in the high C 3 C 2 is 



WOEK OF PISTON ENGINES 361 

added to the receiver volume C 4 C 3 , making the total volume C 4 C 2 . During 
H.P. exhaust (D to E) the steam volume C 4 C 2 in H.P. and rec. is com- 
pressed to volume I 3 E 2 . 

At L.P. admission (N) in low and (E) in high, the volume I 2 I 3 is added, 
making the total volume I 2 E 2 in high, low, and receiver. 

During (7 to Q) in low and (F to G) in high the volume I 2 E 2 in high, low 
and receiver, changes volume until it becomes Q 2 G 2 in high, low, and receiver. 

At H.P. compression, G in the high, the steam divides to Q 2 G 3 in low 
and receiver, while G 3 G 2 remains in high and is compressed to H 3 H 2 , at the 
beginning of admission in the high. The former volume Q 2 G 3 , in low and 
receiver, expands to J 2 J 4 , at the moment cut-off occurs in the low, which 
divides the volume into, J 3 J 4 in receiver, which remains at constant volume 
till high-pressure release, and the second part, J 2 J 3 in the low, which expands 
in that cylinder to K 2 K 3 . 

After low-pressure release the volume in low decreases from K 2 K 3 to 
M 2 M 3 , when the exhaust valve closes and low-pressure compression begins. 

During compression in low, the volume decreases from M 2 M 3 to I 2 I 3 which 
is the volume first spoken of above, which combines with I 3 E 2 , causing the drop 
in the high -pressure diagram from E to F. 

The effects upon pressures, of the various mixings at constant volume 
between high, low, and receiver steam and the intermediate common expan- 
sions and compressions may be set down as follows: 

At C, steam in high, at pressure P c , mixes with steam in receiver at 
pressure Pj, resulting in high and receiver volume at pressure P d . 

From D to E there is compression in high and receiver resulting in 
pressure P e . 

At E steam in high and receiver at pressure P e mixes with steam in low, 
at P n , locating points I in low and F in high at same pressure. 

From (F to G in high) and (/ to Q in low) there is a common compression- 
expansion in high, low, and receiver, the pressures varying inversely as the total 
volume measured between the two displacement curves. At G in there begins 
compression in high alone to H. 

In the low and receiver from Q to J there is an expansion and consequent 
fall in pressure from P Q to Pj. 

After low-pressure cut-off at J the expansion takes place in low-pressure 
cylinder alone, to pressure Pic, when release allows pressure to fall (or rise) 
to exhaust pressure Pi. 

When cut-off in low occurs at / the volume J 3 J 4 is separated off in the 
receiver, where it remains at constant pressure Py until high-pressure release 
at point C. 

At the point M compression in low begins, increasing the pressure in low 
alone from P m to P n . 

There are, it appears, plenty of relations between the various inter- 
mediate and common points, but not enough to fix them unless one be first 
established. One way of securing a starting point is to assume a compression 



362 ENGINEERING THERMODYNAMICS 

pressure P g for the beginning of H.P. compression and draw a compression 
line HG through it, produced to some pressure line af, cutting low-pressure 
compression line at d. Then the H.P. intercept (e-f) must be equal to the 
low-pressure intercept (d-c); this fixes (c) through which a PV = const, line 
intersects the L.P. cut-off volume at J. 

Now knowing by this approximation the pressure at J, the pressure may 
by found at D, E, F, and at G. The pressure now found at G may differ 
considerably from that assumed for the point. If so, a new assumption for 
the pressure at G may be made, based upon the last figure obtained, and 
working around the circuit of pressures, J, D, E, F, and back to G should 
give a result fairly consistent with the assumption. If necessary, a third 
approximation may be made. 

It might be noted that this is much the process that goes on in the receiver 
when the engine is being started, the receiver pressure rising upon each release 
from the high, closer and closer to the limiting pressure that is completely reached 
only after running some time. 

These approximations may be avoided by the following computation, 
representing point pressures by P with subscript and volumes by reference 
to the lower diagram. Pj is the unknown pressure in receiver before high- 
pressure release and after low-pressure cut-off. 

Pressure after mixing at D is then 

fi(C 4 C 3 )+P c (C 3 C 2 ) 

(C 4 C 2 ) *• 

The pressure at F, after mixing is 

Pd We*) (/3 ^ 2) +Pn (/2/3) p ? -(c 4 c 3 ) +p c (c 3 c 2 ) +p n ( PP) 

(PE 2 ) PE 2 ~ fm 

(PE 2 ) 
This pressure multiplied by ( Q 2 p2\ & ives Pg, an d this in turn multiplied 

by J2 jl Wl11 § lve p i- 
Writing this in full, 

Solving for P,, 

[P e (.C*C a )+P.( PP)] (Q»<?) ,, sn 

' (Q 2 G 2 )(J 2 J 4 )-(C 4 C 3 )(Q 2 G 3 )' •••••• y™ 1 -) 



WORK OF PISTON ENGINES 363 

which is in terms of quantities all of which are measurable from the diagram. 
While this formula applies to this particular case only, the manner of obtain- 
ing it is indicative of the process to be followed for other cases. 

When there are three successive cylinders the same constructions can be 
used, the intermediate diagrams taking the position of the low for the com- 
pound case, while the low for the triple may be placed under the high and off-set 
from the intermediate by the volume of the second receiver. In this case 
it is well to repeat the intermediate diagram. Exactly similar constructions 
apply to quadruple expansion with any crank angle relations. 



Prob. 1. By means of graphical construction find the horse-power of a 12- and 18 X24- 
in. single-acting cross-compound engine with 6 per cent clearance in each cylinder, 
if the receiver volume is 5 cu.ft., initial pressure 150 lbs. per square inch gage, back 
pressure 10 lbs. per square inch absolute, speed 125 R.P.M., high-pressure compression 
30 per cent, low pressure 20 per cent, high pressure cut-off 50 per cent, low pressure 
40 per cent, high-pressure crank ahead 70°, logarithmic expansion, and ratio of red 
to crank 4. 

Prob. 2. Consider the above engine to be a tandem rather than a cross-compound 
and draw the new diagrams for solution. 

Prob. 3. A double-acting, 15- and 22 x24-in. compound [engine has the high- 
pressure crank ahead by 60°, and has 5 per cent clearance in the low-pressure cylinder, 
10 per cent in the high, and a receiver 4 times as large as the high-pressure cylinder. 
What will be the horse-power when the speed is 125 R.P.M., initial pressure 150 lbs. per 
square inch absolute, back pressure 5 lbs. per square inch absolute, high-pressure cut- 
off \, low-pressure f, high-pressure compression 20 per cent, low-pressure 30 per cent, 
and ratio of rod to crank 5. Determine graphically the horse-power in each cylinder. 

Prob. 4. Consider the engine of Prob. 3 to be a tandem compound and repeat 
the solution. 



23. Mean Effective Pressure, Engine Power, and Work Distribution and 
their Variation, with Valve Movement and Initial Pressure. Diagram Dis- 
tortion and Diagram Factors. Mechanical Efficiency. The indicated power 
developed by a steam engine is dependent upon three principal factors — piston 
displacement, speed, and mean effective pressure. The first, piston displacement, 
is dimensional in character, and, fixed for a given engine. Speed is limited by 
steam and inertia stresses, with which the present treatment is not concerned, 
or by losses due to fluid friction in steam passages, a subject that will be 
further considered under steam flow. Mean effective pressure is a third factor 
which is to be investigated, most conveniently by the methods laid down in 
the foregoing sections. 

In these formulas for mean effective pressure, it will be observed that 
the terms entering are (a) initial pressure, (6) back pressure, (c) cut-off or 
ratio of expansion, (d) clearance, and (e) compression, for the single-cylinder 
engine. It is desirable to learn in what way the mean effective pressure 
varies upon changing any one of these factors. 



364 ENGINEERING THERMODYNAMICS 

Referring to Section 5, Eq. (262) for logarithmic expansion 

1+c 



(m.e.p.) = (in. pr.) 
-(bk.pr.) 



Z+(Z+c)log e 



>e Z+c 
l-X+(X+c)log« 



(mean forward pressure) 

X+c~ 



(mean back pressure) 



(582) 



it is seen that the mean effective pressure is the difference between a mean 
forward pressure and a mean back pressure. The former depends on 
initial pressure, cut-off, and clearance, and the latter on back pressure, 



































1 


1 1 II 1 1 1 1 1 1 


























































M 


EAN 


FORWARD RRE$SURE 






























M 


.F.P. 
























M 


.F.P. 




















: 


M 


F. 


P. 


























WITH VARYING INITIAL PRESSURE 
CONSTANT CUT OFF=.25 
'< CLEAR. = .10 


— 1 


00 
90 
80 

70 


— 
























—100 






WITH VARYING CLEARANCE 

CONSTANT IN PR.= 1 LBS. 

" CUT OFF = .25 






90 

80 
70 
GO 
50 
40 
30 
20 










































/s- 


<^ 














SO 






























































^ 






























































s 


*- 












7$. 
















































































V / 
















































































ft 


* 




WITH 


/ARYING CUT OFF 




40 
30 
20 






















































30 




T 






<« CLEAR. =-.10 | 

" ZERO CLEARANCE 






















































/ 






























































/ 


















































JO 


















In 


itial ] 


J ress 




/ 


1 


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100 


120 


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WITlH VARYING COMPRESSION 
CONSTANT BACK PRESS =1 5 LBS. 
1 << CLEARANCE =. 10 










- 








WITH VARYING BACK 
PRESSURE 
CONSTANT COMP.=. 35 
"1 CLEAR.= .10 




10 


: 












WITH 


VARYING CLEARANCE 
ANT BACK PRESS .-15 L 


3Sr 














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15 20 25 



.1 .2 



.4 .5 .6 .7 



1.0 .05 .10 .15 .20 



Fig. 104. — Curve to Show Variation of Mean Forward and Mean Back Pressure for Logarithmic 
Expansion and Compression in a Single Cylinder Engine with Clearance. 

compression, and clearance. To study the effect of varying these terms it 
is most convenient to draw curves such as are shown in Fig. 104, and examine 
mean forward pressure and mean backward pressure separately. 

Mean forward pressure is seen by inspection to vary in direct proportion 
to initial pressure. Cut-off, when short, gives a low mean forward pressure, 
but it is to be noted that zero cut-off will not give zero mean effective 
pressure so long as there is clearance, due to expansion of steam in the 
clearance space. Increasing the length of cut-off, or period of admission, 
increases mean forward pressure, but not in direct proportion, the (m.f.p.) 
approaching initial pressure as a limit as complete admission is approached. 



WORK OF PISTON ENGINES 



365 



Clearance has the tendency as it increases, to increase the mean forward 
pressure, though not to a great extent, as indicated by the curve Fig. 104. 

Mean back pressure is usually small as compared to initial pressure, though 
a great loss of power may be caused by an increase of back pressure or com- 
pression. Back pressure enters as a direct factor, hence the straight line through 
the origin in the figure. So long as compression is zero, back pressure and 
mean back pressure are equal. When compression is not zero, there must 
be some clearance, and the ratio of (mean bk.pr.) to (bk.pr.) depends on both 
clearance and compression, being greater for greater compressions and for 
smaller clearances. 



100 



































































































































































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M 


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( 




•( 


5 


.10 


.1 


5 


.20 


.25 




















































































































































































































































M 


.E 


F. 
























M 


.E.P. 






































































































































































































































































INITIAL PRE 


SSURE 


100 LBS.ABS 


























































CLEARANCE 10 
CUT OFr .25 
COMPRESSION .35 


































































































































































































































































































































































, 












Back .Press. 
















, Compression 





























10 15 20 25 



.3 .4 



.7 .8 .9 1.0 



Fig. 105. — Curves to Show Variation of Mean Effective Pressure for Logarithmic Expansion 
and Compression in a Single Cylinder Engine with Clearance. 



The mean effective pressures obtained by subtracting mean back from 
mean forward pressures in Fig. 104 are shown in curve form in Fig. 105. 

The multiple expansion engine can not be so simply regarded. In a general 
way each cylinder may be said to be a simple engine, and subject to variations 
of mean effective pressure due to change in its own initial pressure and back 
pressure, clearance, cut-off and compression, which is true. At the same time 
these factors are interrelated in a way that does not exist in the simple 
engine. Consider, for instance, the high-pressure cylinder of a compound 
engine with infinite receiver, with clearance. An increase of high-pressure 
compression tends first to raise the mean back pressure according to the 
reasoning on simple engine, but at the same time the change has decreased 
the volume of steam passing to receiver. No change having been made 



366 ENGINEERING THERMODYNAMICS 

in the low-pressure cylinder, the volume admitted to it will remain the same 
as before, and the receiver pressure will fall, decreasing mean back pressure 
by a greater amount than compression increased it, and mean forward being 
the same as before, the increase of high-pressure compression has increased 
the mean effective pressure of the high-pressure cylinder. The only effect 
upon the low-pressure cylinder is that resulting from lowering its initial 
pressure, i.e., the receiver pressure. This results in a decrease of low-pressure 
mean effective pressure. Computation will show that the assumed increase 
of high-pressure compression decreases low-pressure work more than it 
increases high-pressure work, or in other words, decreased mean effective 
pressure referred to the low. 

It is impracticable to describe all results of changing each of the 
variables for the multiple-expansion engine. Initial pressure and cut-off 
in the respective stages have, however, a marked influence upon receiver 
pressures and work distribution which should be noted. Power regulation 
is nearly always accomplished by varying initial pressure, i.e., throttling, 
or by changing cut-off in one or more cylinders. 

The effect of decreasing initial pressure is to decrease the pressures on 
the entire expansion line and in all no clearance cycles, to decrease absolute 
receiver pressures in direct proportion with the initial pressure. Since back 
pressure remains constant, the result is, for these no-clearance cycles, that 
the mean effective pressures of all but the low-pressure cylinder are decreased 
in direct proportion to the initial pressure, while that of the low-pressure is 
decreased in a greater proportion. The same is true only approximately 
with cycles having clearance and compression. 

The conditions giving equal work division have been treated in connection 
with the individual cycles, it may here be noted in a more general way that 
if high-pressure cut-off is shortened, the supply capacity of that cylinder is 
decreased, while that of the next cylinder remains unchanged. The result 
is that the decreased supply volume of steam will be allowed to expand to 
a lower pressure before it can fill the demand of the next cylinder than it 
did previously, i.e., the receiver pressure is lowered. Similarly shortening 
cut-off in the second cylinder will tend to increase receiver pressure. To 
maintain constant work division, there must be a certain relation between 
cut-offs of the successive cylinders, which relation can only be determined 
after all conditions are known, but then can be definitely computed and 
plotted for reference in operation. 

So far, in discussing the steam engine, "cycles only have been treated. 
These cycles are of such a nature that they can be only approached in 
practice, but since all conclusions have been arrived at through reasoning- 
based on assumed laws or hypotheses, the term hypothetical may be applied 
to all these cycles. It is desirable to compare the actual pressure-volume 
diagram, taken from the indicator card of a steam engine, and the hypothetical 
diagram most nearly corresponding with the conditions. 

In Fig. 106 is shown in full lines a pressure-volume diagram which has 






WORK OF PISTON ENGINES 



367 



been produced from an actual indicator card taken from a simple non-con- 
densing, four-valve engine having 5 per cent clearance. 

Finding the highest pressure on the admission line A'B' and the lowest 
pressure on the exhaust line D'E', these pressures are regarded as (initial 
pressure) and (back pressure) and a hypothetical diagram constructed cor- 
responding to Cycle III, with cut-off and compression at the same fraction of 
stroke as in the actual engine. 

The first difference between the hypothetical and actual PV diagrams 
is that the point of release C is not at the end of stroke, as was assumed for the 
hypothetical release, C, a difference which is intentional, since it requires time 
for pressure to fall after release to the exhaust pressure. This same fact may 
cause the corner of the diagram to be rounded instead of sharp as at D. 
Similarly, the point of admission F' is before the end of the return stroke has 



60 



JO 



20 

























A 






K 


R 




















\ 
















V 

r 








B'\ 




V 






















^ 


u 
























c^- 


\ 


—J 

H-j- 
< i 

\jn> 




^. 


E' 














C'N 




f 
















i 
i 

i 



Fig. 106. — Diagram to Illustrate Diagram Factors. 

been reached, and for a similar reason the corner A' may be rounded, though 
if release and admission are made sufficiently early the corners D' and A' will be 
sharp, approaching the hypothetical, H and A. 

These differences, however, have little effect upon the area of the actual 
diagram, which is seen to be much smaller than the hypothetical. This 
deficiency of area is the net result of a large number of influences, only a few 
of which can be fully explained in connection with the pressure volume 
discussion. 

Beginning with the point of admission, F f , the line F'A'B' represents 
the period of admission. The rounding at A' has been explained; the inclina- 
tion of the line from A' toward B' is due in part to the frictional loss of 
pressure as the fluid passes at high velocity through ports and passages from 
steam chest to cylinder. As the stroke progresses, the linear velocity of the 
piston increases toward mid-stroke, requiring higher velocities in steam 
passages. The greater consequent friction causes pressure to fall in the 
cylinder. The resistance of pipes and valves leading to the engine have 



368 ENGINEERING THERMODYNAMICS 



an effect on the slope of this line. As cut-off is approached, this pressure 
fall becomes more rapid, due to the partial closure of the admission valve. 

From B', the point of cut-off, to C f , the point of release, is the period 
of expansion, during which the pressures are much lower than during the 
hypothetical expansion line BC, due principally to the lower pressure at the 
point of cut-off B' than at B. Hence, the frictional fall in pressure during 
admission has a marked effect upon the work done during expansion. The 
curve B'C rarely follows the law PV = const, exactly, though it commonly 
gives approximately the same work area. During the first part of expansion, 
the actual pressure commonly falls below that indicated by this curve, but 
rises to or above it before the expansion is complete. This is largely due 
to condensation of steam on the cylinder walls at high pressures, and its 
reevaporation at lower pressure, to be studied in connection with a thermal 
analysis of the cycle. The curve of expansion may also depart from this 
very considerably, due to leakage, either inwardly, through the admission 
valve, or by piston from a region of higher pressure, or outwardly, through 
exhaust valve, or by piston into a region of lower pressure, or by drain, 
indicator, or relief valves, or by stuffing-boxes. 

! From the opening of the exhaust valve at the point of release, C, till 
its closure at compression E', is the period of exhaust. Pressures during this 
period, as during admission, are affected by frictional losses in the passages 
for steam, in this case the pressure in the cylinder being greater than that 
in exhaust pipe due to friction, by an increasing amount, as the velocity of 
the piston increases toward mid-stroke. Thus the line DE f rises above the 
line DE until the partial closure of the exhaust near the point of compres- 
sion causes it to rise more rapidly. 

At the point of compression E' the exhaust valve is completely closed 
and the period of compression continues up to admission at F' '. Leakage, 
condensation, and reevaporation affect this line in much the same way as 
they do the expansion, and often to a more marked degree, due to the fact 
that the volume in cylinder is smaller during compression than during 
expansion, and a given weight condensed, reevaporated, or added or removed 
by leakage will cause a greater change in pressure in the small weight 
present than if the change in weight had occurred to a large body of steam. 

In the compound engine all these effects are present in each cylinder in 
greater or less degree. In addition, there are losses of pressure or of volume 
in the receivers themselves between cylinders, due to friction or conden- 
sation, and where especially provided for, reevaporation by means of 
reheating receivers. The effect of these changes in receivers is to cause a 
loss of work between cylinders, and to make the discharge volume of one 
cylinder greater or less than the supply volume of the next, while these 
were assumed to be equal in the hypothetical cases. 

The effect of all of these differences between the actual and hypothetical 
diagrams is to make the actual indicated work of the cylinder something less 
than that represented by the hypothetical diagram. Since these effects are 






WORK OF PISTON ENGINES 369 

not subject to numerical calculations from data ordinarily obtainable, they 
are commonly represented by a single coefficient or diagram factor which is 
a ratio, derived from experiment, between the actual work and that indicated 
by hypothesis. 

It is at once evident that there may be more than one hypothetical 
diagram to which a certain engine performance may be referred as a standard 
of comparison. When the heat analysis of the steam engine is taken up, 
a standard for comparison will be found there which is of great use. For 
determination of probable mean effective pressure, however, no method of 
calculation has been devised which gives better results than the computa- 
tion of the hypothetical mean effective pressure from one of the standard 
hypothetical diagrams, and multiplying this by a diagram factor obtained 
by experiment from a similar engine, under as nearly the same conditions 
as can be obtained. 

Such diagram factors are frequently tabulated in reference books on the 
steam engine, giving values for the factor for various types and sizes, under 
various conditions of running. Unfortunately, however, the exact standard 
to which these are referred is not stated. In this text it will be assumed, 
unless otherwise stated, that the diagram factor for an actual engine is the 
ratio of the mean effective pressure of the actual engine to that computed 
for Cycle I, without clearance or compression, logarithmic law, with cut-off 
at the same fraction of stroke as usual, initial pressure equal to maximum 
during admission in actual, and back pressure equal to minimum during exhaust 
of the actual engine. 

This is selected as the most convenient standard of comparison for mean 
effective pressures, as it is frequently impossible to ascertain the clearance 
in cases where data are supplied. When it is possible to do so, however, closer 
approximation may be made to the probable performance by comparing 
the actual with that hypothetical diagram most nearly approaching the cycle, 
using same clearance, cut-off, and compression as are found in the actual. 

Commercial cut-off is a term frequently used to refer to the ratio of the 
volume AK to the displacement, Fig. 106, in which the point K is found on 
the initial pressure line AB, by extending upward from the true point of cut-off 
B' a curve PV = const. 

While the diagram factor represents the ratio of indicated horse-power 
to hypothetical, the output of power at the shaft or pulley of engine is less 
than that indicated in the cylinders, by that amount necessary to overcome 
mechanical friction among engine parts. If this power output at shaft or 
pulley of engine is termed brake horse-power (B.H.P) then the ratio of this to 
indicated horse-power is called the mechanical efficiency, E m , of the engine 

Em = Qgy \ (583) 

The difference between indicated and shaft horse-power is the power 
consumed by friction (F.H.P.). Friction under running conditions consists 



370 



ENGINEERING THERMODYNAMICS 



of two parts, one proportional to load, and the other constant and inde- 
pendent of load, or 



(F.H.P.) =iV[(const.) x(m.e.p.) + (const. ) 2 ], 

where N is speed, revolutions per minute. But NX (const.) (m.e.p.) = (I.H.P.)i£i 

and 

(F.H.P.) = (I.H.P.)i£i+iV(const.)2, (584) 

where K\ and (const.)2 are constants to be determined for the engine, whose 
values will change as the conditions of the engine bearing-surfaces or lubri- 
cation alters. This value for (F.H.P.) may be used to evaluate E m , 



' 



E m 



(I.H.P.)- (F.H.P.) 
(I.H.P.) 



1-Kt- 



N (const.): 
(I.H.P.) 



(585) 



but indicated horse-power divided by speed is proportional to mean effective 
pressure, so that 



E m = l—K\ 



K, 



(m.e.p.) 



(586) 



1.00 

.90 

O .80 

c 

% - 60 

■a - 50 

o 

eS 

•§ - 30 
o 

S - 20 

.10 

n 












































































































































































































/ 








































/ 








































/ 








































/ 























































































































,2 4 6 8 10 12 .14 16 18 20 22 24 26 28 30 32 34 36 38 40 

Mean Effective Pressure 
Fig. 107. — Diagram to Show Relation of Mechanical Efficiency and Mean Effective Pressure. 

From this expression, speed has been eliminated, which agrees with general 
observation, that mechanical efficiency does not vary materially with speed. 
Values of the constants K\ and K2 may be ascertained if (m.e.p.) and E m 
are known for two reliable tests covering a sufficient range, by inserting their 
values forming two simultaneous equations. 

The numerical values of K\ found in common practice are between .02 
and .05, and for K2 between 1.3 and 2, in some cases passing out of this range. 
In Fig. 107 is shown the form of mechanical efficiency curve when plotted 
on (m.e.p.) as abscissas, using i£i = .04, i£ 2 = 1.6. It may be noted that at 



WORK OF PISTON ENGINES 371 

higher (m.e.p.) the curve does not approach unity, but the value (1—Ki) 
as a limit. The mechanical efficiency becomes zero for this case, at a mean 
effective pressure of about 1.67 lbs. per square inch, which is that just 
sufficient to keep the engine running under no load. For a given speed and size 
of cylinders, the abscissas may be converted into a scale of indicated 
horse-power. 

Prob. 1. Assuming a back pressure of 10 lbs. per square inch absolute, a clearance 
of S per cent, a cut-off of 40 per cent, and compression of 20 per cent, show how 
(m.e.p.) varies with initial pressure over a range of 200 lbs., starting at 25 lbs. 

Prob. 2. For an initial pressure of 150 lbs. per square inch absolute, show how 
(m.e.p.) varies with back pressure over a range of 30 lbs., starting at \ lb. per square 
inch absolute, keeping other quantities as in Prob. 1. 

Prob. 3. For values of initial pressure, back pressure, etc., as given in Probs. 1 
and 2, show how (m.e.p.) varies with clearance from 1 per cent to 15 per cent. 

Prob. 4. For values of initial pressure, etc., as given in Probst 1 and 2, show how 
(m.e.p.) will vary with cut-off from to 1. 

Prob. 5. For values of initial pressure, etc., as given in Probs. 1 and 2, show how 
(m.e.p.) will vary with compression for values from to 50 per cent. 

Prob. 6. A certain engine developing 675 I.H.P. at a speed of 151 R.P.M., 
delivered at the shaft 606 H.P. measured by an absorption dynamometer. A second 
test at 100 R.P.M. gave 150 I.H.P., and 114 shaft H.P. If this engine is to deliver 
500 H.P. at the shaft at a speed of 150 R.P.M. , what will be the I.H.P.and the mechan- 
ical efficiency? 

Prob. 7. A compound Corliss engine, 25 and 52 ins. diameters, 60 ins. stroke, 
double-acting, was designed for 650 I.H P. at 63 R.P.M. It was found that at this 
speed and I.H.P. the mechanical efficiency was 91 per cent. When running with no 
load, the cylinders indicated 38.1 I.H.P. at 65 R.P.M. Find the probable mechanical 
efficiency when developing 300 I.H.P. at a speed of 64 R.P.M. , 



24. Consumption of the Steam Engine and its Variation with Valve Move- 
ment and Initial Pressure. Best Cut-off as Affected by Condensation and 
Leakage. The weight of steam used by a steam engine per hour divided by 
the indicated horse-power is said to be the water rate or steam consumption 
of that engine. It is almost needless to say that this is not a constant for 
a given engine, since it will change with any change of initial pressure, back 
pressure, or valve setting, leakage, or temperature conditions. Since there 
are at least two other uses of terms water rate or consumption, this may be 
termed the actual water rate, or actual consumption, the latter being a more 
general term which may refer to the weight of fluid used per hour per 
indicated horse-power, whatever the fluid may be, steam, air, carbon dioxide, 
or any other expansive fluid. The present discussion has special reference 
to steam. 

From the hypothetical diagram, by computations such as are described 
for the various foregoing cycles, may be obtained a quantity representing 
the weight of fluid required to develop one horse-power for one hour, by the 



372 ENGINEERING THERMODYNAMICS 

performance of the hypothetical cycle. This may be termed the hypothetical 
consumption or for steam cycles the hypothetical water rate. 

By the use of the actual indicator card, may be obtained, by methods still 
to be described, the weight of fluid accounted for by volumes and pressures 
known to exist in the cylinder, this being called the indicated consumption of 
the engine or indicated water rate if the fluid be steam. 

The heat analysis of the steam-engine cycle will lead to another standard 
of comparison which is of the greatest importance as a basis of determining 
how nearly the actual performance approaches the best that could be 
obtained if the engine were to use all available energy possessed by the steam. 
At present the object is to compare the actual and indicated performance with 
that hypothetically possible with cylinders of the known size. Accordingly 
attention will be confined first to hypothetical consumption, and the quantities 
upon which it is dependent. 

For Cycle III, which is the most general for the single-expansion engine, 
logarithmic law, the expression for consumption in pounds fluid per hour per 
indicated horse-power, found in Section 5, Eq. (267), is as follows: 

Hypothetical consumption, lbs. per hr. per I . H.P. 
13,750 



(m.e.p.) 



CS+O-CW©*)]*. 



in which the value of mean effective pressure itself depends upon (in.pr.), 
(bk.pr.), c, Z, and X. The density of the fluid at initial pressure, d 1} is to be 
ascertained from tables of the properties of steam or of whatever fluid is used. 

In Fig. 108 are the results of computations on the hypothetical steam con- 
sumption, using mean effective pressures as plotted in Fig. 105. For each curve, 
conditions are assumed to be as stated on the face of the diagram, varying only 
one of the factors at a time. 

Other conditions remaining unchanged, it may be noted that consumption 
decreases for an increase of initial pressure, though not rapidly in the higher 
pressure range. 

Cut-off has a marked effect upon consumption, the minimum occurring 
when cut-off is such as to give complete expansion. This occurs when 

1-f-c (in.pr.) 



Z'+c (bk.pr.)' 
or 

Z ' = (1+e) S£prT ) - C ' (587) 

which may be termed hypothetically best cut-off. In the case assumed in the 
diagram, 

Z' = (l+.l)^|-.l = .065. 



WORK OF PISTON ENGINES 



373 



If clearance be varied, maintaining constant compression and cut-off, large 
clearance will give high consumption due to an excessive quantity of fluid 
required to fill the clearance space. Extremely small clearance leads to a high 
pressure at the end of compression, causing a loss of mean-effective pressure, 
and consequent high consumption. Between, the consumption has a minimum 
point, which is dependent for its location on both cut-off and compression. 

Decreasing back pressure has a beneficial effect upon mean effective pressure 
and consumption. This would be still more marked in the figure if a case had 
been selected with a very short cut-off. 

Compression, throughout the ordinary range of practice, has but slight effect 
upon consumption, indicated by the flat middle portion of the curve in Fig. 





| 


1 1 


| 






































| 




















- 





Hyp 


Steam 


Cons 


. 










L 


H 


.s 


c. 




















1 


,0 


H.S 


C. 
























I 






































































35 
























































































\ 












































" J 


























30 










\ 


















30 




























m 


































\ 












































; 
























25 












V 










































,r 




































\ 


\ 












-° 






















































20 
















V. 












HI 






















































; 






















- 






















































15 


: 


















1 




L 


































1 
j 
















: 


















l 






; 




1 


















10 








1 
















10 
i 


: 
















Initjal 


Press. 




. 


1 












, Cut off 




I, , 


, , 




,,(.. 






, Clearance 


) i 


20 


40 


GO 


80 


100 


120 


( 


) 


l 


.2 


.:'. 


a .5 :q 


7 


.8 


.9 HO 





• I 


5 


.10 


.1 


5 | .20 


.25 








| 










































































- 


40 


H 


s.'c 






















30 


H 


.s 


c. 








































































- 




















































35 




























25 


I 




















































: 


























; 


























INITIAL PRESSURE 100 LBS. ABS 
BACK PRESSURE 15 '< " 
CLEARANCE -10 
CUT OFF .25 
COMPRESSION .35 




30 




























20 


: 




















































- 


























25 


- 


























15 


: 


























: 


























: 




















































?n 


- 




















_*_ 






10 


=>, 






















































-_ 
































1 












































15 










1 
| 
















: 




















































: 










1 
















- 








1 












































10 




li- 




M 


1 i 


1 


AJ- 


Bac 


tPre 


ss 




- 


1 






T~ 






Cprhpressi 


on 





























10 15 20 25 



.1.2 .3 .4 .5 



.9 1.0 



Fig. 108. — Curves to Show the Variation of Hypothetical Steam Consumption of Simple 
Engines, Logarithmic Expansion and Compression. 

108. Very small or zero compression permits too much high-pressure steam to 
be admitted to the clearance space without doing work, and excessively large 
clearance causes pressures during compression to rise very high, thereby de- 
creasing mean effective pressure; hence this curve of consumption rises at 
both ends. 

Hypothetically, the best attainable consumption for given initial and back 
pressures is obtained when both expansion and compression are complete. 

The indicated consumption, or, as it is frequently called for the steam engine, 
" steam accounted for by the indicator card " "or " indicated water rate," is 
determined from the indicator card as follows. Let Fig. 109 represent an indicator 
diagram. The points of cut-off and compression are located from the form of 



374 



ENGINEERING THERMODYNAMICS 



the line, at the highest point on the expansion line and the lowest point on the 
compression line respectively. The fraction of the card lengths completed at 
cut-off, 



Z = 



AB 
2D' 



and the fraction of card length from point of compression to end of stroke, 

■AC 

X = AU 



are determined, the pressure at cut-off and compression measured by the 
proper vertical scale, and the corresponding densities, 81, and I2 respectively, 
are ascertained from steam tables for dry saturated steam. Clearance, CI, is 
known or ascertained by the form of the compression curve (Chap. I, Section 12). 



CutOff 




Atm, 



D V 



Fig. 109. — Diagram to Illustrate Method of Determining Indicated Water Rate of Steam 

Engine. 

At the point of cut-off, the weight of dry saturated steam present in the 
cylinder is D(Z-j-c) 81, and at compression the weight present is D(X+c)l 2 , 
on the assumption that the steam in the cylinder is of density l\ and 82 at 
these two instants. Accepting this assumption, the weight of steam used per 
cycle is 



Wt. steam per cycle = w = [(Z+c) 81— (X-\-c)§2\D. 

The work per cycle 

If = 144D(m.e.p.), 

and for n cycles per minute the indicated horse-power is 

THP = 144nZ)(m.e.p .) 
33,000 



(588) 



WORK OF PISTON ENGINES 375 

The indicated consumption is then, in pounds per hour per I.H.P. 
wn 60 X 33,000 XD[(Z+c)%i-(X+c)o 2 ]n 



60 



I.H.P. 144n£>(m.e.p.) 



Ind. consumption, lbs. per hr. per I.H.P. 

13,750 



(m.e.p.) 



[(Z+c)8i-(X+c)8 2 ], (589) 



which is the expression used to find indicated consumption for either simple- 
or multiple-expansion engines. In applying this to the multiple-expansion 
engine the terms Z, X and c are found for any one cylinder, and the mean effective 
pressure is referred to that cylinder. There may be, therefore, as many computa- 
tions as there are expansion stages. For a compound engine, for instance, 
indicated consumption according to high-pressure card is found by inserting 
in formula Z, X and c for the high-pressure card, Bi and $2 for corresponding 
pressures, and for (m.e.p.) use 

(m.e.p. ref. to H.P.) = (m.e.p.)#+(m.e.p.)z,— ^. . . . (590) 

JJh 

If the computation is done by means of events on the low-pressure card, 
the (m.e.p.) must be referred to the low. 

(m.e.p. ref. to L.P.) = (m.e.p.)^rr + (m.e.p.)i (591) 

In general for a multiple-expansion engine 

(m.e.p. ref. to cvl. A) = 2(m.e.p.W-- (592) 

Da 

It is often difficult and sometimes impossible to determine the point of 
cut-off and of compression on the indicator card. The expansion and compres- 
sion lines are of very nearly hyperbolic in form and are usually recognizable. 
The highest point on the hyperbolic portion of the expansion line is regarded as 
cut-off, and the lowest point on the hyperbolic portion of the compression line, 
as the point of compression. It must be understood that by reason of the 
condensation and re-evaporation of steam in cylinders the weight of steam 
proper is not constant throughout the stroke, so that calculations like the 
above will give different values for every different pair of points chosen. The 
most correct results are obtained when steam is just dry and these points are 
at release and compression most nearly. 



376 ENGINEERING THERMODYNAMICS 

When under test of actual engines the steam used is condensed and weighed 
and the indicated horse-power determined, then the actual steam consumption 
or water rate can be found by dividing the weight of water used per hour in 
the form of steam by the indicated horse-power. This actual water rate is 
always greater than the water rate computed from the equation for indicated 
consumption. The reasons for the difference have been traced to (a) leakage 
in the engine, whereby steam weighed has not performed its share of work, to 
(6) initial condensation, whereby steam supplied became water before it could 
do any work, (c) variations in the water content of the steam by evaporation 
or condensation during the cycle, whereby the expansion and compression laws 
vary in unpredictable ways, affecting the work. 

Estimation of probable water rate or steam consumption of engines cannot, 
therefore, be made with precision except for engines similar to those which have 
been tested, in all the essential factors, including, of course, their condition, 
and for which the deficiencies between actual and indicated consumptions 
have been determined. This difference is termed the missing water, and end- 
less values for it have been found by experiments, but no value is of any use 
except when it is found as a function of the essential variable conditions that 
cause it. No one has as yet found these variables which fix the form for an 
empiric formula for missing water nor the constants which would make such a 
formula useful, though some earnest attempts have been made. This is no 
criticism of the students of the problem, but proof of its elusive nature, and the 
reason is probably to be found in the utter impossibility of expressing by a 
formula the leakage of an engine in unknown condition, or the effect of its 
condition and local situation on involuntary steam condensation and evapora- 
tion. It is well, however, to review some of these attempts to evaluate 
missing water so that steam consumption of engines may be estimated. 
After studying the many tests, especially those of Willans, Perry announced 
the following for non-condensing engines, in which the expansion is but little 

Missing wate r = m ___£ (593 , 

Indicated steam ' dV N' 

where d is the diameter of the cylinder in inches and N the number of revolutions 
per minute. This indicates that the missing steam or missing water has been 
found to increase with the amount of expansion and decrease with diameter of 
cylinder and the speed. Thermal and leakage conditions are met by the use 
of difference values of m, for there are given 

m = 5 for well-jacketed, well-drained cylinders of good construction with 
four poppet valves, that is, with minimum leakage and condensation. 

m = 30 or more for badly drained unjacketed engines with slid, valves, that 
is, with high leakage and condensation possibilities. 

w = 15 in average cases. 



WORK OF PISTON ENGINES 377 

For condensing engines Perry introduces another variable — the initial pres- 
sure pounds per square inch absolute, p giving 



120(l+|) 



Missing water^ 
Indicated steam dVnpi ' 

It might seem as if such rules as these were useless, but they are not, especially 
when a given engine or line of engines is being studied or two different engines 
compared; in such cases actual conditions are being analyzed rather than 
predictions made, and the analysis will always permit later prediction of con- 
siderable exactness, if the constants are fixed in a formula of the right empiric 
form. Similar study by Heck has resulted in a different formula involving 
different variables and constants, but all on the assumption that the dis- 
crepancies are due to initial condensation. He proposes an expression equi- 
valent to 



Missing steam .27 S(x2— x\) 



r^P- } ' (595) 



Indicated steam ^/_/y \ piZ 

in which iV = R.P.M. of the engine; 
d = diameter in inches; 
L = stroke in feet; 

*S = the ratio of cylinder displacement surface in square feet to dis- 
placement in cubic feet. 

„ 2X_ 4 _+x 12 L 2 48 

« = / 7\ 9 = T "T" 



W" Ld 



The term (#2— #i) is a constant supposed to take into account the amount 
of initial condensation dependable on the difference between cylinder wall 
and live-steam temperature and is to be taken from a table found by trial as 
the difference between the x for the high pressure and x for the low pressure, 
both absolute, see Table XIV at the end of the Chapter. 

In discussing the hypothetical diagrams, it was found that best economy was 
obtained with a cut-off which gives complete expansion. For other than 
hypothetical diagrams this is not true, which may be explained most easily by 
reference to the curves of indicated, and actual consumption, and missing 
steam, Fig. 110. 

The curve ABC is the hypothetical consumption or water rate for a certain 
steam engine. Its point of best economy occurs at such a cut-off, B, that expan- 



378 



ENGINEERING THEEMODYNAMICS 



sion is complete. The curve GHI is computed by Heck's formula for missing 
water. The curve falls off for greater cut-offs. Adding ordinates of these two 
curves, the curve DEF for probable consumption is found. The minimum 
point in this curve, E, corresponds to a longer cut-off than that of ABC. 
Since cut-off B gave complete expansion, cut-off E must give incomplete expan- 
sion. In other words, due to missing steam, the condition which really gives 
least steam consumption per hour per indicated horse-power corresponds to a 
release pressure, which is greater than the back pressure. 

It should be noted that the minimum point mentioned above will not be 
best cut-off, for the output of the engine is not indicated, but brake horse-power. 



30 



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10 



Per Cent Cut off 

Fig. 110.— Diagram to Show Displacement of Best Cut-off Due to Effect of Missing Water 
from Point B for the Hypothetical Cycle to Some Greater Value E. 

In Fig. Ill on cut-off as abscissa are plotted (EFG) consumption pounds 
per hour per I.H.P., and for the case assumed, (OD) the curve of mechanical 
efficiency, based on cut-off, 



(lbs, stea m per hr.) # B.H.P. _ (lbs. steam per hr.) 

i.h7pT~ - "lilp.~ _ B.H.P. 



or, in other words, 

Consumption, lbs, per hr. per I.H.P. 

E m 



Consumption, lbs. per hr. per B.H.P. (596) 



Due to the increasing value of E m for greater cut-offs, the minimum point B 
corresponds to a cut-off still longer than for the curve EFG, which itself was 
found in Fig. 110 to give a longer cut-off than that of the hypothetical curve. 

Hence the best cut-off for economy of steam, where the net power at the 
shaft is regarded as the output, will be such as to give incomplete expansion, 
or a release pressure above back pressure, this effect being caused by both 
missing steam and by frictional losses. 

Prediction of actual consumption of steam engines as a general proposition 
is almost hopeless if any degree of accuracy worth while is desired, though the 



WORK OF PISTON ENGINES 



379 



effect on steam consumption of changing the value of any one variable can be 
pretty well determined by the previous discussion qualitatively, that is^ in kind, 
though not quantitatively in amount. Probably the best attempt is that 
of Hrabak in German, which takes the form of a large number of tables 
developed from actual tests though not for engines of every class. These tables 
are quite extensive, being in fact published as a separate book and any abstrac- 
tion is of no value. 

There is, however, a sort of case of steam consumption prediction that can be 
carried out with surprising precision and that is for the series of sizes or line 
of engines manufactured by one establishment all of one class, each with 
about the same class of workmanship and degree of fit, and hence having 
leakage and cylinder condensation characteristics that vary consistently through- 
out the whole range. For such as these tables and curves of missing water 







\A 








Mech 


F,f- 










10 












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ler I.H.P. 




















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r Hou 


V 






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.1 .2 

Per Cent Cut off 



Fig. 111. — Diagram to Show Relation of Steam per Hour per I.H.P. and per B.H.P. to Cut-off. 



can be made up and by the best builders are, for making guarantees of steam 
consumption for any service conditions that their engines are able to meet. 

The practice of one firm making what is probably the best line of stationary 
engines in this country is of sufficient interest to warrant description. The 
primary data are curves of indicated water rate plotted to mean effective 
pressure for clearances of three or four per cent, and that mean effective pressure 
is chosen in any one specific case that will give the horse-power desired at the 
fixed speed for some one set of cylinder sizes available. To this indicated 
water rate a quantity is added constituting the missing water which is made 
up of several parts as follows: The first is an addition representing condensa- 
tion which is plotted in curve form as a function of (a) boiler pressure, (6) 
superheat in the steam, (c) piston speed, (d) the class of engine simple, com- 
pound or triple, with jacketed or unjacketed cylinders, and for cylinder ratios 
from 4 to 1, to 6 to 1 in the case of compounds. It is therefore a most complex 
quantity, the nature of the variations in which can only be indicated here. 



380' ENGINEEEING THERMODYNAMICS 

For example, increase of piston speed decreases the condensation loss as 
does multiple expansion, and also jacketing, while increase of superheat in the 
steam also decreases it, but superheat has less effect in triple than in com- 
pounds and less in compound than in simple engines. 

The next factor of correction is that covering leakage losses, also additive 
to indicated water rate and which with it and the condensation loss make 
up the probable steam consumption. The leakage decreases regularly with 
increase of piston speed, is less for large than for small engines, the change 
being rather fast from 50 to 200 horse-power and much slower later, being 
scarcely anything at all over 2000 horse-power. 

Example 1. What cut-off will give the lowest indicated water rate for a 9x12- 
in. engine, with 5 per cent clearance and no compression when running non-condensing 
on an initial pressure of 100 lbs. per square inch gage, and what will be the value 
of this water rate? What steam will be used per hour per brake horse-power 
hypothetically? From Eq. (587) 

_. ,„ ,bk.pr. 
Z'= 1+c— — -c, 
m.pr. 

15 
= (1 +.05)— -.05=8.7 per cent, 
llo 






and 

(m.e.p.) =115 .087 X (.087 +.05) loge^ -15=27.2 lbs. sq.in. 

Hence 



Steam per hour per I.H.P. =%^p 137 '- .05 iXj^ X. 262 = 17.2 lbs. 

From the curve of Fig. 107, assuming it to apply to the engine, for this value of 
(m.e.p.) mechanical efficiency is 90 per cent, hence from Eq. (596) the weight of 
steam per shaft horse-power per hour will be 19.1 pounds. 

Prob. 1. Draw diagram similar to Fig. 108 for following case: 

Initial pressure, 135 lbs. per square inch gage, back pressure 10 lbs. per square 
inch absolute, clearance 5 per cent, cut-off 30 per cent, compression 25 per cent. 

Prob. 2. From indicator diagram shown in Fig. 106 find the indicated water rate 
of the engine from which it was taken. 

Prob. 3. The indicated water rate of a 9Xl2-in. jacketed engine when running 
non-condensing at a speed of 250 R.P.M. with an initial pressure of 100 lbs. per 
square inch gage and \ cut-off is 50 lbs. Using Perry's formula what will be the 
probable actual steam used by engine per horse-power hour ' 

Prob. 4. A 24 x48-in. engine in good condition is found to have an indicated water 
rate of 25 lbs. when cut-off is i, initial pressure 100 lbs. per square inch gage, back 
pressure 10 lbs. per square inch absolute, and speed of 125 R.P.M. What will be the 
missing water, and the rate as found by Perry's formula and by Heck's? 



WORK OF PISTON ENGINES 381 

Prob. 5. What will be the probable amount of steam used per hour by a 36x48- 
in. engine with 5 per cent clearance running at 100 R.P.M. on" an initial pressure of 
150 lbs. per square inch gage a back pressure of 5 lbs. per square inch absolute, \ cut- 
off and 10 per cent compression? 

Prob. 6. How will the amount of steam of Prob. 5 compare with that used by 
a 15X22x36-in engine with 5 per cent clearance in each cylinder, running at 100 
R.P.M. on same pressure range with \ cut-off in high-pressure cylinder, \ cut-off in 
low, and 10 per cent compression in each cylinder? 

25. Variation of Steam Consumption with Engine Load. The Willans 
Line. Most Economical Load for More than One Engine and Best Load 
Division. However valuable it may be to the user of steam engines to have 
an engine that is extremely economical at its best load which, it should be 
noted, may have any relation to its rated horse-power, it is more important 
usually that the form of the economy load curve should be as flat as possible 
and -always is this case when the engine must operate under a wide range of 
load. This being the case it is important to examine the real performance 
curves of some typical engines all of which have certain characteristic 
similarities as well as differences. 

From the discussion of hypothetical and indicated water rates it appears 
that the curve of steam consumption (vertical) to engine load (horizontal) 
is always concave upward and always has a minimum point, not at the maxi- 
mum load. Actual consumption curves are similar in general form, but as 
has been pointed out, the load at which the water rate is least corresponds 
to some greater mean effective pressure than that for the hypothetical, so 
the whole curve is displaced upwards and to the right by reason of cylinder 
condensation and leakage losses. This displacement may be so great as to 
prevent the curve rising again beyond the minimum point, in which case the 
least steam consumption corresponds to the greatest load. Just what form 
the actual water rate-load curve will take depends largely on the form of valve 
gear and type of governing method in use, by throttling initial pressure with 
a fixed cut-off or, by varying cut-off without changing initial pressure, with or 
without corresponding changes in the other valve periods. 

Whenever the control of power is by throttling of the supply steam the 
curve is found to be almost exactly an hyperbola, so that (water rate X horse- 
power) plotted to horse-power is a straight line which being characteristic 
is much used in practical work and is known as the Willans line. All other 
engines, that is, those that govern on the cut-off, have Willans lines that 
are nearly straight, such curvature as exists being expressed by a second 
degree equation instead of one of the first degree. 

Equations for Willans lines can always be found for the working range 
of load, that is, from about half to full load, though not for the entire range, 
except in unusual cases, and these equations are of very great value in pre- 
dicting the best division of load between units, which is a fundamental step 
in deciding, how many and what sizes of engine to use in carrying a given 
load in industrial power plants. 



382 



ENGINEERING THERMODYNAMICS 







































































































































































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WORK OF PISTON ENGINES 



383 



Before taking up the derivation of equations some actual test curves will 
be examined and a number of these are grouped in Fig. 112 for engines of 
various sizes, simple and compound, up to 10,000 H.P., on which vertical 
distances represent pounds of steam per hour, per I.H.P. and horizontal 
LH.P. To show the essential similarity of the curves for engines' of different 
size more clearly, these are re-plotted in Fig. 113 to a new load scale based 
on best load of each, which is taken as unity. This is evidently a function of 
mean effective pressure, just what sort of function does not matter here. In 























































































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Percentage of mosi economical load 



150 



175 



2oo 



Fig. 113. — Typical Water Rate — Lead Curves for Steam Engines Plotted to Practional Loads. 



every case the Willans line is also plotted in Fig. 112, each line being num- 
bered to correspond to its water rate curve. 

As there is a corresponding similarity of form for the water rate and 
Willans line of steam turbines, though the reasons for it will be developed 
later, it must be understood that the mathematical analysis that follows applies 
to both turbine and piston steam engines, and finally it makes no difference 
what units are used for load, whether I.H.P., or B.H.P. or K.W. of a direct- 
connected electric generator. 

In Fig. 114 is shown the water-rate curve to a K.W. base for the 10,000 K.W. 
Curtis steam turbine at the Chicago Edison, Fiske Street Station for which the 
following equation fits exactly: 



Y 17.02 



+ 10.54+. 156P, 



384 



ENGINEERING THERMODYNAMICS 



where Y — pounds of steam per hour ^ 1000; 

P = load (in this case in K.W.)-f-1000; 

Y 

p = pounds of steam per K. W. hr. 

A similar equation fits fairly well the curve of Fig. 115, representing the 
7000 H.P. piston engines of the Interborough Railway, Fifty-ninth Street 
station, as well as the combined piston engine and low-pressure steam turbine 



































































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4500 5000 



rooo 9000 liooo 

1000 P 5= Kilowatts 



13000 



15000 



Fig. 114.— Performance of a 10G00-K.W. Steam Turbine. 



taking its exhaust steam, in the same station, but with different numerical 
constants, as below: 

Piston engine, -= = — - ^- + .6+1.85P, 

Y 89 4 
Combined piston engine and turbine, p = ~B 2.90+.713P. 

A third case of smaller size is shown in Fig. 116, representing the per- 
formance of a 1000-K.W. Corliss piston engine driving a generator for which 
the equation is 



Y = 9& 
P P 



-.8+8.3P. 



WORK OF PISTON ENGINES 



385 



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10000 11000 12000 13000 1J000 15000 16000 
Kilowatts = 1000 P 



Fig. 115.— Performance of a 7000-H.P. Piston Engine alone and with a Low-pressure 

Steam Turbine. 



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Fig. 116.— Performance of a 1000-K.W. Steam Turbine. 



386 ENGINEERING THERMODYNAMICS 

These illustrations could be multiplied indefinitely, but those given will 
suffice to establish the fact that the two following equations are fundamental 
over the working range of any steam engine of whatever type: 

Water rate line, lp = j^+B+CP (597) 

Water per hour, Willans line, Y = A+BP+CP 2 , (598 

in which F is the weight of steam per hour and P the engine load whether 
expressed in indicated or brake horse-power, or in kilowatts. 

At the most economical load the water rate is a minimum, so that 



dP 



(8—i(w.) 



whence the most economical load is 



P'=4d- ■ < 5 ") 



Where the Willans line is straight, C = 0, and the most economical load 
is the greatest load. 

Two engines carrying the same load must divide it and some one pro- 
portion may be best. To find out, consider first any number of similar engines, 
that is, engines that have the same constants A, B, and C, denoting each case 
by subscripts. Then 

Let P = total load; 

li Pi, P2, P3, etc. = individual engine loads; 
11 Y = total water per hour; 
" Fi, F2, Ys = water per hour for each engine. 

Then 

F=Fi + F 2 +F 3 +. . . + Y n 
= nA+B(P l +P 2 +Ps+. . • P w )+C(Pi 2 +P 2 2 +P 3 2 +. . -+P„ 2 ) 
= nA+5P+C(P! 2 +P 2 2 +P3 2 +. • .+P« 2 ). 

Only the last term is variable and this is a minimum when 

Pi = P2 = P3 = P n - 

Therefore for similar engines, the best division of load is an equal division. 



WORK OF PISTON ENGINES 387 

When the engines are dissimilar it is convenient to first consider the case 
of straight Willans lines for which C = 0. Then for two such engines 

Y = A 1 +A 2 +B 1 P 1 +B 2 P2 

= (A 1 +A 2 )+B 1 (P-P 2 )+B 2 P 2 
= (A 1 +A 2 )+B 1 P+(B 2 -B 1 )P 2 . 

At any given load P the first two terms together will be constant, and the 
water per hour will be least when the last term is least. As neither factor 
can be zero, this will occur when P 2 is least. 

Therefore for two dissimilar engines the best division of load is that which 
puts the greatest possible share on the one with the smaller value of B, in its equation, 
provided each has a straight Willans line. 

Two dissimilar engines of whatever characteristics yield the equation, 

Y = A l +A 2 +B l P l +B 2 P 2 +C 1 P 1 2 +C 2 P 2 2 

= (A 1 +A 2 +B,P+C 1 P 2 ) 

+ (B 2 -2PC 1 -B 1 )P 2 

+ (C 1 +C 2 )P 2 2 . 

Differentiation with respect to P 2 , and solving for P 2 , the load for the second 
engine that makes the whole steam consumption least, gives, 

P2 = 2(^T^) + (Cl+^) (600) 

= constant + constant X P. 

Therefore, the load division must be linear and Eq. (600) gives the numerical 
value, when any two engines share a given load. 

This sort of analysis can be carried much further by those interested, but 
space forbids any extension here. It is proper to point out, however, that 
by means of it the proper switch-in points for each unit in a large power station 
can be accurately found, to give most economical operation on an increasing 
station load. 

26. Graphical Solution of Problems on Horse-power and Cylinder Sizes . 
The diagram for mean effective pressures in terms of initial and back 
pressure, clearance, compression and cut-off, Fig. 117, facilitates the solution 
of Eq. (262) in Section 5. The mean effective pressure is the difference 
between mean forward and mean back pressure. The former is dependent 
upon clearance, cut-off and initial pressure. In the example shown on the 
figure by letters and dotted lines, clearance is assumed 5 per cent, shown at 
A. Project horizontally to the point F, on the contour line for the assumed 
cut-off, 12 per cent. Project downward to the logarithmic scale for " mean 



388 



ENGINEERING THERMODYNAMICS 













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3 CM 



WORK OF PISTON ENGINES 389 

forward pressure in terms of initial pressure " to the point G. On the scale 
for " initial pressure " find the point H, representing the assumed initial pres- 
sure, 115 lbs. absolute. Through G and H a straight line is passed to the point 
K on the scale for " mean forward pressure," where the value is read, 
m.f.p. = 49.5 lbs. absolute. 

Mean back pressure is similarly dependent upon clearance, compression 
and back pressure, and the same process is followed out by the points A, B, 
C, D and E, reading the mean back pressure, 3.2 lbs. absolute at the point E. 
Then by subtraction, 

(m.e.p.) = (m.f.p.)-(m.b.p.) =49.5-3.2 = 46.3 lbs. 

Fig. 118 is arranged to show what conditions must be fulfilled in order to 
obtain equal work with complete expansion in both cylinders in a compound 
engine, finite receiver, logarithmic law, no clearance, Cycle VII, when low- 
pressure admission and high-pressure exhaust are not simultaneous. This is 
discussed in Section 11, and the diagram represents graphically the conditions 
expressed in Eqs. (376), (377), (378), (379). 

To illustrate its use assume that in an engine operating on such a cycle, 
the volume of receiver is 1.5 times the high-pressure displacement, 1.5 = ?/, then 

1 

— = .667. Locate the points on the scale at bottom of Fig. 118, corresponding 

to this value. Project upward to the curve marked "ratio of cut-offs " and at 
the side, C, read ratio of cut-offs 

f? = .572. 

Next extending the line A B to its intersection D, with the curve GH, the point 
D is found. From D project horizontally to the contour line representing the 
given ratio of initial to back pressure. In this case, initial pressure is assumed 
ten times back pressure. Thus the point E is located. Directly above E at 
the top of the. sheet is read the cylinder ratio, at F, 

Re = ^ = 2 A. 

If cylinder ratio and initial and final pressures are the fundamental data of 
the problem, the ratio of cut-offs and ratio of high-pressure displacements 
to receiver volume may be found by reversing the order. 



390 



ENGINEERING THERMODYNAMICS 









































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WORK OF PISTON ENGINES 391 



GENERAL PROBLEMS ON CHAPTER III. 

Prob. 1. How much steam will be required to run a 14xl8-in. double-acting 
engine with no clearance at a speed of 200 R.P.M. when initial pressure is 100 lbs. 
per square in. gage, back pressure 28 ins. of mercury (barometer reading 30 ins.), and 
cut-off is J? What will be horse-power under these conditions? 

Note: 8 for 100 lbs. =.26, for 28 ins. =.0029. 

Prob. 2. Draw the indicator cards and combined diagram for a compound steam 
engine without receiver, and with 3 per cent clearance in low pressure and 5 per cent in 
high, when initial pressure is 100 lbs. per square inch absolute, back pressure 10 lbs. 
per square inch absolute, high-pressure cut-off J, high-pressure compression tV> and 
low pressure compression 3. 

Probl 3. A simple double-acting engine, 18x24 ins., is running at 100 R.P.M. 
on compressed air, the gage pressure of which is 80 lbs. The exhaust is to atmosphere. 
If the clearance is 6 per cent and cut-off J, and compression 10 per cent, what horse- 
power is being developed, the expansion being adiabatic, and how long can engine be 
run at rated load on 1000 cu.ft. of the compressed air? 

Prob. 4. Will the work be equally distributed in a 12xl8x24-in. engine with 
infinite receiver and no clearance when cut-off is | in high pressure cylinder, and f in 
low, expansion being logarithmic, initial pressure 150 lbs. per square inch absolute 
and back pressure atmosphere? What wnll be work in each cylinder? 

Prob. 5. The receiver of a 15X20x22 in. engine is 4 times as large as high- 
pressure cylinder. What will be the horse-power, steam used per hour, and variation 
in receiver pressure for this engine, if clearance be considered, zero and initial pressure 
is 125 lbs. per square inch gage, back pressure 5 lbs. per square inch absolute, cut-offs 
\ and f in high- and low-pressure cylinders respectively, and piston speed is 550 ft. per 
minute? 

Note: 8 for 125 lbs. =.315, for 5 lbs. =.014. 

Prob. 6. At no load an engine having 7 per cent clearance, cuts off at 4 per cent 
of its stroke, while at full load it cuts off at 65 per cent of its stroke. At no load, 
compression is 40 per cent and at full load 5 per cent. What percentage of full-load 
horse-power is required to overcome friction, and what percentage of steam used at 
full load, is used on friction load, if initial pressure is constant at 100 lbs. per square 
inch gage, back pressure constant at 5 lbs. per square inch absolute, and expansion 
is logarithmic? 

Note: 8 for 100 lbs. =.262, for 5 lbs. =.014. 

Prob. 7. The initial pressure on which engine is to run is 115 lbs. per square inch 
gage, and steam is superheated and known to give a value of 8 = 1.3. For an engine 
in which clearance may be neglected, work is to be equal, and expansion complete 
in both cylinders, when back pressure is 10 lbs. per square inch absolute. What must 
be the cut-offs and cylinder ratio to accomplish this when receiver is 3^ times high 
pressure cylinder volume? 

Prob. 8. A 12-in. and 18x24 ins. double-acting engine with zero clearance and 
infinite receiver operates on an initial pressure of 150 lbs. per square inch gage, and 



392 ENGINEERING THERMODYNAMICS 

a back pressure of 5 lbs. per square inch absolute. What will be the release and receiver 
pressures, horse-power, and steam consumption when speed is 150 R.P M.> expansion 
logarithmic, and cut-off § in each cylinder? 

Note: 8 for 150 lbs. =.367, for 5 lbs. = .014. 

Prob. 9. If a third cylinder 24 ins. in diameter were added to engine of Prob. 8 
and cut-off in this made J, how would horse-power, steam consumption, receiver and 
release pressures change? 

Prob. 10. What would have to be size of a single cylinder to give same horse-power 
at same revolutions and piston speed as that of engine of Prob. 8 under same conditions 
of pressure and cut-off? 

Prob. 11. With the high-pressure cut-off at \, and low and intermediate cut-offs 
at fV, what will be the horse-power, water rate and receiver pressures of a 30 X 48 X 77 X72- 
in. engine running at 102 R.P.M. on an initial pressure of 175 lbs. per square inch gage 
and a back pressure of 26 & ins. of mercury (barometer reading 30 ins.), if the 
receiver be considered infinite and expansion logarithmic, clearance zero? What change 
in intermediate and low-pressure cut-offs would be required to give equal work distribu- 
tion? 

Note: 8 for 175 lbs. =.419, for 26 ins. =.0058. 

Prob. 12. If it had been intended to have all the cut-offs of the engine of Prob. 
11, equal to J, what should have been the size of the intermediate and low-pressure 
cylinders to give equal work for same pressure range and same high-pressure cylinder? 
Prob. 13. To attain complete expansion in all cylinders of the engine of Prob. 11, 
with the initial and back pressures as there given, what cut-offs would be required 
and what receiver pressures would result? 

Prob. 14. A compound locomotive has no receiver, the high-pressure clearance 
is 8 per cent, and low-pressure clearance 5 per cent. The cylinders are 22 and 
33x48 ins., high-pressure cut-off f, high- and low-pressure compression each 10 per 
cent, initial pressure 175 lbs. per square inch gage, back pressure one atmosphere, and 
expansion and compression logarithmic. What will be the horse-power at a speed 
of 40 miles per hour, the engine having 7-ft. driving wheels? At this speed, how 
long will a tank capacity of 45,000 gallons last? 
Note: 8 for 175 lbs. =.419, for 15 lbs. =.038. 

Prob. 15. A superheater has been installed on engine of Prob. 14 and expansion 
and compression, now follow the law PV S =c, when s = 1.2. What effect will this have 
on the horse-power and steam consumption? 

Prob. 16. What will be the maximum receiver pressure work done in each cylinder 
and total work for a cross-compound engine 36 and 66 X48 ins., running at 100 R.P.M. 
on compressed air of 100 lbs. per square inch gage pressure, exhausting to atmosphere 
if the high pressure cut-off is f, clearance 6 per cent, compression 20 per cent, low- 
pressure cut-off is |, clearance 4 per cent, compression 15 per cent, and receiver volume 
is 105 cu.ft.? 

Prob. 17. A manufacturer gives the horse-power of a 42x64x60-in. engine as 
2020, when run at 70 R.P.M. on an initial pressure of 110 lbs. per square inch gage, 
atmospheric back pressure, and .4 cut-off in high-pressure cylinder. How does this 
value compare with that found on assumption of 5 per cent clearance in high, 4 per 
cent in low, and complete expansion and compression is each cylinder? 

Prob. 18. A mine hoisting engine is operated on compressed air with a pressure of 
150 lbs. per square inch absolute and exhausts to atmosphere. The cylinder sizes are 
26X48x36 ins., and clearance is 5 per cent in each. At the start the high-pressure 



WORK OF PISTON ENGINES 393 

cut-off is 1 and low pressure \, while normally both cut-offs are \. The exhaust 
from high-pressure cylinder is into a large receiver which may be considered infinite. 
The compression is zero at all times. Considering the exponent of expansion to be 
1.4, what Will be the horse-power under the two conditions of cut-off given, for a speed 
of 100 R.P.M.? 

Prob. 19. What must be ratio of cylinders in the case of a compound engine with 
infinite receiver, to give equal work distribution complete expansion and com- 
pression if the least clearance which may be attained is 5 per cent in the high- 
pressure cylinder, and 3 per cent in the low-pressure. The engine is to run non- 
condensing on an initial pressure of 125 lbs. per square inch gage, with expansion 
exponent equal to 1.3? What must be the cut-offs and compressions to satisfy these 
conditions? 

Prob. 20. Assuming 7 per cent clearance in high-pressure cylinder and 5 per cent 
in low, infinite receiver, and no compression, how will the manufacturer's rating of 
2100 H.P. check, for a 36X6x48-in. engine running at 85 R.P.M. on an initial 
pressure of 110 lbs. per square inch gage, and a back pressure of 26 ins. vacuum, with 
.3 cut-off in high pressure cylinder? 

Prob. 21. For a 25X40x36-in. engine, with 5 per cent clearance, t cut-off 
and 20 per cent compression in each cylinder, what will be horse-power for an initial 
pressure of 100 lbs. per square inch gage, and a back pressure of 17.5 lbs. per square inch 
absolute, with logarithmic expansion and compression? 

Prob. 22. What must be the cylinder ratio and cut-off to give complete expansion 
in a no-clearance, 14 and 22 x24-in. engine with no receiver and logarithmic expansion, 
when initial pressure is 100 lbs. per square inch gage, and back pressure 10 lbs. per 
square inch absolute? What will be the horse-power and steam used for these conditions 
at a speed of 150 R.P.M.? 

Note: 8 for 100 lbs. =.262, for 10 lbs. =.026. 

Prob. 23. A 24X20x24 in. engine with no receiver or clearance, runs on com- 
pressed air of 120 lbs. per square inch gage pressure, and exhausts to atmosphere. 
When running at a speed of 125 R.P.M., with high-pressure cut-off \, what horse- 
power will be developed and how many cubic feet of compressed air per minute will 
be required to run the engine, the expansion being adiabatic? Will the work be equally 
divided between the two cylinders? 

Prob. 24. It is desired to run the above engine as economically as possible. What 
change in cut-off will be required, and will this cause a decrease or increase in horse- 
power and how much? How will the quantity of air needed be affected? 

Prob. 25. A mill operates a cross-compound engine with a receiver 3 times as large 
as high-pressure cylinder, on an initial pressure of 125 lbs. per square inch gage, and a 
back pressure of 10 lbs. per square inch absolute. The engine may be considered as 
without clearance, and the expansion as logarithmic. As normally run the cut-off in 
high-pressure cylinder is | and in low, \. It has been found that steam is worth 25 
cents a thousand pounds. What must be charged per horse-power day (10 hours) 
to pay for steam if the missing water follows Heck's formula? 

Note 8 for 125 =.315, for 10 = .026. 

Prob. 26. By installing a superheater the value of s in Prob. 25 could be changed 
to 1.3. The cost of st am would then be 30 cents a thousand pounds. From the effect 
on value of s alone would the installation of the superheater pay? 

Prob. 27. When a 26X48x36-in. cross-compound engine with a receiver volume 
of 35 cu.ft. and zero clearance, is being operated on steam of 125 lbs. per square inch 



394 ENGINEERING THERMODYNAMICS 

gage initial pressure, and atmospheric exhaust, is the work distribution equal, when 
high-pressure cut-off is f and low-pressure cut-off §? For these cut-offs what is 
fluctuation in receiver pressure and what steam will be used per horse-power hour? 

Note: 8 for 125 =.315, for 15 lbs. =.038. 

Prob. 28. To operate engine of Prob. 27 under most economical conditions, what 
values must be given to the cut-offs, and what values will result for receiver pressures, 
horse-power, and steam used per hour? 

Prob. 29. What will be the horse-power and steam used by a 20x30x36-in. 
engine with infinite receiver and no clearance, if expansion be such, that s = 1.25, 
high-pressure cut-off i, low-pressure cut-off |, initial pressure 100 lbs. per square inch 
gage, back pressure 3 lbs. per square inch absolute, and speed 100 R.P.M. 

Note: 8 for 100 lbs. =.262, for 3 lbs. =.0085. 

Prob. 30. The following engine with infinite receiver and no-clearance, runs on 
steam which expands according to the logarithmic law. Cylinders 9, and 13x18 
ins., initial pressure 125 lbs. per square inch gage, back pressure 5 lbs. per square inch 
absolute, high-pressure cut-off §, low-pressure f, speed 150 R.P.M. What will be 
horse-power and steam consumption hypothetical and probable? 

Note: 8 for 125 lbs. =.315, for 5 lbs. =.014. 

Prob. 31. By graphical means find the (m.e.p.), of a 15X22x30-in. cross- 
compound engine, with 5 per cent clearance in each cylinder, if the receiver volume is 
8 cu.ft., initial pressure 125 lbs. per square inch absolute, back pressure 10 lbs. per 
square inch absolute, high-pressure cut-off f , low-pressure ts, high-pressure compres- 
sion 40 per cent, low-pressure 20 per cent, high-pressure crank following 90°, logarithmic 
expansion. 

Prob. 32. Show by a series of curves, assuming necessary data, the effect on 
(m.e.p.) of cut-off, back pressure, clearance, and compression. 

Prob. 33. Show by curves, how the indicated, and actual water rate, of an 18x24- 
in. engine with 5 per cent clearance, and running at 125 R.P.M. on an initial pressure 
of 125 lbs. per square inch gage, and a back pressure of 10 lbs. per square inch absolute, 
may be expected to vary with cut-off from ^o to f . 



TABLES 



395 



Table XIII 

PISTON POSITIONS FOR ANY CRANK ANGLE 

From Beginning of Stroke Away from Crank Shaft to Find Piston Position from 
Dead-Center Multiply Stroke by Tabular Quantity 



Crank 
Angle. 


J = 4 


-Us 


I 
-=5 


i-« 


^ = 6 


-i-r 


I 
-=8 


I 
-=9 




r 


r 


r 


r 


r 


r 


r 


r 


5 


.0014 * 


.0015 


.0015 


.0016 


.0016 


.0016 


.0017 


.0019 


10 


.0057 


.0059 


.0061 


.0062 


.0063 


.0065 


.0067 


.0076 


15 


.0128 


.0133 


.0137 


.0140 


.0142 


.0146 


.0149 


.0170 


20 


.0228 


.0237 


.0243 


.0248 


.0253 


.0260 


.0265 


.0302 


25 


.0357 


.0368 


.0379 


.0388 


.0394 


.0405 


.0413 


.0468 


30 


.0513 


.0531 


.0545 


.0556 


.0565 


.0581 


.0592 


.0670 


35 


.0698 


.0721 


.0740 


.0754 


.0767 


.0787 


.0801 


.0904 


40 


.0910 


.0939 


.0962 


.0981 


.0997 


*1022 


.1041 


.1170 


45 


.1152 


.1187 


.1215 


.1237 


.1256 


.1286 


.1308 


.1468 


50 


.1416 


.1458 


.1491 


.1518 


.1541 


.1576 


.1607 


.1786 


55 


.1713 


.1759 


.1828 


.1827 


.1853 


.1892 


.1922 


.2132 


60 


.2026 


.2079 


.2122 


.2157 t 


.2186 


.2231 


.2295 


.2500 


65 


.2374 


.2431 


.2477 


.2514 ' 


.2545 


.2594 


.2630 


.2886 


70 


.2730 


.2794 


.2844 


.2885 


.2929 


.2973 


.3013 


.3290 


75 


.3123 


.3187 


.3239 


.3282 


.3317 


.3372 


.3414 


.3705 


80 


.3516 


.3586 


.3642 


.3687 


.3725 


.3784 


.3828 


.4132 


85 


.3944 


.4013 


.4068 


.4113 


.4151 


.4210 


.4254 


.4564 


90 


.4365 


.4437 


.4495 


.4547 


.4580 


.4641 


.4686 


.5000 


95 


.4816 


.4885 


.4940 


.4985 


.5022 


.5081 


.5126 


.5436 


100 


.5253 


.5323 


.5378 


.5424 


.5461 


.5520 


.5564 


.5868 


105 


.5711 


.5775 


.5828 


.5870 


.5905 


.5961 


.6002 


.6294 


110 


.6150 


.6214 


.6265 


.6306 


.6340 


.6393 


.6530 


.6710 


115 


.6600 


.6657 


.6703 


.6740 


.6771 


.6820 


.6856 


.7113 


120 


.7026 


.7080 


.7122 


.7157 


.7186 


.7231 


.7265 


.7500 


125 


.7449 


.7495 


.7533 


.7563 


.7588 


.7628 


.7658 


.7868 


130 


.7844 


.7885 


.7920 


.7947 


.7969 


.8004 


.8030 


.8214 


135 


.8223 


.8258 


.8286 


.8308 


.8327 


.8357 


.8379 


.8535 


140 


.8570 


.8600 


.8623 


.8642 


.8658 


.8682 


.8703 


.8830 


145 


.8889 


.8913 


.8931 


.8946 


.8958 


.8978 


.8993 


.9096 


150 


.9173 


.9191 


.9204 


.9216 


.9226 


.9241 


.9252 


.9330 


155 


.9420 


.9432 


.9452 


.9451 


.9457 


.9468 


.9476 


.9531 


160 


.9625 


.9633 


.9640 


.9645 


.9650 


.9656 


.9661 


.9698 


165 


.9787 


.9792 


.9796 


.9799 


.9802 


.9805 


.9809 


.9829 


170 


.9905 


.9908 


.9909 


.9911 


.9912 


.9913 


.9915 


.9924 


175 


.9976 


.9977 


.9977 


.9977 


.9978 


.9978 


.9979 


.9981 


180 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1 . 0000 



396 



ENGINEERING THERMODYNAMICS 



Table XIV 
VALUES OF x FOR USE IN HECK'S FORMULA FOR MISSING WATER 



Absolute 




Absolute 




Absolute 




Steam Pressure. 


X 


Steam Pressure. 


X 


Steam Pressure. 


X 





170 


70 


297.5 


165 


393 


1 


175 


75 


304 


170 


397 


2 


179 


80 


310 


180 


405 


3 


183 


85 


316 


185 


409 


4 


186 


90 


321.5 


190 


413 


6 


191 


95 


327 


1G5 


416.5 


8 


196 


100 


332 . 5 


200 


420 


10 


200 


105 


338 


210 


427 


15 


210 


110 


343 


220 


43 t 


20 


220 


115 


348 


230 


441 


25 


229 


120 


353 


240 . 


447.5 


30 


238 


125 


358 


250 


454 


35 


246 


130 


362.5 


260 


460.5 


40 


254 


135 


367 


270 


467 


45 


262 


140 


371.5 


280 


473 


50 


269.5 


145 


376 


290 


479 


55 


277 


150 


380.5 


300 


■485 


60 


284 


155 


385 






65 


291 


160 


389 




.& 



Table XV 
SOME ACTUAL ENGINE DIMENSIONS 

Simple 



7X9 


7|X15' 


16 X18 


15JX24 


24 X36 


8X9 


8|X15 


16i X 18 


16 X24 


i 26 X36 


9X9 


12 X15 


17 X18 


18 X24 


26|X36 


5^X10 


13 X15 


17|X18 


20 X24 


28 X36 


6|X10 


14 X15 


18 X18 


22 X24 


14 X42 


8 X10 


14|X15 


19 X18 


24 X24 


15 X42 


9 X10 


15 X15 


20 X18 


16|X27 


16 X42 


10 xio 


16 X15 


29 X19 


17^X27 


18 X42 


11 xio 


17|X15 


12 X20 


10 X30 


20 X'42 


9|X10| 


11 X16 


14 X20 


12 X30 


22 X->2 


mxm 


12 X16 


18 X20 


16 X30 


24 X42 


7|X12 


13 X16 


19 X20 


18 X30 


26 X42' 


8 X12 


14JX16 


28 X20 


18|X30 


28 X-i.i 


8|X12 


15 X16 


21 X20 


20 X30 


18 X48 


9 X12 


15|X16 


22 X20 


24 X30 


20 X48 


10 X12 


16 X16 


12 X21 


22 X33 


22 X48 


11 X12 


17 X16 


13 X21 


24 X33 


24 X48 


1HX12 


18 X16 


18|X21 


10 X36 


26 X48 


12 X12 


18|X17 


20 X21 


12 X36 


28 X48 


12|X12 


23 X17 


20 X22 


14 X36 


24 X54 


13 X12 


26 X17 


18 X24 


16 X36 


26 X54 


14 X12 


10 X18 


10 X24 


18 X36 


28 X54 


10 X14 


11 X18 


12 X24 


20 X36 


28 X60 


11 X14 


15 X18 


141- X 24 


22 X36 





TABLES 



397 



Table XV. — Continued 
Compound 

Notes: 1 to run condensing or non-condensing on initial pressure of 100-150. 

2 to run condensing or non-condensing on initial pressure of 100. 

3 to run condensing or non-condensing on initial pressure of 125. 

4 to run condensing or non-condensing on initial pressure of 90—100. 

5 to run condensing or non-condensing on initial pressure of 110-130. 

6 to run condensing or non-condensing on initial pressure of 140-160. 

7 to run condensing or non-condensing on initial pressure of 125 



4|- 8 

6 -10 

7 -13 

6 -12 

7 -12 

8 -12 
8i-15| 

7 -14 

8 -14 
9^-15 
7|-13| 

9 -15^ 
19 -14 
10 -16 

10 -18 

11 -16 
9 -18 

10 -18 

io -m 

11 -19 

11 -18 

12 -18 
12 -20 



X 6 




X 6 




X 8 




xio 


1 


X10 


1 


xio 


1 


xio 




xio 


1 


xio 


1 


Xll 


2 


X12 


3 


X12 


3 


X12 


1 


X12 


1 


X12 


1 


X12 


1 


X14 


1 


X14 


1 


X14 


3 


X14 


3 


X14 


1 


X14 


1 


X14 


1 



13 -18 
13 -20 
7f-13§ 
9 -15| 
11 -19 
13 -19 
7f-13V 
9 -15| 

io -m 

11 -19 

11 -22 

12 -21 

13 -22 

13 -22* 

14 -22 
141-25 

15 -22 
15£-26£ 

16 -25 
13 -23 
15 -26 
16^-29 

9 -15| 



X14 
X14 
X15 
X15 
X15 
X15 
X16 
X16 
X16 
X16 
X16 
X16 
X16 
X16 
X16 
X16 
X16 
X16 
X16 
X17 
X17 
X17 
X18 



1 
1,2 

3 

3 

4 

5 

3 

3 

3 

3 

1 

3 

1 

3 i 

2 

3 

1 

3 

2 
4,6 
4,6 

4 

3 



11 


-19 X18 


3 


12 


-21 X18 


3 


13 


-22|X18 


3 


14 


-24 X18 


1 


m 


-26|X18 


3 


16 


-24 X18 


1 


16 


-26 X18 




16^ 


-28^X18 


3 


8 


-12 X20 


7 


9 


-14 X20 


7 


16 


-28 X20 


1 


17 


-30 X20 


1 


18 


-28 X20 


1 


19 


-30 X20 


1 


19 


-30 X22 


1 


9 


-15^X21 


3 


12 


-21 X21 


3 


13 


-22^X21 


3 


14| 


-26|X21 


3 


15} 


-281X21 


3 


18| 


-32|X21 


3 


20 


-36 X21 


3 


13 


-23 X22 


5 

i 



14}-26 
18 -32 

io -m 

11 -19 

12 -18 

13 -20 

14 -22 



22 -38 
24 -42 

12 -21 

13 -22 § 
16|-28| 
17|-30| 
14|-25 
15|-26| 
18f-32| 
20 -36 
28|-50 
30 -54 
16|-28| 

17|X30| 



X22 
X22 
X24 
X24 
X24 
X24 
X24 
X24 
X24 
X24 
X24 
X27 
X27 
X27 
X27 
X30 
X30 
X30 
X30 
X30 
X30 
X33 
X33 



22 -38 
24 -42 

18f-32| 
20 -36 
26|-46 
28|-50 
141-25 
151-26| 
18|-32| 
20 -36 
16^-281 
17^-301 
22 -38 
24 -42 
18g— 0Z2 
20 -36 
26^-46 
28|-50 
22 -38 
24 -42 
30 -54 
32±-57 
34 -60 



X33 
X33 
X36 
X36 
X36 
X36 
X42 
X42 
X42 
X42 
X48 
X48 
X48 
X48 
X54 
X54 
X54 
X54 
X60 
X60 
X60 
X60 
X60| 



Triple 

Note: All condensing and to run of initial pressure as given. 



Size. 


V 


Size. 


V 


Size. 


V 


10 -151-26X15 


200 


27 -43 { ^X39 


180 


30-50-82X48 


180 


11 -18 -30X20 


250 


25-41-68X48 


190 


12 -20 -34X24 


180 


18 -281-48X40 


180 


27-45-75X54 


190 


12 -19 -32X24 


190 


22 -37 -63X42 


180 


28-45-72X54 


185 




175 


22 -38 -64X42 


185 


28-46-75X54 


180 


12f-22 -36X24 
14 -23 -28X26 


180 
190 


32^-53 ~{g}x48 


265 


29-47-83X54 
32-52-92X50 


160 
200 


18 -29 -47X30' 
16§-24 -41X30 
18 -30 -50X30 
16 -251-43X30 


200 
180 
200 
190 


35 -57 -[ggX48 

36 -57 -\7gX48 


265 
295 


34-56-100X60 
35-58 IggX 60 
34-57-104X63 


200 
190 
200 


16^-24 -41X30 


180 


28 -45 -72X48 


180 







CHAPTER IV 

HEAT AND MATTER. QUALITATIVE AND QUANTITATIVE RELATIONS 
BETWEEN THE HEAT CONTENT OF SUBSTANCES AND THEIR PHYSICAL- 
CHEMICAL STATE, 

1. Substances and Heat Effects Important in Engineering. It has been 
shown in preceding chapters concerned with work in general and with the deter- 
mination of quantity of work that may be done in the cylinders by, or on 
expansive fluids that 

(a) Fluids originally at low may be put in a high-pressure condition by 
the expenditure of work; 

(b) Fluids under high pressure may do work in losing that pressure. 

That work may be done, fluids under pressure are necessary and that the 
greatest amount of work may be done per unit of fluid the fluid itself must be 
expansive, that is, it must be a gas or a vapor. Gases or vapors under pressure 
are, therefore, prerequisites to the economical use of fluids for the doing of work, 
and that this work may be done at the expense of heat or derived from heat, it 
is only necessary that the heat be used to create the necessary primary con- 
dition of high pressure in vapors and gases. There are two general ways of 
accomplishing this purpose — first, to apply the heat to a boiler supplied with 
liquid and discharging its vapor at any pressure as high as desired or as high 
as may be convenient to manage; second, to apply the heat to a gas confined 
in a chamber, raising its pressure if the chamber be kept at a fixed volume, 
which is an intermittent process, or increasing the fluid volume if the size of the 
chamber be allowed to increase, the fluid pressure being kept constant or not, 
and this latter process may be intermittently or continuously carried out. 

These two processes are fundamental to the steam and gas engines that are 
the characteristic prime movers or power generators of engineering practice, 
utilizing heat energy, and with the exception of water-wheels the sole commer- 
cially useful sources of power of the industrial world. Thus, the heating of 
gases and the evaporation of liquids are two most important thermal processes 
to be examined together with their inverse, cooling and condensation, and 
necessarily associated in practical apparatus with the heating and cooling of 
solid containers or associated liquids. From the power standpoint, the effects 
of heat on solids, liquids, gases and vapors, both without change of state and with 
change of state are fundamental, and the substances to be studied as heat carriers 
do not include the whole known chemical world, but only those that are cheap 
enough to be used in engineering practice or otherwise essential thereto. These 
substances of supreme importance are, of course, air and water, with all their 

398 



HEAT AND MATTER 399 

physical and chemical variations, next the fuels, coal, wood, oil, alcohol and 
combustible gases, together with the chemical elements entering into them 
and the chemical compounds which mixed together may constitute them. 

Probably next in importance from the standpoint of engineering practice 
are the substances and thermal processes entering into mechanical refriger- 
ation and ice making. There are but three substances of commercial importance 
here — ammonia, pure and in dilute aqueous solution, carbonic acid and air. 
The process of heating or cooling solids, liquids, gases and vapors, together 
with solidification of water into ice, evaporation and condensation, fundamental 
to power problems, are also of equal importance here, but there is added an 
additional process of absorption of ammonia vapor in water and its discharge 
from the aqueous solution. 

Many are the practical applications of heat transfer or transmission, some 
of which call into play other substances than those named. In the heating 
of buildings there is first combustion with transfer of heat to wa,ter in boilers, 
flow of the hot water or steam produced to radiators and then a transfer of heat 
to the air of the room ; in feed-water heaters, heat of exhaust steam warms 
water on its way to the boilers ; in economizers, heat of hot flue gases is trans- 
ferred to boiler feed water; in steam superheaters, heat of hot flue gases is trans- 
ferred to steam previously made, to raise its temperature, steam pipes, boiler 
surfaces and engine cylinders transfer heat of steam to the air which is opposed 
by covering and lagging, in steam engine condensers heat of exhaust steam is 
transferred to circulation water; in cooling cold storage rooms and making ice, 
a solution of calcium or sodium chloride in water is circulated through pipes 
and tanks and is itself kept cool in brine coolers in which the brine transfers 
the heat absorbed in the rooms and tanks, to the primary substance ammonia 
or carbonic acid and evaporates it. 

While evaporation and condensation as processes are fundamental to the 
machinery and apparatus of both power and refrigeration, they also are of 
importance in certain other industrial fields. In the concentration of 
solutions to promote crystallization such, for example, as sugar, evapora- 
tion of the solution and condensation of the distillate are primary processes 
as also is the case in making gasolene and kerosene from crude oil, in the making 
of alcohol from a mash, and many other cases found principally in chemical 
manufacture. These are examples of evaporation and condensation in which 
little or no gases are present with the vapor but there are other cases in which 
a gas is present in large proportion, the thermal characteristics of which are 
different as will be seen later. Among these processes are: the humidification 
or moistening of air with water in houses and factories to prevent excessive 
skin evaporation of persons breathing the air, excessive shrinkage of wood-work 
and to facilitate the manufacturing processes like tobacco working and thread 
spinning. Conversely, air may be too moist for the purpose, in which case it 
is dried by cooling it and precipitating its moisture as rain or freezing it out as 
ice, and this is practiced in the Gayley process of operating blast furnaces, where 
excess of moisture will on dissociating absorb heat of coke combustion and reduce 






400 ENGINEERING THERMODYNAMICS 






the iron output per ton of coke, and in the factories where, for example, collodion 
is worked, as in tb^ manufacture of photographic films, with which moisture 
seriously interferes. Of course, humidification of air by water is accomplished 
only by evaporation of water, and evaporation of water is only to be accomplished 
by the absorption of heat, so that humidification of air by blowing it over water 
or spraying water into it must of necessity cool the water, and this is the prin- 
ciple of the cooling tower or cooling pond for keeping down the temperature 
of condenser circulating water, and likewise the principle of the evaporative 
condenser, in which water cooler and steam condenser are combined in one. 
The same process then, may serve to cool water if that is what is wanted, or to 
moisten air, when dry air is harmful, and may also serve to remove moisture 
from solids like sand, crystals, fabrics, vegetable or animal matter to be reduced 
to a dryer or a pulverized state. 

There are some important examples of humidification in which the substances 
are not air and water, and one of these is the humidification of air by gasolene 
or alcohol vapor to secure explosive mixtures for operating gas engines. Here 
the air vaporizes enough of the fuel, humidifying or carburetting itself to serve 
the purpose, sometimes without heat being specifically added and sometimes 
with assistance from the hot exhaust. A somewhat similar action takes place 
in the manufacture of carburetted water gas when the water gas having no 
illuminating value is led to a hot brick checkerwork chamber supplied with a 
hydrocarbon oil, the vapor of which humidifies the gas, the heat of vaporization 
being supplied by the hot walls and regularly renewed as the process is inter- 
mittent. Of course, in this case some of the vapors may really decompose 
into fixed gases, a peculiar property of the hydrocarbon fuels, both liquid and 
gaseous, and frequently leaving residues of tar, or soot, or both. 

Finally, among the important processes there is to be noted that of gasifica- 
tion of solid and liquid fuels in gas producers and vaporizers, a process also 
carried on in blast furnaces in which it is only an accidental accompaniment 
and not the primary process. Some of the actions taking place in gas producers 
are also common to the manufacture of coal gas, and coke, in retorts, beehive 
and by-product ovens. 

From what has been said it should be apparent that engineers are concerned 
not with any speculations concerning the nature of heat but only with the kind 
and quantity of effect that heat addition to, or abstraction from, substances 
may be able to produce and not for all substances either. While this interest 
is more or less closely related to philosophic inquiry, having, for its object the 
development of all embracing generalizations or laws of nature, and to the 
relation of heat to the chemical and physical constitution of matter, subject 
matter of physical chemistry, the differences are marked, and a clearly defined 
field of application of laws to the solution of numerical problems dealing with 
identical processes constitutes the field of engineering thermodynamics. 

It is not possible or desirable to take up and separately treat every single 
engineering problem that may rise, but on the contrary to employ the scientific 
methods of grouping thermal processes or substance effects into types. 



HEAT AND MATTER 401 

Prob. 1. Water is forced by a pump through a feed-water heater and economizer 
to a boiler where it is changed to steam, which in turn passes through a superheater 
to a cylinder from which it is exhausted to a condenser. Which pieces of apparatus 
have to do with heat effects and which with work? Point out similarities and 
differences of process. 

Prob. 2. Air is passed over gasolene in a carburetter; the mixture is compressed, 
burned and allowed to expand in a gas engine cylinder. Which of the above steps have 
to do with heat effects and which with work effects? 

Prob. 3. In certain types of ice machines liquid ammonia is allowed to evaporate, 
the vapor which is formed being compressed and condensed again to liquid. Which of 
these steps is a work phase and which a heat phase? Compare with Problem 1. 

Prob. 4. When a gun is fired what is the heat phase and what is the work 
phase? Arc they separate or coincident? 

Prob. 5. Air is compressed in one cylinder, then it is cooled and compressed to 
higher pressure and forced into a tank. The air in the tank cools down by giving 
up heat to the atmosphere. From the tank it passes through a pipe line to a heater 
and then to an engine from which it is exhausted to the atmosphere. Which steps 
in the cycle may be regarded as heat and which as work phases? Compare with 
Problem 2. 

2. Classification of Heating Processes. Heat Addition and Abstraction 
with, or without Temperature Change. Qualitative Relations. That heat 
will pass from a hot to a less hot body if it gets a chance is axiomatic, so that a 
body acquiring heat may be within range of a hotter one, the connection between 
them being, either immediate, that is they touch each other, or another body 
may connect them acting as a heat carrier, or they may be remote with no more 
provable connection than the hypothetic ether as is the case with the sun and 
earth. A body may gain heat in other ways than by transfer from a hotter 
body, for example, the passage of electrical current through a conductor will 
heat it, the rubbing of two solids together will heat both or perhaps melt one, 
the churning of a liquid will heat it, the mixing of water and sulphuric acid will 
produce a hotter liquid than either of the components before mixture, the absorp- 
tion by water of ammonia gas will heat the liquid. All these and many other 
similar examples that might be cited have been proved by careful investigation 
partly experimental, and partly by calculation based on various hypotheses to 
be examples of transformation of energy, mechanical, electrical, chemical, into 
the heat form. While, therefore, bodies may acquire heat in a great many 
different concrete ways they all fall under two useful divisions : 

(a) By transfer from a hotter body ; 

(b) By transformation into heat of some other energy manifestation. 

One body may be said to be hotter than another when it feels so to the 
sense of touch, provided neither is too hot or too cold for injury to the tissues, 
or more generally, when by contact one takes heat from the other. Thus, 
ideas of heat can scarcely be divorced from conceptions of temperature and the 
definition of one will involve the other. As a matter of fact temperature as 
indicated by any instrument is merely an arbitrary number located by some- 
body on a scale, which is attached to a substance on which heat has some visible 



402 ENGINEERING THERMODYNAMICS 

effect. Temperature is then a purely arbitrary, though generally accepted, 
number indicating some heat content condition on a scale, two points of which 
have been fixed at some other conditions of heat content, and the scale space 
between, divided as convenient. Examination of heat effects qualitatively 
will show how thermometers might be made or heat measured in terms of any 
handy effect, and will also indicate what is likely to happen to any substance 
when it receives or loses heat. Some of the more common heat effects of various 
degrees of importance in engineering work are given below: 

Expansion of Free Solids. Addition of heat to free solids will cause them to 
expand, increasing lengths and volumes. Railroad rails and bridges are longer 
in summer than winter and the sunny side of a building becomes a little higher 
than the shady side. Steam pipes are longer and boilers bigger hot, than cold, 
and the inner shell of brick chimneys must be free from the outer to permit 
it to grow when hot without cracking the outer or main supporting stack body. 
Shafts running hot through lack of lubrication or overloading in comparatively 
cool bearing boxes may be gripped tight enough to twist off the shaft or merely 
score the bearing. 

Stressing of Restrained Solids. A solid being heated may be restrained 
in its tendency to expand, in which case there will be set up stresses in the mate- 
rial which may cause rupture. Just as with mechanically applied loads, bodies 
deform in proportion to stress up to elastic limit, as stated by Hooke's law, so 
if when being heated the tendency to expand be restrained the amount of 
deformation that has been prevented determines the stress. A steam pipe 
rigidly fixed at two points when cold will act as a long column in compression 
and buckle when hot, the buckling probably causing a leak or rupture. If fixed 
hot, it will tend to shorten on cooling and being restrained will break something. 
Cylinders of gas engines and air compressors are generally jacketed with water 
and becoming hot inside, remaining cold outside, the inner skin of the metal 
tends to expand while the outer skin does not. One part is, therefore, in tension 
and the other in compression, often causing cracks when care in designing is 
not taken and sometimes in spite of care in large gas engines. 

Expansion of Free Liquids. Heating of liquids will cause them to expand 
just as do solids, increasing their volume. Thus, alcohol or mercury in glass 
tubes will expand and as these liquids expand more than the glass, a tube which 
was originally full will overflow when hot, or a tube of very small bore attached 
to a bulb of cold liquid will on heating receive some liquid; the movement of 
liquid in the tube if proportional to the heat received will serve as a thermometer. 
If the solid containing the liquid, expanded to the same degree as the liquid 
there would be no movement. Two parts of the same liquid mass may be 
unequally hot and the hotter having expanded will weigh less per cubic foot, 
that is, be of less density. Because of freedom of movement in liquids the lighter 
hot parts will rise and the cooler heavy parts fall, thus setting up a circulation, 
the principle of which is used- in hot water heating systems, the hot water from 
the furnace rising to the top of the house through one pipe and cooling on its 
downward path through radiators and return pipe. In general then, liquids 



HEAT AND MATTER 403 

decrease in density on heating and increase in density on cooling, but a most 
important exception is water, which has a point of maximum density just 
above the freezing-point, and if cooled below this becomes not heavier but lighter. 
Consequently, water to be cooled most rapidly should be cooled at first from 
the top and after reaching this point of maximum density, from the bottom, if it 
is to be frozen. 

. Rise of Pressure in Confined Liquids. When liquids are restrained from 
expanding under heating they suffer a rise of pressure which may burst the 
containing vessel. For this reason, hot water heating systems have at their 
highest point, open tanks, called expansion tanks, which contain more water 
when the system is hot than when cold, all pipes, radiators and furnaces being 
constantly full of water. Should this tank be shut off when the water is cold 
something would burst, or joints leak, before it became very hot. 

Expansion of Free Gases. Just as solids and liquids when free expand under 
heating, so also do gases and on this principle chimneys and house ventilation 
systems are designed. The hot gases in a chimney weigh less per cubic foot 
than cooler atmospheric air; they, therefore, float as does a ship on water, 
the superior density of the water or cold gas causing it to flow under and 
lift the ship or hot gas, respectively. Similarly, hot-air house furnaces and 
ventilating systems having vertical flues supplied with hot air can send it upward 
by simply allowing cold air to flow in below and in turn being heated flow up 
and be replaced. 

Rise of Pressure in Confined Gases. Gases when restrained from expanding 
under heat reception will increase in pressure just as do liquids, only over greater 
ranges, and as does the internal stress increase in solids when heated under 
restraint. It is just this principle which lies at the root of the operation of 
guns and gas engines. Confined gases are rapidly heated by explosive combus- 
tion and the pressure is thus raised sufficiently to drive projectiles or pistons 
in their cylinders. 

Melting of Solids. It has been stated that solids on being heated expand 
but it should be noted that this action cannot proceed indefinitely. Continued 
heating at proper temperatures will cause any solid to melt or fuse, and the pre- 
viously rising temperature will become constant during this change of state. 
Thus, melting or fusion is a process involving a change of state from solid 
to liquid and takes place at constant temperature. The tanks or cans of ice- 
making plants containing ice and water in all proportions retain the same 
temperature until all the water becomes ice, provided there is a stirring or cir- 
culation so that one part communicates freely with the rest and provided also 
the water is pure and contains no salt in solution. Impure substances, such as 
liquid solutions, may suffer a change of temperature at fusion or solidification. 
For pure substances, melting and freezing, or fusion and solidification, are 
constant temperature heat effects, involving changes from solid to liquid, or 
liquid to solid states. 

Boiling of Liquids. Ebullition. Continued heating of solids causes fusion, 
and similarly continued heating of liquids causes boiling, or change of state from 



404 ENGINEERING THERMODYNAMICS 

liquid to vapor, another constant temperature process — just what temperature, 
will depend on the pressure at the time. So constant and convenient is this 
temperature pressure relation, that the altitude of high mountains can be found 
from the temperature at which water boils. The abstraction of heat from a 
vapor will not cool it, but on the contrary cause condensation. Steam boilers 
and ammonia refrigerating coils and coolers are examples of evaporating appara- 
tus, and house heating radiators and steam and ammonia condensers of con- 
densing apparatus. 

Evaporation of Liquids; Humidification of Gases. When dry winds blow 
over water they take up moisture in the vapor form by evaporation at any 
temperature. This sort of evaporation then must be distinguished from ebul- 
lition and is really a heat effect, for without heat being added, liquid cannot 
change into vapor; some of the necessary heat may be supplied by the water 
and some by the air. This process is general between gases and liquids and is 
the active principle of cooling towers, carburetters, driers of solids like wood 
kilns. The chilling of gases that carry vapors causes these to condense in part. 
As a matter of fact it is not necessary for a gas to come into contact to produce 
this sort of evaporation from a liquid, for if the liquid be placed in a vacuum 
some will evaporate, and the pressure finally attained which depends on the tem- 
perature, is the vapor pressure or vapor tension of the substance, and the amount 
that will so evaporate is measured by this pressure and by the rate of removal 
of that which formed previously. 

Evaporation of Solids. Sublimation. Evaporation, it has been shown, may 
take place from a liquid at any temperature, but it may also take place directly 
from the solid, as ice will evaporate directly to vapor either in the presence 
of a gas or alone. Ice placed in a vacuum will evaporate until the vapor tension 
is reached, and it is interesting to note that the pressure of vapors above their 
solids is not necessarily the same as above their liquids at the same temperature, 
though they merge at the freezing-point. This is the case with ice-water- 
water vapor. 

Change of Viscosity. Heating of liquids may have another effect measured 
by their tendency to flow, or their viscosity. Thus, a thick oil will flow easier 
when heated, and so also will any liquid. If, therefore, the time for a given 
quantity to flow through a standard orifice under a given head or pressure be 
measured, this time, which is the measure of viscosity, will be less for any liquid 
hot, than cold, for the same liquid. Viscosity then decreases with heat addition 
and temperature rise. 

Dissociation of Gases. When gases not simple are heated and the heating 
continued to very high temperatures, they will split up into their elements or 
perhaps into other compound gases. This may be called decomposition or, better, 
dissociation, and is another heat effect. Thus, the hydrocarbon C2H4 will 
split up with solid carbon soot C and the other hydrocarbon CH4 and steam 
H2O into hydrogen and oxygen. This is not a constant temperature process, 
but the per cent dissociated increases as the temperature rises. 

Dissociation of Liquids. Similar to the dissociation of gases receiving heat at 



HEAT AND MATTER 405 

high temperature is the decomposition of some liquids in the liquid state, notably 
the fuel and lubricating oils, or hydrocarbons which are compounds of H and C 
in various proportions, each having different properties. Sometimes these 
changes of H and C groupings from the old to the new compounds under the 
influence of heating will be at constant and at other times at varying tempera- 
tures; sometimes the resulting substances remain liquid and sometimes soot 
or C separates out, and this is one of the causes for the dark color of some 
cylinder oils. 

Absorption of Gases in Liquids. Liquids will absorb some gases quite freely; 
thus, water will absorb very large quantities of ammonia, forming aqua ammonia. 
Addition of heat will drive off this gas so that another heat effect is the expul- 
sion of gases in solution. Use is made of this industrially in the absorption 
system of ammonia refrigeration. 

Solubility of Solids in Liquids. The heating of liquids will also affect their 
solubility for solid salts; thus, a saturated solution of brine will deposit crystals 
on heat abstraction and take them back into solution on heat addition. 
Certain scale-forming compounds are thrown down on heating the water in- 
tended for boilers, a fact that is made use of in feed-water heating purifiers; 
for these salts increase of temperature reduces solubility. In general then heat 
addition affects the solubility of liquids for solid salts. 

Chemical Reaction. Combustion. If oxygen and hydrogen, or oxygen and 
carbon, be heated in contact, they will in time attain an ignition temperature at 
which a chemical reaction will take place with heat liberation called combus- 
tion, and which is an exothermic or heat-freeing reaction. Another and 
different sort of reaction will take place if C0 2 and carbon be heated together, 
for these will together form a combustible gas, CO, under a continuation of heat 
reception. This is an endothermic or heat-absorbing reaction. Neither of 
these will take place until by heat addition the reaction temperature, called 
ignition temperature for combustion, has been reached. 

Electrical and Magnetic Effects. Two metals joined together at two separate 
points, one of which is kept cool and the other heated, will be found to carry 
an electric current or constitute a thermo-electric couple. Any conductor 
carrying an electric current will on changing temperature suffer a change of 
resistance so that with constant voltage more or less current will flow; this is 
a second electrical heat effect and like the former is useful only in instru- 
ments indicating temperature condition. A fixed magnet will suffer a change 
of magnetism on heating so that heat may cause magnetic as well as electric 
effects. 

These heat effects on substances as well as some others of not so great engi- 
neering importance may be classified or grouped for further study in a variety 
of ways, each serving some more or less useful purpose. 

Reversible and Non-reversible Processes. There may be reversible and 
non-reversible thermal processes, when the process may or may not be con- 
sidered constantly in a state of equilibrium. For example, as heat is applied 
to boiling water there is a continuous generation of vapor in proportion to the 



406 ENGINEERING THERMODYNAMICS 

heat received; if at any instant the heat application be stopped the evapo- 
ration will cease and if the flow of heat be reversed by abstraction, condensa- 
tion will take place, indicating a state of thermal equilibrium in which the 
effect of the process follows constantly the direction of heat flow and is con- 
stantly proportional to the amount of heat numerically, and in sign, of direction. 
As an example of non-reversible processes none is better than combustion, in 
which the chemical substances receive heat with proportional temperature rise 
until chemical reaction sets in, at which time the reception of heat has no fur- 
ther relation to the temperatures, because of the liberation of heat by com- 
bustion which proceeds of itself and which cannot be reversed by heat 
abstraction. Even though a vigorous heat abstraction at a rate greater than 
it is freed by combustion may stop combustion or put the fire out, no amount 
of heat abstraction or cooling will cause the combined substances to change 
back into the original ones as they existed before combustion. The effect of 
heat in such cases as this is, therefore, non-reversible. 

Constant and Variable Volume or Density. When gases, liquids or solids 
are heated they expand except when prevented forcibly from so doing, and as 
a consequence they suffer a reduction of density with the increase of volume; 
this is, of course, also true of changing liquids to their vapors. It should be 
noted that all such changes of volume against any resistance whatever, occur 
with corresponding performance of some work, so that some thermal processes 
may directly result in the doing of work. Heating accompanied by no volume 
change and during which restraints are applied to keep the volume invariable, 
cannot do any work or suffer any change of density, but always results in change 
of pressure in liquids, gases and vapors and in a corresponding change of internal 
stress in solids. 

Constant and Variable Temperature Processes. Another useful division, and 
that most valuable in the calculation of relations between heat effect and heat 
quantity, recognizes that some of the heating processes and, of course, cooling, 
occur at constant temperature and others with changing temperature. For 
example, the changes of state from liquid to solid, and solid to liquid, or freezing 
and fusion, are constant temperature processes in which, no matter how much 
heat is supplied or abstracted, the temperature of the substance changing state is 
not affected, and the same is true of ebullition and condensation, or the changing 
of state from liquid to vapor, and vapor to liquid. These latter constant- 
temperature processes must not be confused with evaporation, which may 
proceed from either the solid or liquid state at any temperature whether constant 
or not. 

Prob. 1. From the time a fire is lighted under a cold boiler to the time steam 
first comes off, what heat effects take place? 

Prob. 2. What heat effects take place when a piece of ice, the temperature of 
which is 20° F., is thrown onto a piece of red-hot iron? 

Prob. 3. What heat effects must occur before a drop of water may be evaporated 
from the ocean, and fed back into it as snow? 

Prob. 4. What heat changes take place when soot is formed from coal or oil? 



HEAT AND MATTER K)7 

Prob. 5. Tn a gas producer, coal is burned to CO., which Is then reduced fco CO. 
Steam is also fed bo fche producer, and II and o formed from it, Give all the heat 
effects which ooour. 

Prob. 6. By means of what beat effeots have you measured temperature ohanges, 
or have known fchem fco be measured? 

Prob. 7. When the temperature changes from io l> F. fco 20° F., give a list of all 
heat effects you know dial commonly occur for several common substances. l)o the 
same for a change in the reverse direction. 

Prob. 8. \[' a, closed cylinder he filled with water it will burst if the temperature 
he lowered or raised sufficiently. What thermal steps occur in each case? 

Prob. 9. If salt water he lowered sufficiently in temperature, a, cake of fresh ice 

and a rich salt solution will he formed. State I he steps or heal, effects which occur 

during the process. 

3. Thermometry Based on Temperature Change Heat Effects. Ther- 
mometer and Absolute Temperature Scales. 'Those thermal processes in which 
heal, addition or abstraction is followed as a result by a corresponding and more 
or less proportional temperature change, are quite numerous and important 
both iii engineering practice and as furnishing a means for thermometer-mak- 
ing, and temperature definition and measurement. According fco Sir William 

Thomson "every kind of t hormoscope must, he founded on some property 

of matter continuously varying wit h the temperature " and he gives fche fol- 
lowing: 

(a) Density of fluid under constant pressure. 

(/>) Pressure of a fluid under a constant volume envelope. 

(c) Volume of the liquid contained in a solid holder (ordinary mercury or 

Spirit thermometer). 

((/) Vapor pressure of a solid or liquid. 

(e) Shape or size of an elastic solid under constant stress. 

(/) Stress of an elastic solid restrained fco constant, size. 

(g) Density of an elastic solid under constant stress. 

(/>,) Viscosity of a fluid. 

(?') Electric current in a thermo-couple. 

(j) Electric resistance of a, conductor. 

(/»•) Magnetic moment- of a fixed magnet. 

Any, or all of these pressure, volume, shape, size, density, rate of flow, 

magnetic or electrical effects, may be measured, and fcheir measure constitutes 
a measure of temperature indirectly, so that instruments incorporating these 
temperature effects to he measured, are also thermometers. 

Any temperature-indicating device may be called a thermometer, though 
those in use for high temperatures are generally called pyrometers, which 
indicates the somewhat important fact that no thermometer is equally useful 

for all ranges of temperature. Practically all thermometers in use for tempera- 
tures short of a red heat:, depend on certain essential relations between the density 

or volume, the pressure and temperature of a fluid, though metals are used in 
some little-used forms in which change of size is measured, or change of shape 



408 ENGINEEBING THEKMOD YM AMICS 

of a double metallic bar, often brass and iron, consisting of a piece of each 
fastened to the other to form a continuous strip. The two metals are expanded 
by the temperature different amounts causing the strip to bend under heating. 
There are also in use electric forms for all temperatures, and these are the 
only reliable ones for high temperatures, both of the couple and resistance 
types except one dependent on the color of a high temperature body, black 
when cold. That most useful and common class involving the interde- 
pendence of pressure and temperature, or volume and temperature, of a fluid 
is generally found in the form of a glass bulb or its equivalent, to which 
is attached a long, narrow glass tube or stem which may be open or closed at 
the end; open when the changes of fluid volume at constant pressure are to 
be observed and closed when changes of contained fluid pressure at constant 
restrained volume are to be measured as the effect of temperature changes. 
For the fluid there is used most commonly a liquid alone such as mercury, or 
a gas alone such as air; though a gas may be introduced above mercury and 
there may be used a liquid with its vapor above. When the fluid is a liquid, 
such as mercury, in the common thermometer, the stem is closed at the end so 
that the mercury is enclosed in a constant-volume container or as nearly so as 
the expansion or deformation of the glass will permit, which is not filled with 
mercury, but in which a space in the stem is left at a vacuum or filled with a 
gas under pressure, such as nitrogen, to resist evaporation of the mercury at 
high temperatures. Gas-filled mercury thermometers, as the last form is called, 
are so designed that for the whole range of mercury expansion the pressure 
of the gas opposing it does not rise enough to offer material resistance to the 
expansion of mercury or to unduly stress the glass container. It should be 
noted that mercury thermometers do not measure the expansion of mercury 
alone, but the difference between the volume of mercury and the glass envelope, 
but this is of no consequence so long as this difference is in proportion to the 
expansion of the mercury itself, which it is substantially, with proper glass 
composition, when the range is not too great. Such thermometers indicate 
temperature changes by the rise and fall of mercury in the stem, and any numeri- 
cal value that may be convenient can be given to any position of the mercury 
or any change of position. Common acceptance of certain locations of the scale 
number, however, must be recognized as rendering other possible ones unneces- 
sary and so undesirable. Two such scales are recognized, one in use with metric 
units, the centigrade, and the other with measurements in English units, the 
Fahrenheit, both of which must be known and familiar, because of the frequent 
necessity of transformation of numerical values and heat data from one system 
to the other. To permit of the making of a scale, at least two points must be 
fixed with a definite number of divisions between them, each called one degree. 
The two fixed points are first, the position of the mercury when the thermometer 
is in the vapor of boiling pure water at sea level, or under the standard atmos- 
pheric pressure of 29.92" = 760 mm. of mercury absolute pressure, and 
second, the position of the mercury when the thermometer is surrounded by 
melting ice at the same pressure. These are equivalent to the boiling- or con- 



HEAT AND MATTER 



409 



densation, and melting- or freezing-points, of pure water at one atmosphere 
pressure. The two accepted thermometer scales have the following character- 
istics with respect to these fixed points and division between them : 

THERMOMETER SCALES 



Centigrade scale 
Fahrenheit scale . 



Pure Water 
Freezing-point, 
at one atni. pr. 




32 



Pure Water 

P>oiling-point. 

at one atm. pr. 



100 
212 



Number of Equal Divisions 
Between Freezing and Boiling. 



100 
180 



From this it appears that a degree of temperature change is on the centigrade 
scale, y-J-g- of the linear distance between the position of the mercury surface 
at the freezing- and boiling-points of water, and on the Fahrenheit scale, T ^ of 
the same distance. From this the relation between a degree temperature change 
for the two scales can be given. 



One degree temperature 1 _ 180 _ 9 f of one degree temperature 
change centigrade 100 5 j change Fahrenheit; 



or 



One degree temperature 1 _ 5 
change Fahrenheit J 9 



of one degree temperature 
change centigrade. 



It is also possible to set down the relation between scale readings, for when the 
temperature is 0° C, it is 32° F., and when it is 100° C. it is (180+32) =212° F., 
so that 

9 
Temperature Fahrenheit = 32+- (Temperature centigrade), 



or 



Temperature centigrade = ^ (Temperature Fahrenheit — 32) . 



For convenience of numerical work tables are commonly used to transform 
temperatures from one scale to the other and such a transformation is shown 
in a curve, Fig. 119, and in Table XXIX at end of the Chapter. 

By reason of the lack of absolute proportionality between temperature 
and effect, other fixed points are necessary, especially at high temperatures, 
and the following of Table XVI have been adopted by the U. S. Bureau of Stand- 
ards and are considered correct to within 5° C, at 1200° C. 



ENGINEERING THERMODYNAMICS 



/ 



1550 
1500 
1450 



-A 



-A- 



Si 

d 

^ 35 

c3 

* : 



g 



II 

320 



I 

01 

*j)1350 
P 



1300 



1250 
660 

1200 



z 



ff 



140 

CO 
140 



-40 

-58 
-70 
-91 



105 C 



80 120 

215 

240 
235 



t 



560 610 CGO 








7 


1010 




































7 
















s 




s 









460 

850 



60 100 

Degrees Centigrade 



K 



260 3 


10 300 




-u Z 




7 
















/ 




t. 




£ 




2 


7 








f 












. * 




<L _j 













Fig. 119.— Graphical Relation between Centigrade and Fahrenheit Thermometer Scales. 



HEAT AND MATTER 



411 



TABLE XVI 
FIXED TEMPERATURES 

U. S. BUREAU OF STANDARDS 



Temperature, 
°C. 


Temperature, 
o F 


Determined by the Point at which 


232 


449 


Liquid tin solidifies 


327 


621 


Liquid lead solidifies 


419.4 


787 


Liquid zinc solidifies 


444.7 


832.5 


Liquid sulphur boils 


630.5 


1167 


Liquid antimony solidifies 


658 


1216 


Liquid aluminum, 97.7% pure, solidifies 


1064 


1947 


Solid gold melts 


1084 


1983 


Liquid copper solidifies 


1435 


2615 


Solid nickel melts 


1546 


2815 


Solid palladium melts 


1753 


3187 


Solid platinum melts 



Thermometers in which a liquid and its vapor exist together, depend on a 
property to be noted in detail later, the relation of vapor pressure to tempera- 
ture and its independence of the volume of vapor. So long as any vapor exists 
above the liquid the temperature will depend only on the pressure of that vapor 
so that such thermometers will indicate temperature by the pressure measure- 
ment, after experimental determination of this pressure-temperature relation 
of vapors. Conversely, temperature measurements of vapors by mercury ther- 
mometers will lead to pressure values, and at the present time some steam 
plants are introducing mercury thermometers on the boilers and pipe lines, in 
place of the proverbially inaccurate pressure gages. 

Gas thermometer, is the name generally applied to the class in which the 
fluid is a gas, whether air, hydrogen, nitrogen or any other, and whether the 
pressure is measured for a fixed contained volume, or the volume measured 
when acted on by a constant pressure. These gas thermometers are so bulky 
as to be practically useless in ordinary engineering work and are only employed 
as standards for comparison and for tests of extraordinary delicacy in investi- 
gation work. They give much larger indications than mercury thermometers 
because the changes of gas volume under constant pressure are far greater 
than for mercury or any other liquid. Regnault was the first to thoroughly 
investigate air thermometers and reported that the second form, that of constant 
gas volume with measurement of pressure, was most useful. 

Using the centigrade scale, fixing freezing point at 0° C, and making the 
corresponding pressure po, atmospheric at this point, and reading at 100° C. 
another pressure pioo, he found experimentally a relation between these two 
pressures and the temperature corresponding to any other pressure p, as 
given by the empiric formula, 



<=10 o-P=Po_. 

pioo — po 



(601) 



412 ENGINEERING THERMODYNAMICS 

He also determined the pressure at the boiling-point to be related to the pressure 
at the freezing-point, by 

pioo = 1.3665 p , 

which on substitution gives 

* = 10 °-|^ =272 W-- i y ..... (602) 
.3665p \po J v J 

This constant, 272.85, the reciprocal of which is .003665, is, of course, the pressure 
increase factor per degree C. rise of temperature for a gas; held at constant 
volume, received extended investigation and it was found \ that it had about 
the same value applied to the other type of thermometer in which gas volumes 
are measured at constant pressure. This was true even when the pressure 
used was anything from 44 to 149 cm. of mercury, though it is reported 
that for 44 cm. pressure the value 272.98, and for 149 cm. pressure, the value 
272.7, seemed closer. For hydrogen it was found that the constant was sub- 
stantially the same as for air, while for carbonic acid it was 270.64, and while 
the hydrogen thermometers agreed with the air over the whole scale, showing 
proportional effects, this was hardly true of carbonic acid. Such uncertainty 
in the behavior of these thermometers and in the fixing of the constants was 
traced to the glass in some cases, but there still remained differences charge- 
able only to the gases themselves. Comparison of the air with mercury ther- 
mometers showed that there was not a proportional change with the temperature 
and that temperatures on the two, consistently departed. 

Examination of Eq. (602), giving the relation between two temperatures and 
the corresponding gas pressures, will show a most important relation. If in Eq. 
(602), the pressure be supposed to drop to zero and it is assumed that 
the relations between pressure and temperature hold, then when p = 0, 
t—— 272.85. This temperature has received the name of the absolute zero, 
and may be defined as the temperature at which pressure disappears or becomes 
zero at constant volume, and correspondingly, at which the volume also dis- 
appears, since it was found that similar relations existed between volume and 
temperature at constant pressure. Calling temperature on a new scale begin- 
ning 272.85° below the centigrade zero by the name absolute temperatures, 
then 






Absolute temperature 
centigrade 



, Scale temperature 
centigrade 



As this constant or absolute temperature of the centigrade scale zero, is an 
experimental value, it is quite natural to find other values presented by differ- 
ent investigators, some of them using totally different methods. One of these 
methods is based on the temperature change of a gas losing pressure without 
doing work, generally described as the porous plug experiment, and the results 



HEAT AND MATTER 413 

as the Joule-Thomson effect, and another is based on the coefficient of expansion 
of gases being heated. Some of these results agreed exactly with Regnault's 
value for hydrogen between 0° C. and 100° C. for which he gave —273° C.= 
— 491.4° F. Still other investigations continued down to the last few years 
yielded results that tend to change the value slightly to between —491.6° F., 
and —491.7° F., and as yet there is no absolute agreement as to the exact value. 
In engineering problems, however, it is seldom desirable or possible to work 
to such degrees of accuracy as to make the uncertainty of the absolute zero a 
matter of material importance, and for practical purposes the following values 
may be used with sufficient confidence for all but exceptional cases which are 
to be recognized only by experience. 

[Centigrade = 273] 
Absolute Temperature (T) j jp a h re nheit = 460 f + Scale Temperature (t). 

When great accuracy is important it is not possible at present to get a better 
Fahrenheit value than 459.65, the mean of the two known limits of 459.6 and 
459.7, though Marks and Davis in their Steam Tables have adopted 459.64, which 
is very close to the value of 459.63 adopted by Buckingham in his excellent 
Bulletin of U. S. Bureau of Standards and corresponding to 273.13 on the 
centigrade scale. 

These experiments with the gas thermometers, leading to a determination 
of temperature as a function of the pressure change of the gas held at constant 
volume, or its volume change when held at constant pressure, really supply a 
definition of temperature which before meant no more than an arbitrary number, 
and furnished a most valuable addition to the generalization of relations between 
heat content of a body and its temperature or physical state. 

A lack of proportionality between thermometer indication and temperature, 
has already been pointed out, and it is by reason of this that two identical ther- 
mometers, or as nearly so as can be made, with absolute agreement between 
water boiling- and freezing-points, will not agree at all points between, nor will 
the best constructed and calibrated mercury thermometers agree with a similarly 
good gas thermometer. 

The temperature scale now almost universally adopted as standard is that 
of the constant volume hydrogen gas thermometer, on which the degree F. 
is one one-hundred-and-eightieth part of the change in pressure of a fixed 
volume of hydrogen between melting pure ice, and steam above boiling pure 
water, the initial pressure of the gas at 32° being 100 cm. = 39.37 ins. Hg. A 
mercury in glass thermometer indication is, of course, a measure of the proper- 
ties of the mercury and glass used, and its F. degree of temperature is defined 
in parallel with the above as one one-hundred-and-eightieth part of the volume 
of the stem between its indications at the same two fixed points. A comparison 
of the hydrogen thermometer and two different glasses incorporated in mercury 
thermometers is given below, Table XVII, from the Bulletin of the U. S. Bureau 
of Standards, v. 2, No. 3, by H. C. Dickinson, quoting Mahlke, but it must be 



414 



ENGINEEEING THEEMODYNAMICS 



remembered that other glasses will give different results ana even different 
thermometers of the same glass when not similarly treated. 

Table XVII 

FAHRENHEIT TEMPERATURES BY HYDROGEN AND MERCURY 

THERMOMETERS 



Temperature by 

Hydrogen 
Thermometer. 



32 

212 
302 
392 

428 
464 
500 
536 

572 



Difference in 

Reading by 

Mercurv in Jena 

59" Glass. 






+1.3 



Difference in 

Reading by 

Mercury in 16' 

Jena Glass. 






- .18 
+ .072 
+ .39 
+ .83 
+1.79 
+2.4 
+3.53 



Temperature by 

Hydrogen 

Thermometer. 



617 
662 
707 
752 

797 
842 
887 
932 



Difference in 

Reading by 

Mercury in Jena 

59" Glass. 



+ 10.6 
+ 16.6 

+18.7 
+24.6 
+28.2 
+38.3 
+41.4 
+50.0 



Difference in 

Reading by 

Mercury in 16' 

Jena Glass. 



Useful comparisons of air, hydrogen, nitrogen, and other gases, with alcohol 
and mercury, in various kinds of glass, are given in the Landolt-Bornstein- 
Meyerhoffer, and in the Smithsonian Physical Tables, but are seldom needed 
for engineering work. 

One sort of correction that is often necessary in mercury thermometer 
work is that for stem immersion. Thermometers are calibrated as a rule with 
the whole stem immersed in the melting ice or the steam, but are ordinarily 
used with part of the stem exposed and not touching the substance whose tem- 
perature is indicated. For this condition the following correction is recom- 
mended by the same Bureau of Standards Bulletin : 

Stem correction = .000088 n (t-ti)° F 

When n = number of degrees exposed; 

t = temperature indicated Fahrenheit degrees; 

h = mean temperature of emergent stem itself, which must necessarily 
be estimated and most simply by another thermometer next to 
it, and entirely free from the bath. 



Prob. 1. What will be the centigrade scale and absolute temperatures, for the 
following Fahrenheit readings? -25°, 25°, 110°, 140°, 220°, 263° scale, and 300°, 
460°, 540°, 710°, 2000° absolute. 

Prob. 2. What will be the Fahrenheit scale and absolute temperatures, for the follow- 
ing centigrade readings? -20°, 10°, 45°, 80°, 400°, 610° scale, and 200°, 410°, 650°, 
810°, 2500° absolute. 

Prob. 3. By the addition of a certain amount of heat the temperature of a 
quantity of water was raised 160° F. How many degrees C. was it raised? 



HEAT AND MATTER 415 

Prob. 4. To bring water from 0° C. to its boiling-point under a certain pressure 
required a temperature rise of 150° C. What was the rise in Fahrenheit degrees? 

Prob. 5. For each degree rise Fahrenheit, an iron bar will increase .00000648 of 
its length. How much longer will a bar be at 150° C. than at 0° C? At 910° C. 
absolute than at 250° C. absolute? 

Prob. 6. The increase in pressure for S0 2 for a rise of 100° C. is given as .3845 at 
constant volume. What would have been absolute zero found by Regnault had he 
used SO2 rather than air? 

Prob. 7. A thermometer with a scale from 40° F. to 700° F. is placed in a thermome- 
ter well so that the 200° mark is just visible. The temperature as given by the 
thermometer is 450°. If the surrounding temperature is 100° F., what is true tempera- 
ture in the well? 

4. Calorimetry Based on Proportionality of Heat Effects to Heat Quantity. 
Units of Heat and Mechanical Equivalent. Though it is generally recognized 
from philosophic investigations extending over many years, that heat is one 
manifestation of energy capable of being transformed into other forms such 
as mechanical work, electricity or molecular arrangement, and derivable from 
them through transformations, measurements of quantities of heat can be made 
without such knowledge, and were made even when heat was regarded as a 
substance. It was early recognized that equivalence of heat effects proved 
effects proportional to quantity; thus, the melting of one pound of ice can cool 
a pound of hot water through a definite range of temperature, and can cool 
two pounds through half as many degrees, and so on. The condensation of 
a pound of steam can warm a definite weight of water a definite number of 
degrees, or perform a certain number of pound-degrees heating effect in water. 
So that taking the pound- degree of water as a basis the ratio of the heat liberated 
by steam condensation to that absorbed by ice melting can be found. Other 
substances such as iron or oil may suffer a certain number of pound degree 
changes and affect water by another number of pound-degrees. The unit 
of heat quantity might be taken as that which is liberated by the condensation 
of a pound of steam, that absorbed by the freezing of a pound of water, that to 
raise a pound of iron any number of degrees or any other quantity of heat 
effect. The heat unit generally accepted is, in metric measure, the calorie, 
or the amount to raise one kilogramme of pure water one degree centigrade, 
or in English units, the British thermal unit, that necessary to raise one pound 
of water one degree Fahrenheit. Thus, the calorie is the kilogramme degree 
centigrade, and the British thermal unit the pound degree Fahrenheit, and the 
latter is used in engineering, usually abbreviated to B.T.U. There is also 
occasionally used a sort of cross unit called the centigrade heat unit, which is 
the pound degree centigrade. 

The relation between these is given quantitatively by the conversion table 
at the end of this Chapter, Table XXX. 

All the heat measurements are, therefore, made in terms of equivalent 
water heating effects in pound degrees, but it must be understood that a water 
pound degree is not quite constant. Careful observation will show that the 



416 ENGINEERING THERMODYNAMICS 

melting of a pound of ice will not cool the same weight of water from 200° F. 
to 180° F., as it will from 60° F. to 40° F., which indicates that the heat capacity 
of water or the B.T.U. per pound-degree is not constant. It is, therefore, 
necessary to further limit the definition of the heat unit, by fixing on some 
water temperature and temperature change, as the standard, in addition to the 
selection of water as the substance, and the pound and degree as units of capacity. 
Here there has not been as good an agreement as is desirable, some using 
4° C. = 39.4° F. as the standard temperature and the range one-half degree 
both sides; this is the point of maximum water density. Others have used 15° 
C. = 59° F. as the temperature and the range one-half degree both sides; still 
others, one degree rise from freezing point 0° C. or 32° F. There are good 
reasons, however, for the most common present-day practice which will prob- 
ably become universal, for taking as the range and temperatures, freezing- 
point to boiling-point, and dividing by the number of degrees. The heat unit 
so defined is properly named the mean calorie or mean British thermal unit; 
therefore, 



Mean calorie = t™ (amount of heat to raise 1 Kg. water from 0° C. to 100° C). 

Mean B.T.U. =~ (amount of heat to raise 1 lb. water from 32° F. to 212° F.). 
lot) 



In terms of the heat unit thus defined, the amount of heat per degree tem- 
perature change is variable over the scale, but only in work of the most accurate 
character is this difference observed in engineering calculations, but in accurate 
work this difference must not be neglected and care must be exercised in using 
other physical constants in heat units reported by different observers, to be sure 
of the unit they used in reporting them. It is only by experience that judgment 
can be cultivated in the selection of values of constants in heat units reported 
for various standards, or in ignoring differences in standards entirely. The 
great bulk of engineering work involves uncertainties greater than these differ- 
ences and they may, therefore, be ignored generally. 

By various experimental methods, all scientifically carried out and exetnding 
over sixty years, a measured amount of work has been done and entirely con- 
verted into heat, originally by friction of solids and of liquids, for the deter- 
mination of the foot-pounds of work equivalent to one B. T. U.,when the 
conversion is complete, that is, when all the work energy has been converted into 
heat. This thermo-physical constant is the mechanical equivalent of heat. 
Later, indirect methods have been employed for its determination by calcula- 
tion from other constants to which it is related. All of these experiments 
have led to large number of values, so that it is not surprising to find doubt as 
to the correct value and different values are used even by recognized authori- 
ties. The experiments used include: 






HEAT AND MATTER 417 

(a) Compression and expansion of air; Joule. 

(6) Steam engine experiments, comparing heat in supplied and exhausted 

steam; Hirn. 

(c) Expansion and contraction of metals; Edlund and Haga. 

(d) Specific volume of vapor; Perot. 

(e) Boring of metals; Rumford and Hirn. 
(/) Friction of water; Joule and Rowland. 
({/) Friction of mercury; Joule. 

(h) Friction of metals; Hirn, Puluj, Sahulka. 
(i) Crushing of metals; Hirn. 
(j) Heating of magneto electric currents; Joule. 
(k) Heating of disk between magnetic poles; Violle. 

(7) Flow of liquids (water and mercury) under pressure; Hirn, Bartholi. 
(m) Heat developed by wire of known absolute resistance; Quintus Icilius, 

Weber, Lenz, Joule, Webster, Dieterici. 
(n) Diminishing the heat contained in a battery when the current produces 

work; Joule, Favre. 
(o) Heat developed in, and voltage of Daniell cells; Weber, Boscha, Favre 

and Silberman, Joule, 
(p) Combination of electrical heating and mechanical action by stirring 

water; Griffiths. 
(q) Physical constants of gases. 

The results of all of these were studied by Rowland in 1880, who himself 
experimented also, and he concluded that the mechanical equivalent of heat 
was nearly 

778.6 ft.-lbs. = l B.T.U., at latitude of Baltimore, 
or 

774.5 ft.-lbs. = l B.T.U., at latitude of Manchester. 

with the following corrections to be added for other latitudes. 

Latitude 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 

Ft.-lbs 1.62 1.50 1.15 .62 .15 -.75 -1.41 -1.93 -2.30 -2.43 

Since that time other determinations have been made by Reynolds and 
Morby, using mechanical, and Griffiths, Schuste and Gannon, Callendar and 
Barnes, using electrical transformation into heat. Giving these latter deter- 
minations equal weight with those of Joule and Rowland, the average is 

1 small calorie at 20° C. (nitrogen thermometer) =4.181 XI 7 ergs. 



418 ENGINEERING THERMODYNAMICS 

On the discussion of these results by Smith, Marks and Davis accept and use 
the mean of the results of Reynolds, and Morby and Barnes, which is 

1 mean calorie = 4.1834 X10 7 ergs, 

= 3.9683 B.T.U. 
1 mean B.T.U. = 777.52 ft.-lbs., 

when the gravitational constant is 980.665 cm. sec 2 , which corresponds to 32.174 
lbs., and is the value for latitude between 45° and 46°. 

For many years it has been most common to use in engineering calculations, 
the round number 778, and for most problems this round number is still the 
best available figure, but where special accuracy is needed it is likely that no 
closer value can be relied upon than anything between 777.5 and 777.6 for the 
above latitude. 

r Example. To heat a gallon of water from 60° F. to 200° F. requires the heat 
equivalent of how many foot-pounds? 

1 gallon =8.33 lbs.,' 

200° F. -60° F. =140° F. rise, 

8.33 X 140 = 1665 pound-degrees, 

= 1665 B.T.U. 

=778X1665, ft.-lbs. 
= 90,800 ft.-lbs. 

Prob. ,1. A feed-water heater is heating 5000 gallons of water per hour from 
40° F. to 200° F. What would be the equivalent energy in horse-power units? 

Prob. 2. A pound of each of the following fuels has the heating values as given. 
Change them to foot-pounds. 

Average bituminous coal, 14,000 B.T.U. per lb., Average kerosene, 18,000 B.T.U. per lb. 
" small anthracite, 12,000 B.T.U. per lb., " alcohol, 10,000 B.T.U. per lb. 
Average gasolene, 20,000 B.T.U. per lb. 

Prob. 3. A cubic foot of each of the following gases yields on combusture, the 
number of heat units shown. Change them to foot-pounds. 

Natural gas (average) , 880 B.T.U. per cu.ft., Carburetted water gas, 700 B.T.U. per cu.ft. 
Coal gas, 730 B.T.U. per cu.ft., Mond gas, 150 B.T.U. per cu.ft. 

Blast furnace gas, 100 B.T.U. per cu.ft. 

Prob. 4. A pool contains 20,000 cu.ft. of water and must be warmed from 40° F. 
to 70° F. How much work might be done with the equivalent energy? 



HEAT AND MATTER 419 

Prob. 6. How many calories and how many centigrade heat units would be 
required in Prob. 4? 

Prob. 6. In the course of a test a man weighing 200 lbs. goes up a ladder 25 ft. 
high, every 15 minutes. If the test lasted 12 hours how many B.T.U. did he expend? 

Prob. 7. A reservoir contains 300 billion gallons of water which are heated each 
year from 39° F. to 70° F. What is the number of foot-pounds of work equivalent? 

Prob. 8. A pound of water moving at the rate of 450 ft. per second is brought 
to rest, so that all of its energy is turned into heat. What will be the temperature rise? 

Prob. 9. For driving an automobile 30 horse-power is being used. How long 
will a gallon of average gasolene, sp.gr. =.7, last, if 10% of its energy is converted 
into work? 

Prob. 10. Power is being absorbed by a brake on the flywheel of an engine. 
If the engine is developing 50 horse-power how many B.T.U. per minute must be 
carried off to prevent burning of the brake? 

5. Temperature Change Relation to Amount of Heat, for Solids, Liquids, 
Gases, and Vapors, not Changing State. Specific Heats. Provided gases do not 
decompose, vapors condense, liquids freeze or evaporate, and solids melt, under 
addition or abstraction of heat, there will always be the same sort of relation 
between the quantity of heat gained or lost and the temperature change 
for all, differing only in degree. As the reception of heat in each case 
causes a temperature rise proportional to it and to the weight of the sub- 
stances, this constant of proportionality once determined will give numerical 
relations between any temperature change and the corresponding amount of 
heat. Making the weight of the substance unity, which is equivalent to the 
consideration of one pound of substance, the constant of proportionality may 
be defined as the quantity of heat per degree rise, and as thus defined is the 
specific heat of the substance. Accordingly, the quantity of heat for these cases 
is equal to the product of specific heat, temperature rise and weight of substance 
heated. 

The heat, as already explained, may be added in two characteristic ways: 
(a) at constant volume or density, or (6) at constant pressure. It might be 
expected that by reason of the increase of volume and performance of work 
under constant pressure heating, more heat must be added to raise the tempera- 
ture of one pound, one degree, than in the other case where no such work is done, 
and both experimental and thermodynamic investigations confirm this view. 
There are, therefore, two specific heats for all substances, capable of definition : 

(a) The specific heat at constant volume, and 

(6) The specific heat at constant pressure. 
These two specific heats are quite different both for gases and for vapors, which 
suffer considerable expansion under constant pressure heating, but for solids 
and liquids, which expand very little, the difference is very small and is to be 
neglected here. As a matter of fact, there are no cases of common engineering 
practice involving the specific heat of liquids and solids under constant volume, 
and values for the specific heats of liquids and solids are always without further 
definition to be understood as the constant pressure values. 



420 ENGINEERING THERMODYNAMICS 

Let C, be the specific heat of solids and liquids suffering no change of state. 

C p , be the specific heat of gases and vapors at constant pressure and 
suffering no change of state. 

C v , be the specific heat of gases and vapors at constant volume and suf- 
fering no change* of state. 

fa and tij be the maximum and minimum temperatures for the process. 

w, be the weight in pounds. 

Then will the heat added, be given by. the following equation, if the tempera- 
ture rise is exactly proportional to the. quantity of heat, or in other words, 

if the specific heat is constant. 

Q = Cw(t2 — ti), for solids and liquids . (603) 

Q = C v w(t2 — ti) } for gases and vapors (not near condensation) when 

volume is constant. (604) 

Q = C p w(t2 — h), for gases and vapors (not near condensation) when 

pressure is constant. (605) 

When, however, the specific heat is variable, as is the case for many sub- 
stances, probably for all, the above equation cannot be used except when 
the specific heat average value, or mean specific heat is used. If the variation 
is irregular this can be found only graphically, but for some substances the 
variation is regular and integration will give the mean value. It has been 
the custom to relate the specific heat to the temperature above the freez- 
ing-point of water, expressing it as the sum of the value at 32° F., and 
some fraction of the temperature above this point to the first and second 
powers, as in Eq. (606). 

Specific heat at temperature (t) = a+b(t-32)+c(t-32) 2 . (606) 

In this equation a is the specific heat at 32°, while b and c are constants, 
different for different substances, c being generally zero for liquids. 

When this is true, the heat added is related to the temperature above 
32° by a differential expression which can be integrated between limits 






— 32 

[a+b(t-32)+C(t-32) 2 ]dt 

32 



= a[fe-32)-(i 1 -32)]+|[0 2 -32) 2 -0 1 -32) 2 ]+|[fe-32) 3 -0 1 -32) 3 ]. (607) 

Usually the heats are calculated, above 32° so that the heats between any 
two temperatures will be the difference between the heats from 32° to those 
two temperatures. In this case ti = 32°, and, h = t, whence 

B.T.U. per lb., from 32° to t,= L+|(Z-32)+|(*-32) 2 l (t-23). . . (608) 






HEAT AND MATTER 421 

For this range of temperature 32° to t, the quantity of heat may be ex- 
pressed as the product of a mean specific heat and the temperature range 
or 

Heat from 32° to t= (mean sp. heat from 32° to t°) X(*-32). . (609) 
Comparing Eq. (608) with Eq. (609) it follows that 

\iZ^rf^t}- a+ ^-^+^-^ < oio > 

The coefficient of (t — 32) in the mean specific heat expression, is half that in 
the expression for specific heat at t, and the coefficient of (t — 32) 2 , is one-third. 
This makes it easy to change from specific heat at sl given temperature 
above 32°, to the mean specific heat from 32° to the temperature in question. 

The specific heats of some substances are directly measured, but for some 
others, notably the gases, this is too difficult or rather more difficult than cal- 
culation of values from other physical constants to which they are related. 
- It often happens that in engineering work the solution of a practical 
problem requires a specific heat for which no value is available, in which case 
the general law of specific heats, known as the law of Dulong and Petit, for 
definite compounds may be used as given in Eq. (611). 

(Specific heat of solids) X (atomic weight) =6.4. . . . (611) 

This is equivalent to saying that all atoms have the same capacity for 
heat, and while it is known to be not strictly true, it is a useful relation in 
the absence of direct determinations. Some values, experimentally determined 
for the specific heats of solids, are given in Table XXXI at the 'end of this Chap- 
ter, together with values calculated from the atomic weights to show the degree 
of agreement. The atomic weights used are those of the International Com- 
mittee on Atomic Weights (Jour. Am. Chem. Soc, 1910). When the specific 
heat of a solid varies with temperature and several determinations are avail- 
able, only the maximum and minimum are given with the corresponding tem- 
peratures, as these usually suffice for engineering work. 

To illustrate this variability of specific heat of solids, the values deter- 
mined for two samples of iron are given in Figs. 120 and 121, the former 
showing the variation of the mean specific heat as determined by Oberhoffer 
and Harker from 500° F. up, and the latter the amount of heat per pound of 
iron at any temperature above the heat content at 500° F., which is gen- 
erally called its total heat above the base temperature, here 500° F. 

It is extremely probable that the specific heats of liquids all vary irregularly 
with temperature so that the constant values given in Table XXXII at the end 
of the Chapter must be used with caution. This is certainly the case for water, 
and is the cause of the difficulty in fixing the unit of heat, which is best solved 
by the method of means. In Fig. 122 are shown in curve form the values for the 



422 



ENGINEERING THERMODYNAMICS 



specific heats of water at temperatures from 20° F. to 600° F., as accepted 
by Marks and Davis after a critical study of the experimental results of 



-18 



a 



.12 



.10 



(a) 



Oberhoff'er 
Harker 



>" 



/ 



^ 



/' 



Z 



X: 



(b) 



(a) 



1000 2000 

..Temperature in Degrees Fahr. 

Fig. 120.— Mean Specific Heat of Iron above 500° F., Illustrating Irregular Variations not 

Yielding to Algebraic Expression. 

Barnes and Dieterici and adjustment of the differences. The integral curve 
is plotted in Fig. 123 which, therefore, gives the heat of water from 32°F. to any 



o^"300 



o 

d 
2 200 



o 
£100 

3 

H 



{a) 
W 



From Oberhol'i'er Data 
Harker 



^ 



-y 



f 



**- 



Ub) 



/fa) 



500 



1000 



2500 



1500 2000 

Temperature in Degrees Fahr . 

Fig 121.— Total Heat of Iron above 500° F., Illustrating its Approximation to a Straight 
Line Relation in Spite of Wide Variation in Specific Heat Given in Fig. 120. 




temperature up to the highest used in steam practice and which is designated 
in steam tables, summarizing all the properties of water and steam, as the 



HEAT AND MATTEK 



423 















































































































1.15 










































































































| 






















































a 1.10 




















































































• 
























8 

mi.05 
































































































































































1.00 





















































100 



200 300 400 

Temperature Degrees Fahr. 



500 



Fig. 122. — Specific Heat of Water at Various Temperatures. 



600 
















































































* 






























































fa 




































<X> 






































































o 
■d 






































































O 




































»200 





















































































































































































































200 400 

Temperature in Degrees Fahr. 



600 



Fig. 123. — Total Heat of Water from 32° F., to any Temperature, the Heat of the Liquid at 

that Temperature above 32° F. 



424 



ENGINEERING THERMODYNAMICS 






heat of the liquid. For the purpose of comparison, the mean specific heat 
of water is given in Fig. 124 from 32° F. to any temperature which is obtained 
from the heat of the liquid above 32° F. by dividing it by the temperature 
above 32° F. 

In the table of specific heats of liquids there is a column giving the value 
calculated from the atomic weights to show at a glance the degree with 
which liquids satisfy the Dulong and Petit law. 

Variability of specific heat is especially noticeable in liquids that are solu- 
tions with different amounts of dissolved substance, in which case the specific 
heat varies with the density and temperature. Problems of refrigeration 
involve four cases of this kind: (a), calcium, and (b), sodium chloride, 





































































1.06 


















1 














































o 

s 

t 1.04 

.a 


































































W 
d 
m 1.0? 

s, 

1.00 


































































































































.98 



































































200 400 600 

Temperature in Degrees Fahrenheit 

Fig. 124. — Mean Specific Heat of Water from 32° to any Temperature. 

brines, the densities of which vary considerably but which are used with 
but little temperature range, seldom over 20° F. and often not over 5° F., 
(c), anhydrous ammonia and (d), carbonic acid. 

As the density of brines is often reported on the Baume scale and liquid 
fuels always so, a comparison of this with specific gravities is given in Table 
XXXIII in connection with the specific heat tables at the end of this Chapter 
to facilitate calculation. 

One of the best-known solutions so far as accuracy of direct experimental 
data is concerned, is calcium brine, results for which, from 35° C. to 20° C. 
given below, are from U. S. Bureau of Standards Bulletin by Dickinson, 
Mueller and George, for densities from 1.175 to 1.250. For chemically pure 



HEAT AND MATTER 



425 



calcium chloride in water, it was found that the following relation be- 
tween density D, and specific heat C, at 0° C, 



D = 2.8821 - 3.6272C+ 1 .7794C 2 , 



(612) 



and these results plotted in Fig. 125 show the specific heat variation with 
temperature to follow the straight line law very nearly. This being the case 
the mean specific heat for a given temperature range is closely enough the 
arithmetical mean of the specific heat at the two limiting temperatures. To 
the figure are also added dotted, the specific heats for some commercial brines, 
not pure calcium chloride, but carrying magnesium and sodium chloride of 
density 1.2. 

It might be conveniently noted here that the relation between freezing- 
point and density for pure calcium chloride by the same bulletin is given 
in Table XVIII below: 

Table XVIII 
FREEZING-POINT OF CALCIUM CHLORIDE 

U. S. Bureau of Standards 



Density of Solution. 


Per cent CaCl 2 by Wt. 


Freezing-point, 
°C. 


Freezing-point, 
o F 




1.12 


14.88 


- 9 


15.8 




1.14 


16.97 


-13 


8.6 




1.16 


19.07 


-16 


3.2 




1.18 


21.13 


-20 


- 4.0 




1.20 


23.03 


-24 


-11.2 




1.22 


24.89 


-29 


-20.2 




1.24 


26.77 


-34 


-29.2 




1.26 


28.55 


-40 


-40.0 



Other values for the specific heats of brines as commonly used are given 
in Table XIX, the accuracy of which is seriously in doubt and which 
may be checked by more authoritative values at different points where deter- 
minations have been made. 

Anhydrous ammonia liquid, has a variable specific heat with temperature, 
but the experimental values are too few to make its value and law quite certain. 
Several formulas have been proposed, however, that tend to give an impression 
of accuracy not warranted by the facts though quite convenient in preparing 
tables. 

Authority Specific heat of NH3 liquid at t° F. 

Zeuner 1.0135 +.00468 (*-32) (a) 

Dieterici 1.118 +.001156 (*-32) (6) 



Wood 
Ledoux 



1.1352+.00438 ($-32) (c) 
1.0057 +.00203 0-32) (d) 



(613) 



426 



ENGINEERING THERMODYNAMICS 

























.89 


































D=l 


.02 




























,87 
.80 
























































D=l. 


i^^ 
































































n= 


.175, 




























cg 




































D=1.2 


4^ 


3 

2 












,.. 


^^ 


^ 


^ 








.70 




^ 










D 


=1.22. 








0>" 
































D- 


=1.25^ 




















^""b= 


■1^ 




























.65 























-10 10 30 50 70 

Temperature in Degrees Fahr. 

Fig. 125. — Specific Heat of Calcium Chloride Brine of Various Densities D at Temperatures 

-10° F. to +70° F. 



HEAT AND MATTER 



427 



Table XIX 
SPECIFIC HEAT OF SODIUM CHLORIDE BRINE 



Density, B6 


Sp.gr. 


Per cent NaCl 
by Wt. 


Sp. Heat. 


Temp. F. 


Authority. 


1 


1.007 


1 

1.6 
4.9 
5.0 
10.0 
10.3 
10.3 
11.5 
12.3 
15.0 
18.8 
18.8 
20.0 
24.3 
24.5 
25 


.992 

.978 

.995 

.960 

.892 

.892 

.912 

.887 

.871 

.892 

.841 

.854 

.829 

.7916 

.791 

.783 


-0 

64.4 

66-115 

-0 

-0 

59-120 

59 - 194 

61-126 

64.4 

-0 
63-125 
68-192 

64-68 
64 


Common 
Thomsen 






Winkelmann 


5 
10 


1.037 
1.073 


Common 
Common 
Teudt 






Teudt 






Marignac 
Winkelmann 






15 


1.115 


Common 
Teudt 






Teudt 


19 


1.150 


Common 
Winkelmann 






Thomsen 


23 


1.191 


Common 



From these expressions the mean specific heat follows by halving the coefficient 
of (£ — 32) F., and these were determined and plotted to scale, together with some 
direct experimental values of Drewes, in Fig. 126. Giving greatest weight 
to Drewes and Dieterici, a mean curve shown by the solid line is located as 
the best probability of the value for liquid anhydrous and it has the Eq. (614). 

Mean specific heat of anhydrous } _-, 07 , nnoW/-^ (Md.\ 

liquid NH 3 from 32° F. to t° F. J -i- 07 *-***^ **). . . .(614) 

From this value the heat of liquid ammonia above 32° F. has been determined 
and is presented graphically in Fig. 127 from which, and the equation, the 
tabular values at the end of the Chapter were determined. 

Ammonia dissolved in water, giving an aqueous solution as used in the absorp- 
tion refrigerating system, has a nearly constant specific heat so closely approxi- 
mating unity as shown by Thomsen, who gives 

3 per cent NH3 in water solution ,*sp.ht., = .997, at 66° F. 

1.8 per cent NH 3 in water solution, sp.ht., =.999, at 66° F. 

.9 per cent NH3 in water solution, sp.ht., =.999, at 66° F., 

that it is customary in these calculations to ignore any departure from unity, 
the value for water. 

Liquid carbonic acid, another important substance in engineering, especially 
in mechanical refrigeration, is less known as to its specific heat than is ammonia, 
and that is much too uncertain. There is probably nothing better available 
at present for the necessary range than the results of Amagat and Mollier, 
reported by Zeuner for the heat of the liquid, which are reproduced in Fig. 
128, and used in the table at the end of this Chapter. 



428 



ENGINEERING THERMODYNAMICS 



It is, however, with gases that the most complex situation exists with respect 
to specific heats. As has already been pointed out, gases may be heated at 































































1.4 


















































^ 


/-" A 




I 1 - 2 


















/-" 




D 
— C 




C_ 


_ 








— 


D ^ 




^-^ 


— 




ps 


:^B 
F 






^ 








p. 

CG 

|i.o 
3 


F_ 








2 


x 5 ^ 


i."^- 


^1' 








~ E 








B_ 


.* 


^ 


V 




b 




AA - Zeuner 

BB-Ledoux 

CC-Wood 

DD-Dieterici 

EE Drewes 

FF-Mean Used in Book 








.8 


<< 


^ 















































50 



+ 50 
Temperature in Degrees Fahr. 



150 



Fig. 126. — Mean Specific Heat of Liquid Anhydrous Ammonia from —50° F. to 150° .F 



200 



W 

o 
•d 
« 

* 100 



£3 

H 

n 

S o 
I 



E 
-100 



















































































































































































































































































































y 


/' 






















































































































































X 


s* 
















































C^ 




































































































A 








AA- Zeuner 

B- Wood Note: All Curves 
C • Dieterici Practically Coin- 
D- Ledoux cident above 32 ° 
E- Mean 
























^0^ 


























A . 


_J 


^ 


^ 






























^ 






























^ 























































50 



50 100 

Temperature in Degrees Fahr. 



150 



200 



Fig. 127. — Heat of Liquid Anhydrous Ammonia above —50° F. 



constant volume, doing no external work while being heated, or at constant 
pressure, in which latter case work is done by expansion of the gas against the 
resisting constant pressure. Therefore, there must be two different specific 



HEAT AND MATTER 



429 



heats for each gas, one C v at constant pressure and the other C at constant 
volume, the difference between them representing the heat equivalent of the 




Temperature in Degrees Fahr. 
Fig. 128. — Heat of Liquid Carbonic Acid above 32° F. 

work of expansion done during the rise of temperature. Most experimental 
determinations of the specific heats of gases have been made at constant pressure 



430 ENGINEERING- THEEMODYNAMICS 

and the constant volume value found from established relations between it 
and other physical constants. These relations most commonly used are two, 
Eq. (615) connecting the difference with a constant R and the other Eq. (616) 

777.52(C P -C V )=R, (615) 

7?«T (616) 

connecting their ratio to a. constant y. These constants have each 
a special significance that may be noted here and proved later, thus R is the 
ratio of the PV product of a pound of gas to the absolute temperature, and y the 
particular value taken by the general exponent s in PV s = c, when the expansion 
represented takes place with no heat addition or abstraction, i.e., adiabatic, 
it is also a function of the velocity of sound in gases. Table XXXIV at the end 
of this Chapter gives some authentic values, with those adopted here designated 
by heavy type. 

v Variability of specific heats of gases and vapors is most marked and of some 
engineering importance, because so many problems of practice involve highly 
heated gases and vapors, the most common being superheated steam and the 
active gases of combustion in furnaces, gas producers and explosive gas engines. 
In fact, with regard to the latter it may be regarded as quite impossible with 
even a fair degree of accuracy to predict the temperature that will result in the 
gaseous products from the liberation of a given amount of heat of combustion. 
The first fairly creditable results on the variability of the specific heats of gases 
of combustion at high temperatures were announced by Mallard and LeChatelier, 
Vieille and Berthelot, all of whom agree that the specific heat rises, but who 
do not agree as to tlie amount. A general law was proposed by LeChatelier, 
giving the specific heat as a function of temperature by an equation of the 
following form: 

Specific heat at t° F„ (V=C), = C v = a+b(t-32), . . (617) 

in which a = specific heat at constant volume at 32° F. This yields, 

B.T.U.per lb, from 32° 1 n _ L.A, q^l (t vy\ (mk\ 

* F tot°F (V = C) f = ^( 32to ^ a +2^ M 

j Mean specific heat from 1 , __ b , . ( 

J32°F.,to*°F., (7 = C)J ~ c » = a +2 ( ) ( 9) 

The specific heat at constant pressure is obtained by adding a constant 
to the value for constant volume according to 

Cr-C+^f^, (620) 



HEAT AND MATTER 



431 



whence 



R 



Specific heat at t° F., (P = C), = C p = a+^ 7 — +6(*-32) 



(621) 



B.T. U. per lb, from 

32°F„to*°F„ (P = C), 



032to < =[a+^-+|(^-32)J(<- 



777.52 ' 2 
R , b 



Mean specific heat from _ r > _ nJ £*_ i_^.//_oo\ 

32°F., to*°F.,(P = C), " 777.52 + 2^ 6Z)t 



32) . . (622) 
. . . (623) 



The values of these constants have been determined by LeChatelier, Clerk, 
Callender, and Holborn and Austin, from which the following values are 
selected. 

Table XX 

SPECIFIC HEAT CONSTANTS, GASES, 



Gas. 


a 


_j_ R 
a+ 777l2 


b 


b 

2 


Authority. 


C0 2 


.1477 


.1944 


.000097 


.0000484 


LeChatelier 


C0 2 




.2010 


.0000824 


.0000412 


Holborn and Austin 


N 2 


.170 


.2404 


.0000484 


.0000242 


LeChatelier 


. N 2 




.2350 


.000021 


.0000105 


Holborn and Austin to 2606° F. 


N 2 




.2350 


.0000208 


.0000104 


Callender 1544° F. to 2440° F. 


o 2 


.1488 


.2125 


.0000424 


.0000212 


LeChatelier 


H 2 


.3211 




.000122 


.000061 


LeChatelier 


Air 




.2431 


.000135 


.0000675 


Callender (1544° F. to 2440° F.) 



For purposes of comparison the following curves are plotted, showing all 
these results of specific heat at constant volume, at temperature t° F., the total 
heat above 32° F. per pound of gas, and the mean specific heat from 32° F. to 
t° F. in Fig. 129. 

Probably there is now more known of the specific heat of superheated steam 
than of any common gaseous substance, and it is likely that other substances 
will be found in time to have somewhat similar characteristics. Pure computa- 
tion from the laws of perfect gases indicates that the specific heat of gases or 
superheated vapors must be either a constant, or a function of temperature 
only, and this is what prompted the form of the LeChatelier formula. Bold 
experimentation on steam, disregarding the law, or rather appreciating that 
superheated steam is far from a perfect gas, principally by Knobloch and 
Jacob and by Thomas, showed its specific heat to be a function of both pres- 
sure and temperature. Results were obtained that permitted the direct solu- 
tion of problems of heat of superheat, or the heat per pound of vapor at any 
temperature above that at which it was produced, or could exist in contact with 
the liquid from which it came. Critical study of various results by Marks and 
Davis led them to adopt the values of Knobloch and Jacob with slight modifi- 



432 



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HEAT AND MATTER 



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434 ENGINEERING THERMODYNAMICS 

cations ; for which evidence was in existence, raising the specific heats at low 
pressures and temperatures, and their conclusions are adopted in this work. 
In Fig. 130 is shown (A) the Marks and Davis modification of the C p curve 
of Knobloch and Jacobs, the integral of which (C) gives the heat of superheat 
from any temperature of steam generation to actual steam temperature, while 
(B) shows the values for the mean specific heat above the temperature of satura- 
tion for the particular pressure in question. 

When substances of the same class are mixed so that wi, W2, w$, etc., lbs. 
of the different substances having specific heats Ci, C2, C3, etc., or C p \, C P 2, 
C P z, etc., or C c i, C V 2, C v z, etc., then the specific heat of the mixture is given by 

c = CiWi+C 2 W2+CsWs+etc. . . 

Wl + W2 + W3 + etC. ' v 

n C P lWl + Ct,2W2 + (7t,3W3 + etC. 

C P = 1 : : — 7 , (o25) 

Wl + W2+W3 + etC. 

c = C P iWi-\-C P 2W 2 -{-C P 3W3+et c. . v 

v Wi+W2+W3+etc. 

Example. If 5 lbs. of olive oil at a temperature of 100° F., 10 lbs. of petroleum 
at a temperature of 150° F., and 50 lbs. of water at 50° F. are mixed together, what 
will be the resultant temperature and how much heat will be required to heat the mix- 
ture 100° above this temperature? 

Sp. ht. of olive oil - .4, 
Sp. ht. petroleum = .511, 
Sp. ht. water =1.000. 

Let z=the final temp. The heat given up by the substances falling in tempera- 
ture is equal to that gained by those rising, hence 

50(z-50) Xl =5(100 -x) X .4+10(150 -z) X.511, 

50z -2500 =200 -2x +766 -5.11s, 

57.11^=3466, or, z=60.7°F., 

a , , n • , ClWi+C2W 2 +C3Ws 

Sp.ht. of mixture = , from Eq. (611), 

W1+W2+W3 

5X.4+10X.511+50X1 _57.11 _ C7CA 
" 5+10+50 " 65 - 8785 * 

whence the heat required will be 65 X. 8786 = 57 B.T.U. 

Prob. 1. To change a pound of water at 32° F. to steam at 212° F. requires 
1150.4 B.T.U's. If the same amount of heat be given to a cubic foot of each of 
the following substances at 32° F., what will be final temperature in each case? (a) cop- 
per; (6) iron; (c) mercury; (d) clay; (e) stone. 



HEAT AND MATTER 435 

Prob. 2. How many pounds of the following substances could be warmed 10° F. 
by the heat required to raise 100 lbs. of water from 40° F. to 200° F.? 

(a) Ethyl alcohol from 100° F.; 

(b) Sea water from 60° F., (density = 1.045); 

(c) Glycerine from 60° F; 

(d) Tin from 480° F. 

Prob. 3. If 150 lbs. of water at 200° F. are added to a tank containing 200 lbs 
of petroleum at 70° F., what will be the resultant temperature, neglecting any heat 
absorbed or given up by the tank itself? 

Prob. 4. To melt 1 lb. of ice requires 144 B.T.U. How much would this lower 
the temperature of 1 lb. of the following substances (1) at constant pressure; (2) at 
constant volume; (a) air; (b) oxygen; (c) ammonia; (d) hydrogen; (e) nitrogen? 

Prob. 5. What would be the specific heats of the following mixture? Hydrogen 
3 lbs., oxygen 1 lb., nitrogen 7 lbs., carbon dioxide 20 lbs., carbon monoxide 10 lbs.? 

Prob. 6. Air is approximately 77 per cent N2, and 23 per cent O2 by weight. By 
means of the specific heats of the components, find its specific heats at constant pres- 
sure, and at constant volume. 

Prob. 7. By means of the specific heats, find the values of R and y most correct 
at atmospheric temperature (60° F.) for, hydrogen, air, carbon dioxide, carbon monoxide 
and nitrogen. 

Prob. 8. How much water could be heated from 40° F. to 60° F. by the heat 
needed to superheat 10 lbs. of steam at 200 lbs. per square inch absolute to 700° F.? 

Prob. 9. A building containing 250,000 cu. ft. of space is heated by a hot-water 
system. Considering the air to change eight times per hour, how many pounds of 
water per hour must be circulated if the drop in temperature of the water is from 
200° to 100° and the temperature of the outside air is 30° F. while that of the room 
is 60° F. neglecting wall conducted heat? 

Prob. 10. How much heat would be required to warm a pound of liquid C0 2 from 
zero to 80° F.? Compare with water and ammonia. 

6. Volume or Density Variation with Temperature of Solids, Liquids, Gases 
and Vapors, Not Changing State. Coefficients of Expansion. Coefficients 
of Pressure Change for Gases and Vapors. Solids increase in length or in any 
linear dimension, a certain fraction of their original length for each degree 
temperature rise and the expansion is usually assumed to be in proportion 
to temperature rise. The relation between original and final length can be 
set down in an equation involving the coefficient of expansion. 

Let a = coefficient of linear expansion = fractional increase in length per 
degree. 

Zi and t\ = original length or any other linear dimension and the cor- 
responding temperature; 
h and ti = length which l\ becomes after heating and the corresponding 
temperature. 
Then 

Increase in length = l2~li = ahfo — h), .».••. (627) 
New length l 2 = li+ali(t2 — h), 

= Zi[l+a(fc-ii)]. .."..■. (628) 



436 ENGINEERING THERMODYNAMICS 

Solids, of course, expand cubically and the new volume will be to the old, 
as the cubes of the linear dimension. 

Let a = coefficient of volumetric expansion; 
^i= original volume; 
V2 = final volume after heating. 
Then when the temperature rises one degree, 

^ = ^V = (l+a) 3 = l+3a+3a 2 +a 3 = l+ a .... (629) 

If a is small, and it is generally less than yoooo> ^ nen the square and cube can 
be neglected in comparison with the first power, whence 

l+a = l+3a and <x = 3a. 

so that the coefficient of volumetric expansion may be taken as sensibly 
equal to three times the coefficient of linear expansion, and similarly, the 
coefficient of surface expansion as twice the coefficient of linear expansion. 

Liquids, by reason of the fact that they must always be held in solid 
containers, may be said to have no linear expansion, and therefore, although 
the expansion may be one direction only, the amount is due to the total 
change of volume rather than the change of length along the direction of 
freedom to expand. The same is true of gases, so that for gases and liquids 
only coefficients of volumetric expansion are of value and these are given in the 
Tables XXXV, XXXVI, XXXVII, and XXXVIII at the end of this Chapter. 
With liquids and gases it is usual to take the volume at 0° C. or 32° F. and 
29.92 ins. Hg pressure as a standard, and the coefficient gives the increase as 
a fraction of this, per degree departure from the freezing-point. This is the 
universal practice with gases. 

It appears that the coefficients of expansion for solids are quite different 
from one another, ranging from over 15X10 -4 for wax, to.085XlO~ 4 for 
Jena normal glass, a range of over two hundred and sixty times. Determina- 
tions of the value at various temperatures for any one substance indicate a 
variation with temperature, which proves that proportionality of increase of 
dimensions to temperature rise, does not hold true, a fact which has led to 
formulas of the form 

the value of which is dependent on the determination of the constant and veri- 
fication of correctness of form, which has not by any means been conclusively 
done. For most engineering work the constant values nearest the temperature 
range will' suffice except for certain liquids, vapors, and gases. A more marked 
tendency to follow such a law of variation with temperature is found with 
liquids and coefficients for some are given in the standard physical tables. 



HEAT AND MATTEE 437 

The two important liquids, mercury and water, have been separately 
studied in greater detail and the latter exhibits a most important exception 
to the rule. For mercury, according to Broch 

1 ; 2= = t ; 1 (l + .000455^+54Xl0- 12 ^ 2 +602Xl0- 1 ^ 3 ), . . . (630) 

which exhibits a refinement of value only in instrument work such as barometers 
and thermometers. Water, as already mentioned, has its maximum density 
at 39.1° F. and expands with both fall and rise of temperature. Its expansion 
is given by a similar formula by Scheel, as follows: 

^ 2 = yi (l-.03655Xl0" 3 ^+2.625Xl0~ 6 ^ 2 -1.16U 3 ). . . (631) 

Most commonly the expansion of water is not considered in this'way, but by 
comparing densities at varying temperatures, and all sets of physical tables 
contain values which in this work are significant only as affecting the change 
of volume in turning water to steam and such values as are needed are 
incorporated in the steam tables later. 

The study of the expansion of gases and vapors at constant pressure, and 
rise of pressure at constant volume, per degree has perhaps been fairly com- 
plete and is of greatest significance, because from it most of the important laws 
of thermodynamics have been derived. This work may be said to have 
started with the Regnault air and gas thermometer work, already described. 
Some of the authentic values collected in the Landolt, Bornstein, Myerhoffer, 
and Smithsonian Physical Tables, are given at the end of this Chapter, 
where <x v is the coefficient of pressure change at constant volume, and <x p the 
coefficient of expansion, or volume change at constant pressure. 

The remarkable thing about the coefficients for these gases and vapors is the 
approach to constancy for most of the gases, not only of the coefficients of expansion 
for P = c nor the similar constancy of the coefficients of pressure rise for V = c, but 
more remarkable than either of these is the similarity of the two constant coeffi- 
cients. These facts permit of the generalizing of effect when P = c, 
and when V = c, and of the announcement of a law by means of which 
all such problems can be solved instead of applying separate coefficients for 
every substance and every different temperature necessarj^ for solids and 
liquids where, for example, the maximum coefficient was over 260 times as 
great as the least. The average coefficient for all gases, applying both to 
pressures and volumes, is the same as enters into the gas thermometer work 
and its best value is found to be 



« = TT^r^ = -002034, per degree F. 
a = 27VH = - 00 3661, P er degree C. 



(632) 



438 ENGINEERING THERMODYNAMICS 

and approximately 

a = 4^2 = • 00203 > per de S ree F - 
a = 973 =.00366, per degree C. 



(633) 



These are the same as the reciprocals of the absolute temperature of the 
ice-melting point, and are but expressions of conditions for reduction of the 
volume and pressure at the ice-melting temperature to zero by constant 
pressure and constant volume abstraction of heat respectively, and by 
stating the amount of reduction per degree give by implication the number 
of degrees for complete reduction, j 

Example. The rails on a stretch of railroad are laid so that they just touch when 
the temperature is 120° F. How much total space will there be between the rails 
per mile of track at 0° F.? 

For wrought iron a will be nearly the same for Bessemer steel = .00000648. 

Hence the linear reduction in 5280 ft. for a change of 120° F. will be 

5280 X120X. 00000648 =4.1 ft. 

Prob. 1. A steam pipe is 700 ft. long when cold (60° F.), and is anchored at one 
end. How much will the other end move, if steam at a temperature of 560° F. is 
turned into the pipe? 

Prob. 2. A copper sphere is one foot in diameter at 50° F. What must be the 
diameter of a ring through which it will pass at a temperature of 1000° F.? 

Prob. 3. A hollow glass sphere is completely filled with mercury at 0° F. What 
per cent of the mercury will be forced out if the temperature rises to 300° F.? 

Prob. 4. A room 100 ft.XoO ft.XlO ft, is at a temperature of 40° F. The tem- 
perature rises to 70° F. How many cubic feet of air have been forced from the room? 

Prob. 5. The air in a pneumatic tire is at a pressure of 90 lbs. per square inch 
gage and at a temperature of 50° F. Due to friction of the tire on the ground in 
running, the temperat re rises to 110° F. What will be the pressure? 

Prob. 6. A brick lighthouse is approximate y 200 ft. high. Should it be exactly 
this at 0° F., what would it be at 100° F.? 

Prob. 7. Show that if a glass tube is rigidly held at each end by brackets attached 
to an iron tank it will break if the tank is warmed. 

Prob. 8. From Eq. (618) find the density of water at 60° F., 100° F., 212° 
F., and compare with the values in the steam tables. 

Prob. 9. A drum containing C0 2 gas at a pressure of 250 lbs. per square inch 
gage is raised 100° F. above its original temperature. What will be the new pressure? 

7. Pressure, Volume and Temperature Relations for Gases. Perfect and 
Real Gases. Formulating the relations between the pressure change at constant 
volume and the volume change at constant pressure, 

Let P and V be the simultaneous pressure and volume of gas; 
" t be its scale temperature at the same time, F.; 
" T be its absolute temperature at the same time, F. 



HEAT AND MATTER 



439 



Then at constant volume the pressure reached at condition (a) after heating 
from 32° F. is given by 

Pa-P32° = ^ P32X( ^~ 32) - 
Similarly for another temperature t b , the pressure will be 



Whence 



A-M 1+ *S?] 






H 



ta-32 

492 492-32 + * a t a +460 



t b -S2 492-32+fe ^+460' 



492 



or 



Similarly 



P T 

~^ = 7fr, for V constant, 



V T a 

Tr—ffTi f° r P constant, 

V b lb 



(634) 



(635) 



Both Eqs. (634) and (635) are true, for no gas all the time, but very nearly 
true for all, under any range of change, and a hypothetical gas is created for which 
it is exactly true all the time, known as a perfect gas, about which calculations 
can be made as would be impossible for real gases and yet the results of which 
are so close to what would be the result with real gases, as to be good enough 
for engineering practice. Therefore, with a mental reservation as a guard 
against too great confidence in the work, all real gases will be assumed perfect 
and to follow Eqs. (634) and (635) except when experience shows the results 
are too far wrong to be useful. 

These laws, known by the names both of Charles and Gay-Lussac, are closely 
associated with another also doubly named as Boyle's or Mariotte's and like- 
wise an idealization of experimental observations known to be nearly true for 
all gases. This is to the effect that so long as temperatures are kept constant 
the pressures of gases vary inversely as their volume, or that, 



Pa 

P» 






=— , and, P a V a = PbVb = constant, for T constant . (636) 



Study of the PV product, for various gases has revealed a good deal on the 
general properties of matter, especially as to the transition from one state to 
another. This is most clearly shown by curves which may be plotted in two 



440 



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HEAT AND MATTER 441 

ways. To coordinates of pressure and volume a family of equilateral hyper- 
bolas one for each temperature, would represent the true PV = C or isothermal 
relation and any variation in the constancy of the product would be shown by 
its departure from the hyperbola. Still more clearly, however, will the depart- 
ure appear when the product PV is plotted against pressures, constancy of 
product would require all lines to be straight and inconstancy appear by 
departures from straight lines. To illustrate, the data from Young for car- 
bon dioxide are plotted both ways in Fig. 131, from 32° F. to 496° F., the values 
of PV at 32° and 1 atm. are taken as unity on one scale. It appears that up to 
the temperature of 88° F, known as the critical temperature, each isothermal 
plotted to P and 1'V coordinates consists of three distinct parts: 

(a) a curved line sloping to the right and upwards; 

(b) a straight line nearly or exactly horizontal; 

(c) a nearly straight line sloping upward rapidly and to the left. 

In this region then the isothermaJs are discontinuous, and this is caused by 
the liquification or condensation of the gas, during which increase of pressure, 
produces no change of volume, provided the temperature is low enough. It 
also appears that each PV line has a minimum point and these minima joined 
result in a parabola. At the end of this Chapter are given in Table XXXIX the 
values of PV at three different temperatures and various pressures for oxygen, 
hydrogen, carbon dioxide and ammonia, in terms of the values at 32° and 1 
atm. for further comparison and use. Further study along these lines is not 
profitable here and the topic while extremely interesting must be dropped with 
the observation, that except near the point of condensation or liquefaction, 
gases or vapors, which are the same thing except as to nearness to the critical 
state, follow the Boyle law closely enough for engineering purposes. 

None of these approximate laws, Eqs. (634), (635) and (636) can be con- 
sidered as general, because each assumes one of the variables to be constant, but a 
general law inclusive of both of these follows from further investigation of a 
fixed mass of gas suffering all sorts of pressure volume and temperature changes, 
such as occur in the cylinders of compressors and gas engines. A table of 
simultaneous experimental values of pressure, volume, and temperature, for any 
gas will reveal the still more general relation inclusive of the preceding three as 
follows : 

PaVa_P^,_PV_ r .„,*. 

T T ~ T \®^') 

in which C g is approximately constant for any one gas and assumed constant 
for perfect gases in all calculations. For twice the weight of gas at the same 
pressures and temperatures C g would be twice as large, so that taking a constant 
R for one pound, and generally known as the " gas constant," and introducing 
a weight factor w, the general characteristic equation for the perfect gas is, 

PV^wRT (638) 



442 



ENGINEERING THERMODYNAMICS 



This general law may be derived from the three primary laws by imagining 
in Fig. 132, two points, A and B, in any position and representing any two states 
of the gas. Such points can always be joined by three lines, one constant 

































A 




X 


































Y 














B 

















Diagram to derive Law yf=Cg 



Fig. 132. — Curve of Continuous Relation between P, V, and T for Gases. 

pressure A to X, one constant temperature X to Y, and the other constant 
volume Y to B. For these the following relations hold, passing from A to B 



V a =V.. 



T a 



But 

and 
whence 

or 

Passing to B, 



V X =Vy 



T X = T» 



Py 



V a =V y ^ 



PyTa^ 
■La J- y 



PaVa P.F, 



P.-P&. 



But 



HEAT AND MATTER 443 



P b ^V b 



or in general 



PaVa^P V V V = ~°T b ' U = P b V b 



PV 

-=f = constant. = wR 



when the weight of gas is w lbs. 

For numerical work, the values of R must be fixed experimentally by direct 
measurement of simultaneous pressure, volume, and temperature, of a known 
weight of gas or computed from other constants through established relations. 
One such relation already mentioned but not proved is 

R = 777.52(C P -C V ) (639) 

It is extremely unlikely that the values of R found in both ways by a multi- 
tude of observers under all sorts of conditions should agree, and they do not, 
but it is necessary for computation work that a reasonable consistency be attained 
and that judgment in use be cultivated in applying inconsistent data. In the 
latter connection the general rule is to use that value which was determined by 
measurement of quantities most closely related to the one being dealt with. 
Thus, if R is to be used to find the state of a gas as to pressure, volume, and tem- 
perature, that value of R determined from the first method should be selected, but 
the second when specific heats or Joule's equivalent are involved. Of course, a 
consistency could be incorporated for a perfect gas, but engineers deal with real 
gases and must be on guard against false results obtained by too many hypoth- 
eses or generalizations contrary to the facts. Accordingly, two values of R are 
given in Table XL, at the end of this chapter, one obtained from measure- 
ments of specific heats at constant pressure and determinations of the ratio 
of specific heats unfortunately not always at the same temperature and gen- 
erally by different people, and the other by direct measure of gas volume at 
standard 32° F. temperature and 1 atm. pressure. These measurements are 
separately reported in Sections (5) and (8), respectively. 

If a gas in condition A, Fig. 133, expand in any way to condition B, then 
it has been shown that 

P a V a s ^P b V b s , 

in which s has any value and which becomes numerically fixed only when the 
process and substance are more definitely defined. Comparing the temperatures 
at any two points A and B, it follows that 

m PaVa -, m P b V b 

T a = — 5-, and T b = — — , 
wR wR ' 



444 
whence 

But 
and 

whence 



ENGINEERING THERMODYNAMICS 



n P b V b 



T a PaV a ' 
PaVaVt-^PoVtV*- 1 , 



aV a \Vj ' 



P b V 

p, 






(640) 



















A 
































































B 



















(641) 



Fig. 133. — Expansion or Compression of Gas between A and B, Causing a Change of Tem- 
perature. 

and 

Yj, = /Tayh 

V a \tJ > 

Eqs. (640) and (641) give the relation between temperatures and volume,? 
But 

Va _ TgP j) _ TbPg 
V b PaTb T a Pt> 

which, substituted in above, gives 



or 



n = my- l /AV- 1 
T a \tJ \pJ ' 



HEAT AND MATTER 445 

and 



s-l 



or 

Pi 

P, 



H8)"' (642) 

Hrf 1 (643) 



Eqs. (629) and (630), give the relation between pressures and temperatures. 
It is convenient to set down the volume and pressure relations again to 
complete the set of three pairs of most important gas equations. 



Va 



=(W\ (644) 

HS - (645) 



These are perfectly general for any expansion or compression of any gas, but 
are of value in calculations only when s is fixed either by the gas itself or by 
the thermal process as will be seen later. 

Example. A pound of air has a volume of 7.064 cu.ft. at a pressure of two atmos-. 
pheres and a temperature of 100° F. Find the value of R for air from the data; 
also the final volume and temperature if expansion occurs so that s--*1.4 until the 
pressure becomes \ an atmosphere. 

PV=wRT, or 2116x2x7.064 = lX#X560, or #=53.38, 

s-l .4 

Ti /PA * M 

s-feD =(4) -"•■ 

.\ T 2 = T x + 1 .49 = — - = 352 abs. = - 108° F. 
1.49 



v l \P,) 



1 

S =2.7, or V 2 =2.7 Vi = 19.1 cu.ft. 



Prob. 1. A perfect gas is heated in such a way that the pressure is held constant. 
If the original volume was 10 cu.ft. and the temperature rose from 100° F. to 400° 
F., what was the new volume? 

Prob. 2. The above gas was under a pressure of 100 lbs. per square inch gage at 
the beginning of the heating. If the volume had been held constant what would have 
been the pressure rise? 

Prob. 3. A quantity of air, 5 lbs. in weight, was found to have a volume of 50 cu.ft. 
and a temperature of 60° F. What was the pressure? 

Prob. 4. A cylinder holding 12 cu.ft. has a pressure of 250 lbs. per square 
inch gage, and the temperature is 50° F. What would be the weight of its contents 
were it filled with (a) C0 2 ; (b) NH 3 ; (c) Oxygen; (d) Hydrogen? 



L>p. 


C v . 


3.409 


2.412 


.217 


.1535 


.2175 


.1551 


.2438 


.1727 



446 ENGINEEEING THERMODYNAMICS 

Prob. 5. At a pressure of 14.696 lbs. per square inch and a temperature of 
melting ice, one pound of air has a volume of 12.387 cu.ft. From the data find the 
value of R for air. The specific heats of air are given by one authority as C p = .2375 
and C» = .1685. Find R from the data and see how the two values obtained compare. 

Prob. 6. At the same pressure and temperature as in Prob. 5, a pound of the 
following substances has a volume as shown. From the data and the values of 
specific heats, find R by the two methods. 

Substance. Cu.ft. per lb. 

Hydrogen 178.93 

Carbon dioxide .... 8.15 

Oxygen 11.21 

Nitrogen 12.77 

Prob. 7. 5 cu.ft. of gas at a pressure of 3 atmospheres absolute and a tempera- 
ture of 50° F. expand to atmospheric pressure. What will be the final volume and 
temperature, if s = 1.35? 

Prob. 8. 1000 cu.ft. of gas at atmospheric pressure and 60° F. are compressed 
into a tank of 100 cu.ft. capacity. What will be the pressure in the tank and the 
temperature of the gas at the end of the process, if the gas is C0 2 and the com- 
pression adiabatic? 

Prob. 9. What will be the final volume, pressure and temperature, if a pound of 
air at atmospheric pressure (14.7 lbs. per square inch) and a temperature of 60° F. 
be compressed adiabatically until its absolute temperature is six times its original 
value? 

8. Gas Density and Specific Volume and its Relation to Molecular Weight 
and Gas Constant. The density of a gas is best stated for engineering 
purposes as the weight of a cubic foot, but as this becomes less on rise of 
temperature or decrease of pressure it is necessary to fix a standard condition 
for reporting this important physical constant. It is best to take one atmosphere 
760 mm. or 29.92 ins. of mercury as the pressure, and 0° C.=32°F. as the 
standard temperature, though it is in some places customary in dealing with 
commercial gases, such for example as those used for illumination, to take the 
temperature at 60° F. and illuminating gas at this condition is often known 
among gas men as standard gas. In this work, however, the freezing-point 
and standard atmosphere will be understood where not specifically mentioned, 
as the conditions for reporting gas density and its reciprocal, the specific volume 
of gases or the cubic feet per pound. The chart, Fig. 134, shows the relation 
of volume and density at any pressure and temperature to the volume and 
density under standard conditions. 

These constants have been pretty accurately determined by many investi- 
gators, whose figures, to be sure, do not agree absolutely, as is always the 
case in experimental work, but the disagreement is found only in the last 
significant figures. Some selected values of reliable origin are reported at 
the end of this Chapter in Table XLI for the important gases and these 
may be used in computation work. 



HEAT AND MATTER 



447 



It often happens in dealing with gases and especially superheated vapors 
that a value is needed for which no determination is available, so that general 











L7 


16 


15 


Pressure in J 
14 


Pounds Per Sq. 
13 12 


In. i 


U>s. 
11 




10 










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/ 


1 


i 


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f 


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1 


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— 



I 

1.2 



10 



I 

I 

1.1 



20 



l.p5 

I 



30 40 ]50 GO 70 

Temperature i Degrees Fahr. 

I I I lJM*> | I 



80 



90 



1' ' 



I I 

.8 



I I 



I I I I I 



'J 



Upper Scale = Ratio 

Inner Scale = Ratio 



Density at 32 & 29.92" 



Density at any T & P 
Volume at 32° F 



Lower Scale 

Outer Scale = Ratio 



Volume at 32 & 29.92 " 
Ratio yolume at anv T & P 
Density at 32° F. 



l.;8 
I I 



Volume at any T Density at any T 

Fig. 134. — Equivalent Gas Densities At Different Pressures and Temperatures. 

laws of density or specific volumes of substances are necessary to permit the 
needed constant to be estimated. These relations may be applied to vapors 



448 ENGINEERING THERMODYNAMICS 

as well as to gases even though the standard conditions are those for the 
liquid state, on the assumption that all gases and vapors will expand under 
temperature, or contract under pressure rise, to the same degree, retaining 
the same relative relations between all substances as exist at the standard 
atmosphere and freezing-point. A vapor thus reported below its point of 
condensation and assumed to have reached that condition frOm one of higher 
temperature at which it exists as vapor is often called steam gas, or alcohol 
gas, for example in the case of water and alcohol. 

Such general relations between the densities of gases as are so desirable 
and useful in practical work have been found by studying the manner in which 
gases chemically combine with respect to the volume relations before and after 
the reaction. Following several experimenters, who reported observed rela- 
tions, Gay-Lussac stated a general law, as follows : 

When two or more gaseous substances combine to form a compound, the vol- 
umes of the combining gases bear a simple ratio to each other and also to 
that of the compound when it is also a gas. 

He also attempted to derive some relation between this law and Dalton's atomic 
combining law, which states that, in combining chemically, a simple numerical 
relation exists between the number of atoms of different elements which unite 
to form a compound. This was not successful, but Avagadro later found the 
expected relation by assuming that it is a particle, or a number of atoms, or 
a molecule, that is important in combining, and the law stated is as follows: 

Equal volumes of different gases measured at the same pressure and tempera- 
ture contain the same number of molecules. 

It is possible by analysis of these two laws to get a relation between the volumes 
of gases and the weights of their molecules because the molecular relation of 
Avagadro, combines with the combining law of Gay-Lussac to define the rela- 
tion between the number of combining molecules. At the same time the weight 
relations in chemical reactions, based on atomic weights, may be put into a 
similar molecular form, since the weight of any one substance entering is the 
product of the number of its molecules present and the weight of the molecule. 
Applying the relation between the number of molecules derived previously, 
there is fixed a significance for the weight of the molecule which for simple gases 
like hydrogen and oxygen is twice the atomic weight and for compound gases, 
like methane and carbon dioxide, is equal to the atomic weight. Applying 
this to the Avagadro law, the weights of equal volumes of different gases must 
be proportional to their molecular weights, as equal volumes of all contain the 
same number of molecules. 

Putting this in symbolic form and comparing any gas with hydrogen, as to 
its density, because it is the lightest gas of all and has well determined charac- 



. HEAT AND MATTER 449 

teristics, requires the following symbols, denoting hydrogen values by the 
subscript h. 

Let m = molecular weight of a gas, 

" 8 = density in lbs. per cu.ft. — 



then 
and 



§1 _m\ 
82 rri2 

81 m\ 

8// m H ' 



(646) 
(647) 



But as the molecular weight of hydrogen is for engineering purposes equal to 2 
closely enough and hydrogen weighs .00562 lb. per cu.ft. = 8#, at 32° F and 
29.92 ins. Hg, 

Lbs. per cu.ft. = 81 = .00281 mi (648) 

To permit of evaluation of Eq. (648) it is necessary that there be available 
a table of molecular weights of gases and the atomic weights of elements from 
which they are derived, and the values given at the end of this Chapter in Table 
XLII are derived from the international table. As atomic weights are 
purely relative they may be worked out on the basis of any one as unity, and 
originally chemists used hydrogen as unity, but for good reasons that are of no 
importance here, the custom has changed to yq the value for oxygen as unity. 
These atomic weights are not whole numbers but nearly so, therefore, for con- 
venience and sufficient accuracy the nearest whole number will be used in 
this work and hydrogen be taken as unity except where experience shows it 
to be undesirable. 

The reciprocal expression to Eq. (648) can be set down, giving the specific 
volume of a gas or its cubic feet per pound at 32° F. and 29.92 ins. Hg., as 
follows: 

Cu.ft.perlb:=|=^ 1 2 Ii m 1 =^ (649) 

This is a most important and useful conclusion as applied to gases and vapors 
for which no better values are available, and in words it may be stated as follows : 

The cubic feet per pound of any gas or vapor at 32° and 29.92 ins. Hg, is 

equal to 355.87 divided by its molecular weight, 
or 

The molecular weight of any gas or vapor in pounds, will occupy a volume of 

355.87 cu.ft. at 32° and 29.92 ins. Hg. 

The approach t© truth of these general laws is measured by the values 
given for specific volume and density at the end of this Chapter (a) experiment- 
ally derived and, (6) as derived from the hydrogen value by the law. 



450 ENGINEERING THERMODYNAMICS 

Another and very useful relation of a similar nature is derivable from what 

has been established, connecting the gas constant R with molecular weights. 

w 
The general law PV=wRT when put in the density form by making 8 = ^- 

becomes 

f = RT (650) 

Whence, comparing gases with each other and wUh hydrogen at the same 
pressure and temperature 

Pi 

pTsTiS' (651) 



T-=p-, and §1=—-— . ....... (652) 

which indicate that the densities of gases are inversely proportional to the 
gas constants, or the density of any gas is equal to the density of hydrogen times 
the gas constant for hydrogen divided by its own. 

Inserting the values of density at 32° and 29.92 ins. Hg and of the gas con- 
stant for hydrogen, it follows that for any gas 

4 2QQ8 
Lbs. percu.ft. = 8i= — ~, . ...... (653) 

■» K 

the reciprocal of which gives the specific volume at 32° F. and 29.92 ins. Hg, or 

Cu.ft. P<* lb. = 71=^-^ (654) 

Example. 1. Explanation and use of Chart, Fig. 134. This diagram is for the 
purpose of finding the cubic feet per pound, or pounds per cubic foot, of a gas at 32° 
F. and a pressure of 29.92 ins. of Hg, if its volume or weight per cubic foot be known 
at any pressure and temperature. The curves depend upon the fact that the pounds 
per cubic foot (B) vary directly as the pressure and inversely as the temperature. 
That is 

T 29.92 

on no = nmr> . 

492 P 



, ^,y.92 = OTP~ A 



The line of least slope is so drawn that for any temperature on the horizontal scale 

its value when divided by 492 may be read on the vertical scale. The group of lines 

with the greater slope is so drawn that for any value on the vertical scale this quantity 

29 92 
times — - — may be used on the horizontal scale. That is, the vertical scale gives the 






HEAT AND MATTER , 451 

ratio of densities as affected by temperature for constant pressure, while horizontal 
scale gives the ratio as affected by both temperature and pressure. A reciprocal 
scale is given in each case for volume calculations. 

To find the pounds per cubic foot of gas at 32° F. and 29.92 ins. of mercury when 
its value is known for 90° and 13 lbs. per sq.in. On the temperature scale, pass 
vertically until the temperature line is reached, then horizontally until the curve 
for 13 lbs. absolute is reached. The value on the scale below is found to be 1.265, 
so that the density under the standard conditions is 1.265 of the value under known 
conditions. Had it been required to find the cubic feet per pound the process would be 
precisely the same, the value being taken from the lower scale, which for the example 
reads .79, or, the cubic feet per pound under standard conditions is 79 per cent of 
the value under conditions assumed. 

Example 2. By means of the molecular weight find the density of nitrogen at 

32° F. and 29.92 ins. Hg, and the cubic feet per pound for these conditions. 

From Eq. (646) 

Si mi . 28 X. 00562 

— = — or S x = . 

o H m H 2 

Hence 8 for nitrogen = .07868 pounds per cu.ft. and, 

h^6-8 = 12 ' 709 - CU - ft - perlb - 

Prob. 1. Taking the density of air from the table, find the value of R for air, by 
means of Eq. (653) and compare its value with that found in Section 7. 

Prob. 2. Compare the density of carbon monoxide when referred to 32° F. and 
60° F. as the standard temperature, as found both ways. 

Prob. 3. By means of their molecular weights find the density of oxygen, nitrogen 
and carbon dioxide at 32° F. and 29.92 ins. Hg. 

Prob. 4. What are the cubic feet per pound of acetylene, methane and ammonia 
at 32° F. and 29.92 ins. Hg? 

Prob. 5. An authority gives the following values for R. Compare the densities 
found by this means with the densities for the same substance found by the use of 
the molecular weights. 

Oxygen 48.1 

Hydrogen 764.0 

Carbon monoxide 55.0 

Prob. 6. What will be the volume and density under standard conditions, of a 
gas which contains 12 cu.ft. per pound at a temperature of 70° F. and a pressure of 
16 lbs. per square inch absolute? 

Prob. 7. What will be the difference in volume and density of a gas when con- 
sidered at 60° and 29.92 ins. of Hg, and at 32° F. and 29.92 ins. of Hg? 

9. Pressure and Temperature Relations for Vapor of Liquids or Solids. 
Vaporization, Sublimation and Fusion Curves. Boiling- and Freezing-points 
for Pure Liquids and Dilute Solutions. Saturated and Superheated Vapors. 

Substances may exist in one of three states, solid, liquid or gas, the latter being 
generally called vapor when, at ordinary temperatures the common state is that 



452 ENGINEERING THERMODYNAMICS 

of liquid or solid, or when the substance examined is near the point of lique- 
faction or condensation, and just which state shall prevail at any time depends 
on thermal conditions. Within the same space the substance may exist in two" of 
these three states or even all three at the same time under certain special condi- 
tions. These conditions may be such as to gradually or rapidly make that part 
in one state, turn in to another state, or may be such as to maintain the relative 
amounts of the substance in each state constant; conditions of the latter sort are 
known as conditions of equilibrium. These are experimental conclusions, but 
as in other cases they have been concentrated into general laws of which they 
are but special cases. The study of the conditions of equilibrium, whether of 
physical state or chemical constitution, is the principal function of physical 
chemistry, in the pursuit of which the Gibbs phase rule is a controlling prin- 
ciple. According to this rule each possible state is called a phase, and the 
number of variables that determine which phase shall prevail or how many 
phases may exist at the same time in equilibrium for one chemical substance 
like water, is given by the following relation, which is but one of the conclusions 
of this general principle of equilibrium. 

Number of undefined variables = 3— number of phases. 

Now it is experimentally known that if water be introduced into a vacuum 
chamber some of it will evaporate to vapor and that, therefore, water and its 
vapor may coexist or the number of phases is two, but this does not state how 
or when equilibrium will be attained. The rule above, however, indicates that 
for this case there can be but one undefined or independent variable and, of 
course, since the pressure rises more when the temperature is high than when 
low, the two variables are pressure and temperature, of which accordingly only 
one is free or independent, so that fixing one fixes the other. In other words 
when a vapor and its liquid are together the former will condense or the latter 
evaporate until either pressure or temperature is fixed, and fixing the one the 
other cannot change, so that the conditions of equilibrium are indicated by 
a curve to coordinates P and T, on one side of which is the vapor state and 
on the other that of liquid. Such a curve is the vapor pressure-temperature curve 
of the substance, sometimes called its vapor tension curve, and much experi- 
mental information exists on this physical property of substances, all obtained 
by direct measurement of simultaneous pressures and temperatures of a vapor 
above its liquid, carefully controlled so that the pressure or the temperature is 
at any time uniform throughout. 

The conditions of equilibrium between vapor and liquid, defined by the vapor 
tension curve extend for each substance over a considerable range of pressure 
and temperature, but not indefinitely, nor is the range the same for each. At 
the high-pressure and temperature end a peculiar interruption takes place due 
to the expansive effect of the temperature on the liquid and the compressive 
effect of the pressure on the vapor, the former making liquid less dense and the 
latter making vapor more dense, the two densities become equal at some 
pressure and temperature. The point at which this occurs is the " critical point " 
at which the equilibrium between liquid and vapor that previously existed, 



HEAT AND MATTEE 



453 



ends and there is no longer any difference between vapor and liquid. This 
point is a most important one in any discussion of the properties of matter, 
and while difficult to exactly locate, has received much experimental attention, 
and some of the best values are given below in Table XXI for the pressure, 
density, and temperature defining it, for the substances important in engineering 



Table XXI 
THE CRITICAL POINT 








Symbol. 


Critical Temp. 


Critical Pres- 
sures. 


Critical 

Density 

Water 

at 

4°C=1. 


Authority. 


Criti- 
cal vol. 
Cu.ft. 
per Lb. 




Substance. 


0° C. 


0° F. 


Atm. 


Lbs. 

per 

Sq.in. 


Authority. 


Hydrogen 


H 2 
2 

N 2 

NH 3 

NH; 

co 2 
C02 

H20 

H 2 

H2O 

H2O 
H2O 
H 2 


-243.5 
-II8.1 

-146.i 

+130.0 
+ 131.0 

+ 31.35 
+ 30.921 

+358.1 
+364.3 
+365.0 

+374. 

+374.6 

+374.5 


-390.1 
-180.4 

-232.8 

266. 
267.8 

88.43 
87.67 

676.4 
687.7 
689. 

705.2 

706.3 

706.1 


20 
50i 

35.i 

115. 
113. 

72.9 

77.i 

194.61 
200.5 


294 
735 

515 

1690 
1660 

1070 

1130 

2859 
2944 

3200 
3200 


.652 
.442 

.464 
.452 

.429 


Olszewski 

1 Wroblewski 

2 Dewar 

1 Olszewski 

2 Wroblewski 
Dewar 

Vincent and 
Chappuis 
Amagat 

1 Andrews 

2 Cailletet and 

Mathias 
Nadejdini 
Batteli 
Cailletet and 

Colardeau 
Traube and 

Teichner 
Holborn and 

Baumann 
Marks 


26.8 
13. 




Nitrogen 

Ammonia 

Ammonia 

Carbon dioxide. . . 
Carbon dioxide. . . 






Nadejdini 
Batteli d 




Water 


Water 




Water 









To illustrate this discussion there is presented the vapor tension curves of 
water, ammonia and carbon dioxide to a large scale in chart form derived 
from the tabular data both at the end of this Chapter, while a small scale dia- 
gram for water is given in Fig. 135. These data are partly direct experimental 
determinations and partly corrections obtained by passing a smooth 
curve representing an empiric equation of relation between pressure and tem- 
perature, through the major part of the more reliable experimental points. 
These pressure-temperature points are very accurately located for y/ater, the 
first good determinations having been made by Regnault in 1862 and the last 
by Holborn and Baumann of the German Bureau of Standards in the last year. 
The data presented are those of Regnault corrected by various investigations 
by means of curve plotting, and empiric equations by Wiebe, Thiessen and 
Schule, and those of various later observers, including Battelli, Holborn, Hen- 
ning, Baumann, Ramsay and Young, Cailletet and Colardeau, somes eparately, 
but all together as unified by Marks and Davis in their most excellent steam 



454 



ENGINEERING THERMODYNAMICS 






tables, and later by Marks alone for the highest temperatures 400° F. to the 
critical point, which he accepts as being located at 706.1° F. and 3200 lbs. square 



2400 



,2000 



P. 

f lfrK) 

o 



1200 



80C 



400 



































































































1 
















































1 






















































































































































/ 


















































/ 


















































/ 


































































































/ 


















































/ 


















































/ 


















































/ 


















































/ 
















































/ 


















































/ 


















































/ 


















































/ 


















































I 
















































/ 


















































/ 


















































/ 


















































/ 
















































/ 


















































/ 


















































/ 
















































I 










































AY 


ate 


r 








V 


apo 


r 








































/ 


1 
















































/ 


















































/ 

















































































































































































































































































































































































150 



600 



300 450 

Temperature in Degrees Fahr. 

Fig. 135. — Vapor of Water, Pressure-temperature Curve over Liquid (Water). 

inch. In calculations the values of Marks and Davis, and Marks, will be 
accepted and used. 



HEAT AND MATTER 



455 



Carbon dioxide and ammonia are by no means as well known as steam, 
and the original data plotted, while representing the best values obtainable, must 
be accepted with some uncertainty. A smooth curve Figs. (139) and (140) 
has been drawn for each through the points at locations that seem most fair, 
for both these substances and the values obtained from it are to be used in 
calculations; these curves have been located by the same method as used by 
Marks in his recent paper and described herein later. The equalized values 
are given in the separate table at the end of the Chapter with others for latent 






.1 








































/ 


»— 1 


































































CO 

5-05 














Ice 




























.5 






















CQ 

cd 






















t» 

CQ 

CD 

Ph 
















/ v * 


tpor 




-i 


>0 




-35 




-1 







+15 




+40 



Temperature in Degrees Fahr. 
Fig. 136.— Vapor of Water, Pressure-temperature Curve over Solid (Ice). 

heats and volumes, but while consistent each with the other are probably but 
little more correct than values reported by others which are inconsistent. 

The curves and the equivalent tabular data are most useful in practical 
work, as they indicate the temperature at which the vapor exists for a given 
pressure, either as formed during evaporation or as disappearing during con- 
densation, or the other way round, they indicate the pressure which must be 
maintained to evaporate or condense at a given temperature. 

Just as the vapor-liquid curves indicate the conditions of equilibrium between 



456 



ENGINEERING THERMODYNAMICS 



vapor and its liquid, dividing the two states and fixing the transition pressure 
or temperature from one to the other, so also does a similar situation exist with 
respect to the vapor-solid relations. In this case the curve is that of " sub- 
limation " and indicates the pressure that will be developed above the solid 
by direct vaporization at a given temperature in a closed chamber. In Fig. 
136 is plotted a curve of sublimation of vapor-ice, based on Juhlin's data, 
Table XXII, which indicates that the line divides the states of ice from that 
of vapor so that at a constant pressure, decrease of temperature will cause 
vapor to pass directly to ice and at constant temperature a lowering of pres- 
sure will cause ice to pass directly to. vapor. 

Table XXII 
JUHLIN'S DATA ON VAPOR PRESSURE OF ICE 



Temperature. 


Pressure. 


C. 


F. 


Min. Hg. 


Lbs. sq.in. 


-50 


-58. 


.050 


.001 


-40 


-40. 


.121 


.0023 


-30 


-22. 


.312 


.006 


-20 


- 4. 


.806 


.0156 


-15 


5. 


1.279 


.0247 


-10 


14. 


1.999 


.0386 


- 8 


17.6 


2.379 


.0459 


- 6 


21.2 


2.821 


.0544 


- 4 


24.8 


3.334 


.0643 


- 2 


28.4 


3.925 


.0758 


- 


32. 


4.602 


.0888 



Likewise the liquid, water, may pass to the solid, ice, by lowering temperature 
at a fixed pressure as indicated in Fig. 137, plotted from data by Tamman, 
Table XXIII, and which becomes then the curve of "fusion" 

Table XXIII 

TAMMAN'S DATA ON FUSION PRESSURE AND 
TEMPERATURE OF WATER-ICE 



Temperature. 


Pressure. 


C. 


F. 


Kg. sq.cm. 


Lbs. sq.in. 





32. 


1 


1423 


- 2.5 


27.5 


336 


4779. 


- 5. 


23. 


615 


8747.4 


- 7.5 


18.5 


890 


13658.8 


-10.0 


14. 


1155 


16428. 


-12,5 


9.5 


1410 


20055. 


-15. 


5. 


1625 


23113. 


-17.5 


.5 


1835 


26100. 


-20. 


- 4. 


2042 


27044. 


-22.1 


- 7.8 


2200 


31291. 



HEAT AND MATTER 



457 













































































































































\ 






















\ 
















































\w 


ater 




























































Ice 




































































































\ 






















\ 




8 


( 


) 


£ 


1 


1( 




2< 


1 


3 


I 



Fig, 



Temperature in Degrees Fahrenheit 
137. — Water — Ice, Pressure- temperature Curve. 



458 



ENGINEERING THEEMODYNAMICS 



These three curves plotted to the same scale meet at a point located at a pressure 
of 4.6 mm. Hg. = .18 ins. Hg., and temperature +.0076° C. =32.01° F., ordi- 
narily taken at 32° F., which point is named the triple point, as indicated in 
Fig. 138. The fact that the vapor pressure for water extends below freezing- 
point and parallels more or less that of ice indicates the condition of supercooled 





























.2 






• 








/ 
























/ 






















W 


iter / 














02 

6< 






























Ice 




















w 

u 

o> 
Pn 

CO 

o 

Ph 

CO 












Vapor 


















/ 


f 


^-Tri] 


)le Poir 


t 












u 

CO 
CO 

CD 
u 
Ph 






// 




















/' 





































































50 

Temperature Degrees Fahr. 

P'ig. 138. — Water Vapor — Water — Ice, Combined Curves of Pressure-temperature Rela- 
tion. The Tri-ple Point. 



water, one of unstable equilibrium instantly dispelled by the introduction of 
a little ice at the proper stable state for this temperature. 

Ordinary engineering work is not concerned with the entire range indicated 
in Fig. 138 for any substance, but with the higher temperature ranges for some 
and the low for others, with transition from solid to liquid state for metals 
and similar solids and the transition from liquid to vapor for a great many, of 
which water" comes first in importance, then the refrigerating fluids, ammonia 



HEAT AND MATTER 459 

and carbon dioxide, and last certain fuels like alcohol and the petroleum oils 
with their distillates and derivatives. 

Melting-points, or the fusion temperature of such solids as are important, 
are usually given for only one pressure, the standard atmosphere, as in ordinary 
practice these substances are melted only at atmosphere pressure, and some 
such values are given at the end of the Chapter in Table XLIII. 

This is not the case, however, for boiling-points, which must be defined 
a little more closely before discussion. The vapor pressure curves indicate 
that as the temperature of a liquid rises, the pressure rises also if the substance 
is enclosed, but if the pressure were relieved by opening the chamber to a region 
of lower pressure and kept constant, then the temperature would no longer 
rise and boiling or ebullition would take place. The boiling-point then is the 
highest temperature to which the liquid and its vapor could rise under the 
existing pressure. When not otherwise defined the term boiling-point must 
be taken to mean the temperature of ebullition . for atmospheric pressure of 
29.92 ins. Hg, and values for several substances are given at the end of this 
Chapter in Table XLIV. 

Vapor having the temperature required by the pressure of the pressure- 
temperature curve is known as saturated vapor, and this may be defined as 
vapor having the lowest temperature at which it could exist as vapor, under 
the given pressure. Vapors may, however, be superheated, that is, have 
higher temperatures than saturated vapors at the same pressure, but cannot 
so exist for long in the presence of liquid. Superheating of vapors, therefore, 
implies isolation from the liquid, and the amount of superheat is the number of 
degrees excess of temperature possessed by the vapor over the saturation 
temperature for the pressure. In steam power plant work, especially with 
turbines, it is now customary to use steam with from 75° F. to 150° F. of 
superheat, and it might be noted that all so-called gases like oxygen and 
nitrogen are but superheated vapors with a great amount of superheat. 

It has already been mentioned that the saturated vapor pressure-temperature 
curve of direct experiment is seldom accurate as found, but must be corrected 
by empiric equations or smooth average curves, and many investigators have 
sought algebraic expressions for them. These equations are quite useful also 
in another way, since they permit of more exact evaluation of the rate of change 
of pressure with temperature, which in the form of a differential coefficient 
is found to be a factor in other physical constants. One of these formulas 
for steam as adopted by Marks and Davis in the calculation of their tables 
is given in Eq. (655), the form of which, was suggested by Thiessen: 

0+459.6) log^ = 5.409y-212°)-3.7.lXl0- 10 [(689-0 4 -477 4 ], . (655) 

in which t = temperature F.; and p = pressure lbs. sq.in. 

This represents the truth to within a small fraction of one per cent up to 400° 
F., but having been found inaccurate above that point Professor Marks has 



460 ENGINEERING THERMODYNAMICS 

very recently developed a new one, based on Holborn and Baumann's 
high temperature measurements, which fits the entire range, its agreement 
with the new data being one-tenth of 1 per cent, and with the old below 400° 
F., about one-fifth of 1 per cent, maximum mean error. It appears to be the 
best ever found and in developing it the methods of the physical chemists have 
been followed, according to which a pressure is expressed as a fraction of the 
critical pressure and a temperature a fraction of the critical temperature. 
This gives a relation between reduced pressures and temperatures and makes use 
of the principle of corresponding states according to which bodies having the same 
reduced pressure and temperature, or existing at the same fraction of their 
critical are said to be in equivalent states. The new Marks formula is given 
in Eqs. (656) and (657), the former containing symbols for the critical 

pressure p c 

temperature T c abs. 
in pressure pounds per square inch, and temperature absolute F. 



and the latter giving to them their numerical values, 



log £*= 3.006854 (y-l) 1 + . 0505476-^+. 629547 (^--.7875 VI, . (656) 

log p = 10.515354- 4873.71 7 7 - 1 -.004050962 7 +.0000013929647 72 . . (657) 

As the method used in arriving at this formula is so rational and scientific, 
it has been adopted for a new determination, from old data to be sure, of the 
relations between p and T for ammonia and carbon dioxide, so important as 
substances in refrigeration, especially the former. According to this method 
if p c and T c are the critical pressures and temperatures, both absolute, and 
p and T those corresponding to any other point, then according to Van der 
Waals, 

logJ=/(J-l) (658) 

Accordingly, the logarithm of the critical divided by any other pressure, is 
to be plotted against the quantity [(critical temperatur edivided by the tem- 
perature corresponding to the pressure) — 1], and the form of curve permits of 
the determination of the function, after which the values of the critical 
point are inserted. This has been done for NH3 and CO2 with the result for 
NH 3 

' log J = .045+2.75(J- c -l)+.325(J- c -iy, .... (659) 



which on inserting the critical constants, 
p c =114 atm. = 1675.8 lbs. per square inch 
7^ = 727.4° F. absolute 



which are the Vincent and Chap- 
puis values, 



HEAT AND MATTER 



461 



becomes, 



log p = 5.60422-1527.547 7 " 1 -17196ir 



For CO2 it was found that 

log^ = .038+2.65(p-l)+1.8(p-l)' 

which on inserting the critical constants, 



(660) 



. . (661) 



p c = 77 atm. = 1131.9 lbs. per square inch 
T c = 547.27° F. abs. 



which are Andrews' values, 






becomes, 

logp = 7.46581-4405.7657 7 - 1 + 1617501.366T- 2 -257086165.87067 7 - 3 . (662) 

Curves showing the relation of reduced and actual temperatures and pressures 
are given in Fig. 139 for ammonia and in Fig. 140 for carbon doixide. 

For the past half century far more time and effort have been devoted to 
making other formulas of relation of p to T for saturated vapor not only for 
steam, but also for other vapors, than would have sufficed for accurate exper- 
imental determination, and as these help not at all they are omitted here. Equa- 
tions of physical relations can be no better than the data on which they are based, 
and for the substances ammonia and carbonic acid the charts or formulas must 
be used with a good deal of suspicion. 

In all engineering calculations requiring one of these constants even for steam 
no one is justified in using a formula; the nearest tabular or chart value must 
be employed and it will be as accurate as the work requires. Time is at 
least as important as accuracy, if not more so, for if too much time is required 
to make a calculation in commercial work, it will not be made because of the 
cost, indirect and approximate methods being substituted. 

It is sometimes useful in checking the boiling-point of some substance 
little known, to employ a relation between boiling-points of different substances 
at the same pressure applied to a substance well-known. 

Let T a and T b be absolute temperatures of boiling for substances A and B under 

same pressure; 
Ta and T b ' be absolute temperature of boiling for substances A and B under 

some other pressure. 
Then, 

^ = ^+c(7V-n). ......... (663) 

Such equations as this are useful in finding the saturation curve of other sub- 
stances from that for water, which is now so well established, when enough 
points are known for the other substances to establish the constant c. Also 

Ta 

the ratio — ~ plotted against the temperature difference TV — 7& should give 



462 



ENGINEERING THERMODYNAMICS 



8 1.6 



1.4 



1.2 



1.0 



^> 

o 

o .6 

a 

'§.4 

O 
Hi 




From Values of Wood o 

" " " Dieterici • 

" " Regnault + 

" » Ledoux ' 



.1 



.5 



.2 .3 .* .o 

(Critical Temperature Divided by any other Temperature)—! 



,6 




60 110 160 210 

Temperatures in Degrees Fahr. 

Fig. 139. — Ammonia Pressure-temperature Relations, for Saturated Vapor. 



HEAT AND MATTER 



463 



1.6 



cS 
M 
o 















































































































































































































































































































































































































/• 














































































































































































iyr 










































OS 










































Oj/ 








From Cailletet's Data • 
" Regnault's ^ 

" Stewart's Interpolation of Zeuner's Data + 
" Zeuner's Tabulation of Mollier's Formula 
based on Amagat's Data o 

































^* 
















k 


K 

























































































.1 .2 .3 .4 .5 

(.Critical Temperature Divided by any other Temperature)-! 




—100 



50 50 

Temperatures in Degrees Fahr. 



Fig. 140. — Carbon Dioxide Pressure-temperature Relations for Saturated Vapor. 



464 



ENGINEERING THERMODYNAMICS 



a straight line, and if the line is not straight the experimental values may be 
wrong or the law untrue. This procedure has been followed in Fig. 141, in 
checking the curves for CO2 and NH 3 against those for water, but it is impos- 
sible to say whether the discrepancies for CO2 are due to a failure of the law or 
bad experimental values, probably both, as the law holds poorly for water itself. 



250 



200 



'I 

■4-1 

Sl50 

I 



100 



50 

























/ 








































/ 






































1 


1 














1 ° 


























CD 

d 




















c 


1 




















c 


3 















































































.6 

Values of 



T B ' 



Fig. 141.— Curves for C0 2 and NH 3 to Check the Linear Relation Eq. (663). 



All of the preceding refers, of course, to pure substances, but in practical 
work there are frequently encountered problems on solutions where large 
differences may exist compared to the pure liquids. Thus, for salts in water, it 
is well known that addition of a salt lowers the freezing-point, that more salt 
lowers it more, and it was first thought that the depression was in proportion 
to the amount dissolved. This being found to be untrue, recourse was had 



HEAT AND MATTER 



465 



again to molecular relations by Raoult, who announced the general law that 
the molecular depression of the freezing-point is a constant. 



Molecular lowering of freezing-point E' 



Mm 



w 



= const., 



(664) 



in which 

At = depression of freezing-point in degrees F. ; 
w = weight dissolved in 100 parts of solvent; 
m = molecular weight of substance dissolved. 

From Eq. (664) the freezing-point for brines may be found as follows: 



Freezing-point of aqueous solutions = 32° — (const.) X 



w 



m 



(665) 



As examples of the degree of constancy of the " constant " the following values 
Table XXIV, taken from Smithsonian Tables are given: 



Table XXIV 
LOWERING OF FREEZING POINTS 



Salt. 


g. Mol. 
1000 g. H 2 0* 


Molecular 
Lowering. 


Authorities. 




NaCl 


.004 
.01 


3.7 
3.67 












.022 


3.55 


Jones 






.049 


3.51 


Loomis 






.108 
.232 
.429 


3.48 
3.42 
3.37 


Abegg 
Roozeboon 






.7 


3.43 






NH 4 C1 


01 


3 6 








.02 


3.56 








.035 
.1 


3.5 
3.43 


Loomis 






.2 


3.4 








.4 


3.39 






CaCl 2 


.01 
.05 


5.1 

4.85 












.1 


4.79 








.508 


5.33 








.946 

2.432 

3.469 

3.829 

.048 

.153 


5.3 
8.2 
11.5 
14.4 
5.2 
4.91 


Arrhenius 

Jones-Getman 

Jones-Chambers 

Loomis 

Roozeboon 






.331 


5.15 








.612 


5.47 








.788 


6 34 







466 ENGINEERING THERMODYNAMICS 

Just as the pressure of dissolved substances in liquids lowers the freezing- 
point, so also does it lower the vapor pressure at a given temperature or raise 
the boiling-point at a given pressure. Investigation shows that a similar 
formula expresses the general relation: 

Molecular rise of boiling-point = E = = constant = 5.2, . (666) 

when water is the solvent. 

From Eq. (666) the rise of the boiling-point is found to be 

TYl 

Rise of boiling-point = 5.2 — (667) 

When liquids are mixed, such as is the case with all fuel oils and with 
denatured alcohol, the situation is different than with salts in solution, and 
these cases fall into two separate classes: (a) liquids infinitely miscible like 
alcohol and water or like the various distillates of petroleum with each other, 
and (6) those not miscible, like gasolene and water. 

The vapor pressure for miscible liquid mixtures is a function of the pressure 
of each separately and of the molecular per cent of one in the other when there 
are two. This rule, which can be symbolized, is no use in engineering work, 
because in those cases where such mixtures must be dealt with there will be 
generally more than two liquids, the vapor pressure characteristic and molec- 
ular per cent of each, or at least some of which will be unknown. 

When, however, the two liquids in contact or in fact any number are 
non-miscible they behave in a very simple manner with respect to each other, 
in fact are quite independent in action. Each liquid will evaporate until its 
own vapor pressure is established for the temperature, as if the other were not 
there, and the vapor pressure for the mixture will be the sum of all the separate 
ones. On the other hand the boiling-point will be the temperature at which 
all the vapor pressures together make up the pressure of say the atmosphere, 
and this is necessarily lower than the highest and may be lower than the 
lowest value for a single constituent. This action plays a part in vaporizers 
and carburettors using alcohol and petroleum products. To permit of some 
approximations, however, a few vapor tension curves for hydrocarbons and alco- 
hols are given later in the Section on vapor-gas mixtures, and data on the vapor 
pressure and temperature relations of ammonia-water solution are given in the 
section on the solution of gases in liquids. 

Example 1. Through how many degrees has ammonia vapor at a pressure of 
50 lbs. per square inch absolute been superheated, when it is at the temperature at 
which steam is formed under a pressure of 100 lbs. per square inch absolute? 

From the curve of pressure and temperature of steam the temperature is 328° F. 
for the pressure of 100 lbs. From the similar curve for ammonia vaporization occurs 
under a pressure of 50 lbs. at a temperature of 22° F. Hence, superheat = 328 — 
22=306°F. 



HEAT AND MATTER 467 

Prob. 1. Three tanks contain the following liquids together: water, ammonia, 
and carbon dioxide respectively, and at a temperature of 30° F. What pressure 
exists in each tank? If the temperature rises to 70° F. how much will the pressure 
rise in each? 

Prob. 2. The pressure exerted by water vapor in the atmosphere when saturated, 
is that due to the temperature and is independent of the pressure of the air. The 
total pressure read by a barometer is the sum of the air pressure and the water vapor 
pressure. What is the pressure due to each under a saturated condition for tem- 
peratures of 50° F., 100° F., 150° F., and 200° F., the barometer in each case being 
29.92 inches of Hg? 

Prob. 3. In order to secure a sufficiently high rate of heat transfer the steam in 
a radiator must be at a much higher temperature than the room to be warmed. If it is to 
be 150° above room temperature what must be its pressure for room temperatures 
of 50° F., 60° F., 70° F., 80° F., and 125° F.? 

Prob. 4. In one type of ice machine ammonia gas is condensed at a high pressure 
and evaporated at a low pressure. What is the least pressure at which gas may be 
condensed with cooling water of 70° F., and what is the highest pressure which may 
be carried in the evaporating coils to maintain a temperature in them of 0° F.? 

Prob. 5. Should carbon dioxide be substituted in the above machine what pressures 
would there be in the condensing coils, and in the evaporating coils? 

Prob. 6. How many degrees of superheat have the vapors of water, ammonia and 
carbon dioxide at a pressure of 15 atmospheres and a temperature of 500° F.? 

Prob. 7. Change the following pressures in pounds per square inch absolute \o 
reduced pressures for water, ammonia, and carbon dioxide, 15 lbs., 50 lbs., 100 lbs., 
500 lbs. 

Prob. 8. At the temperature of melting ice what will be the vapor pressure of ammonia 
and carbon dioxide? At the temperature of melting tin what will be the pressure of 
water vapor? At this same temperature how many degrees of superheat would 
ammonia vapor under 100 lbs. pressure have, and how many degrees superheat would 
carbon dioxide vapor have under 1000 lbs. pressure? 

Prob. 9. If 10 lbs. of common salt, NaCl, be dissolved in 100 lbs. of water, what 
will be the boiling point of the solution at atmospheric pressure, what the freezing-point? 

10. Change of State with Amount of Heat at Constant Temperature. Latent 
Heats of Fusion and Vaporization. Total Heats of Vapors. Relation of Spe- 
cific Volume of Liquid and of Vapor to the Latent Heat. As previously explained, 
a liquid boils or is converted into a vapor at constant temperature when the 
pressure on the surface is constant. Then during the change of state the amount of 
heat added is indirect proportion to the amount of vapor formed. The amount of 
vapor to convert a pound of liquid into vapor at any one steady tempera- 
ture, is the latent heat of vaporization some values for which are given at the 
end of this chapter in Table XLV, and it must be understood that this 
latent heat is also the amount given up by the condensation of a pound of vapor. 
Latent heat is not the same for different pressures or temperatures of vapori- 
zation but is intimately associated with the volume change in the transition 
from the liquid to the vapor state. That this should be so, is clear on purely 
rational grounds because there is necessarily external mechanical work done 



468 ENGINEERING THERMODYNAMICS 

in converting the liquid to the vapor, since this is accompanied by a change 
of volume against the resisting pressure at which the conversion takes place. 
Thus, if 

Vv = specific volume of the vapor in cubic feet per pound; 
V L = specific volume of the liquid in cubic feet per pound ; 
P = pressure of vaporization lbs. per sq.ft. absolute. 

Then 

f Mechanical external work done dur- 1 rt/rT T7 s ., „ .„,,„, 

. ... f1lK \ =P(V V -V L ) it.lbs (668) 

[ ing vaporization ot 1 lb. J 

Of course, at high temperatures the volume of a pound of liquid is greater 
than at low because of its expansion with temperature rise, and under the cor- 
responding higher pressures the volume of a pound of vapor is less, because 
of the compressional effect of the pressure, than at low pressures, so that as 
pressures and temperatures rise the difference Vv—V L becomes less and dis- 
appears at the critical point where it is zero. The latent heat being thus asso- 
ciated with a factor that becomes less in the higher ranges of temperatures 
and pressure may be expected, likewise to become less unless some other factor 
tends to increase. All the energy of vaporization making up the latent heat 
may be said to be used up in (a) doing external work as above, or (6) overcom- 
ing attraction of the molecules for each other. As at the critical point there 
is no molecular change and no external work, the latent heat becomes zero 
at this point. 

This relation between latent heat and volume change was formulated by 
Clausius and Clapeyron, but Eq. (669) is generally known as Clapeyron's 
equation : 

Let L = latent heat ; 
" J = mechanical equivalent of heat = 778, or better 777.52, in such 

cases as this; 
" T = absolute temperature of vaporization; 

" -7^ = rate of increase of vapor pressure per degree change of corre- 
sponding temperature. 
Then 

T dP 

%(V V -V L ) (656) 



777.52 dT 

This formula is used to calculate latent heat from the specific volumes of vapor 
and liquid and from the curvature of the saturation curve when they are known, 
but as these volumes are especially difficult to measure, direct experimental 
determination of the latent heat should be depended upon to get numerical 
values wherever possible. The formula will then be useful for the inverse 
process of calculating specific volumes from latent heats or as a means of 



HEAT AND MATTER 469 

checking experimental values of both, one against the other. It is, however, 

dP 
just as useful to calculate latent heats from the specific volumes, and — of 

the vapor curve, when the latent heats are less positively determined than 
the volumes or densities. 

Another simpler relation of a similar general character exists and is useful 
in estimating latent heats approximately for some little known substances 
like, for example, the liquid fuels, and in the use of which accurate physical 
data are badly needed. Despretz announced that 



Vy~V L 



is nearly constant for all substances, and this was simplified by Ramsay and 
Trouton on the assumption, first, that the volume of the liquid is very small 
at ordinary temperatures and may be neglected, in comparison with the volume 
of the vapor, and second, that the volume of the vapor is inversely proportional 
to the molecular weight m and directly proportinal to absolute temperatures 
so that (Trouton' s law) 



m-m = constant = C 



(670) 



or 

T T 

L = c— 

m 

the constant c is given the following values by Young: 

C0 2 c = 21.3 

NH 3 c = 23.6 

Hydrocarbons c = 20 . 21 

Water and alcohols c = 26 -0 

For such substances as water and steam, the properties of which must be 
accurately known, general laws like the above are of no value compared with, 
direct experimental determination except as checks on its results, and even 
these checks are less accurate than others that are known. 

These experimental data are quite numerous for water, but as generally 
made include the heat of liquid water from some lower temperature to the 
boiling-point. The amount of heat necessary to warm a pound of liquid from 
temperature 32° F. to some boiling-point, and to there convert it entirely into 
vapor is designated as the total heat of the dry saturated vapor above the origi- 
nal temperature. This is, of course, also equal to the heat given up by the con- 
densation of a pound of dry saturated vapor at its temperature of existence and 
by the subsequent cooling of the water to some base temperature taken univer- 
sally now as 32° F. in engineering calculations. 



470 ENGINEERING THERMODYNAMICS 

From observations by Regnault and formulated by him in 1863 the present 
knowledge of the total heat of water may be said to date. He gave the 
expression, Eq. (671), in which the first term is the latent heat at 32° and 
one atmospheric pressure: 

Total heat per pound dry saturated steam = #= 1091.7 + .305(*-32). (671) 

This was long used as the basis of steam calculations, but is now to be discarded 
in the light of more recent experimental data, the best of it based on indirect 
measurements by Grindley, Griessmann, Peake, who observed the behavior of 
steam issuing from an orifice, together with the results of Knobloch and Jacob 
and Thomas on specific heats of superheated steam, and in addition on direct 
measurements by Dieterici, Smith, Griffiths, Henning, Joly. All this work 
has been recently reviewed and analyzed by Davis, who accepts 1150.3 
B.T.U. as the most probable value of the total heat under the standard atmos- 
phere and the following formula as representing total heats from 212° up to 
400° F. 

i7 = 1150.3 + .3745(£-212) - .0005500 -212) 2 . . . (672) 

The Davis curve containing all the important experimental points and the 
accepted line, extended dotted from 212° to 32°, is presented in Fig. 142. 

From the total heats given by this formula the latent heat is obtainable 
by subtraction, according to the relation, 

Latent heat (L) = total heat of vapor above 32° F. (H)— heat of 

liquid from 32° F. to boiling point (h), (673) 

in which the heat of the liquid is computed from a mean curve between Dieterici's 
and Regnault's values, having the equation h = .9983 -.0000288 0—32) + 
.0002133 — 32) 2 . This is the basis of the values for latent and total heats in the 
Marks and Davis steam tables referred to, and accepted as the best obtain- 
able to-day. From these tables a pair of charts for latent heat and total heat of 
dry saturated steam are given at the end of this Chapter. 

The specific volume and density of dry saturated steam, given in the charts 
and table are calculated, as this seems to promise more exact results than direct 
experiment, the method of calculation involving three steps : 

d 7) 
(a) From the pressure- temperature equation the ratio of — is found by 

(XI. 

differentiation as follows: 

logp = 10.515354-4873.717 7 ~ 1 -.004050962 7 +.0000013929647 12 , 
whence 

fp= (^^^-•00405096+.0000027859287 7 ) p. 



HEAT AND MATTER 



471 



(b) From the latent heats the difference between specific volume of vapor 
and liquid, (V V -V L ) is calculated by substituting (a) in Clapeyron's equation. 

(c) From the Landolt, Bornstein, Myerhoffer tables for density of water 
the volume V L is taken, whence by addition the volume of the vapor V v is found 
V v =(Vv-V L ) + V L . 

For ammonia and carbonic acid there are no data available on total heats 
by either direct measure or by the orifice expansion properties, and very few 




32° 50° eS° 86° 104° 122° 140° 158° 176° 194° 212 c 230° 248° 266° 284° 302° 320° 338° 356° 374° 392° 

•Temperature in. Degrees E-. 

Fig. 142.— Total Heat of Dry Saturated Steam above 32° F. (Davis). 

determinations of the latent heat itself, so that the process that has proved so 
satisfactory with steam cannot be directly followed with these substances. 
Accordingly, a process of adjustment has been used, working from both ends, 
beginning with the pressure temperature relations on the one hand and specific 
volumes of liquid and vapor on the other, the latent heat is determined by 
Clapeyron's equation and where this does not agree with authentic values an 
adjustment of both latent heat and specific volume is made. 



472 ENGINEERING THERMODYNAMICS 

This process is materially assisted by the so-called Cailletet and Mathias 
law of mean diameter of the curves of density of liquid and vapor, which are 
given in Figs. 143, 144 and 145, for water, ammonia and carbon dioxide, on 
which the points are marked to indicate the source of information. 

On each of these curves the line BD is the line of mean density, its abscissa 
being given by the following general equation, 

s = i(^+^-)=a+bt+c? (674) 

Of course, this mean density line passes through the critical volume B. For 
these three cases this Eq. (674) is found to have the form, 



For water s = 28.7 -.0150 -300)-. 0000150 -300) 2 . (a) 

For ammonia s = 20 -.0220 -30). (b) 

For carbonic acid.. s = 33.1-.O2190+2O)-.OOO160-2O) 2 . (c) 



(675) 



A more exact equation for water has been determined by Marks and Davis 
in their steam tables and is 

s = 28.424-.O165O0-32O)-.OOOO1320-32O) 2 . . . . (676) 

From the smooth curve, which has the above equation, the volumes and densi- 
ties of liquid and vapor that are accepted have been derived, and are presented 
in chart form on a large scale and in tabular form at the end of the Chapter, 
the values for water being those of Marks and Davis. 

du 
From these volume differences and the — relation the latent heats have 

dT 

been calculated and the newly calculated points are compared with experimental 

values in Fig. 146. 

The total heats are obtained by adding to the latent heat the heat of 
liquid above 32° from —50° F. up to the critical point for CO2 and to 
150° F. for NH3, which include the working range for refrigeration. 
These liquid heats have already been determined in Section 5 in discussing 
specific heats. 

Charts and tables at the end of this Chapter give the final values of total heat, 
heat of liquid, latent heat, specific volume and density of dry saturated 
vapor based on large-scale plottings, without equations beyond that for the 
pressure-temperature relations for saturated vapor, and the results are be- 
lieved to be as reliable as it is possible to get them without more experimental 
data. 

The properties of dry-saturated steam are given in Table XL VII, and 
charts, A, B, C, D, E, F; the properties of superheated steam, in Table XL VIII; 
dry-saturated ammonia vapor in Table XLIX, and Charts G, H, I, J, K, L; 
and dry-saturated carbon dioxide vapor in Table L, and Charts M, N, 
O, P, Q, R. 



HEAT AND MATTER 



473 




.'3 .1 .06 

Upper Scale = Lbs. per Cu.Ft 



.025 
Lower Scale = Cu.Ft.per Lb. 



700 



J3 

fr 600 



u 
S500 



400 











































N 








































\ 
\ 


\ 






































> 


\ 








































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.5 



1.5 



Volume of Vapor in Cu.Ft. 
Fig. 143. — Specific Volume and Density of Liquid and Dry Saturated Vapor of Water. 



474 



ENGINEERING THERMODYNAMICS 































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D oh 
































































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5 
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Upper Scale 


10 

.1 

= Lbs. per 


1 15 
075 

Da. Ft. 








20 
.05 








25 
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30 

Lower Scale 




Cu 


1 .35 
03 

. Ft. per 


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ft 




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§>100 

P 

a 
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1 

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8 






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a 










































































































































































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-50 























































































5 10 15 

Volume of Vapor in Cubic Feet per Pounds. 



30 



Fig. 144. — Specific Volume and Density of Liquid and Dry Saturated Vapor of Ammonia. 



HEAT AND MATTER 



475 




.1 .05 .04 

Upper Scale = Lbs. pec Cu. Ft 



I 45 

.0*5 .02 

Lower Scale =-Cu.Ft.per Lb. 























































Crit 


ical 


Poir 


t 


































80 






M 


k 














































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bo 














































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3 




























































































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-20 




















































. 


1 










i 










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Volume of Tapor i» Cu.Ft.per Pouri^ 



Fig. 145. — Specific Volume and Density of Liquid and Dry Saturated Vapor of Carbon 

Dioxide. 



476 



ENGINEERING THERMODYNAMICS 


















sirx'a 










































































p. 

w 

w a 

£ ffi o 

* n 2 s £3 

^ 0) 05 O 02 

O fl C -H - 

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Vo 












































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2 



eirra 



HEAT AND MATTER 



477 



The volumes of dry-saturated steam determined from the tables when 
compared with their pressures show that there exists an approximate 
relation of the form for steam, 



V(Vv- V L ) l064e = constant = 497, 



(677) 



when pressures are in pounds per square inch and volumes are in cubic feet. 
This curve plotted to PV coordinates is called the saturation curve for the vapor. 
It is useful in approximate calculations of the work that would be done by steam 
expanding so that it remains dry and saturated or the work required to compress 
vapor such as ammonia under the same conditions. But as the specific volume 
of liquid is generally negligible it may be written as one of the general class 



PV S = constant, 



(678) 



for which s = 1.0646 and constant = 497. 

This curve supplies a means for computing the work for wet vapors (not too wet) 
as well as dry, provided only that they at no time become superheated or change 
their quality, by using for V some fraction of the true specific volume repre- 
senting the dryness. The very fact that a great volume of vapor may be 
formed from an insignificant volume of liquid makes the saturation curve a 
useful standard of comparison with actual expansion and compression lines for 
wet vapors. 




Fig. 147. — Comparison of Steam Expansion Line of an Indicator Card with the Satura- 
tion Line for both Dry Saturated Vapor and for Vapor Constantly Wet at the Initial Value. 

Plotting the saturation curve beside an actual cylinder expansion or 
compression curve will show the quality of vapor at all times and also give a 
measure of evaporation and condensation taking place during the process. In 
Fig. 147 is shown a set of diagrams taken from a simple Corliss engine, 18X24 
ins. with 4 per cent clearance, a 2J-inch piston rod and tail rod, running at 150 
R.P.M., to which have been added lines of zero pressure and volume by the 
method explained in Chapter I. The discharge from the condenser per hour 
for a constant load of the value to give the above cards was 2600 lbs. Allow- 



478 ENGINEERING THERMODYNAMICS 

ing for the rods, the displacement volumes of each end of the cylinder will be 
5990 cu.ins., and since the clearance volume is 4 per cent, the steam volume 
will be 239.6 cu.ins. From the left-hand card it will be seen that the cut-off 
was at point C, 16.5 per cent of stroke, hence the volume at C is (.165 X 
5990) +239.6 = 1228 cu.ins. It will also be seen from the card, that the pres- 
sure at C was 73.5 lbs. per square inch absolute. From the curves or the 
tables at the end of the Chapter 1 cu.ft. of dry steam at this pressure weighs 

1228 
.1688 lb. and hence the weight of steam in this end of cylinder was — — X.1668 

1728 

or .1185 lb. at cut-off. From the card it will also be seen that at the end of 

the exhaust stroke, denoted by the point D, the pressure was 30 lbs., at which 

the weight of 1 cu.ft. of dry steam is .0728 lb., hence the weight of steam in the 

239 6 
cylinder was ——^-X. 0728 = .01010 lb., and the amount admitted was .1185 — 

.0101 = .1084 lb. 

In as much as the two ends of cylinder are identical and as the cards from 
both ends are practically the same, it may be assumed that the same weight 
of steam was in each end, or that .1084X2 = .2168 lb. are accounted for by the 
card per revolution, or .2168X150X60 = 1950 lbs. per hour. There is then 
the difference to otherwise account for, of 2600—1950 = 650 lbs. per hour, 
which can only have been lost by condensation. 2600 lbs. per hour is 2600 
-i- (150X60X2) = .1442 lb. per stroke, which with the .0101 lb. left from pre- 
vious stroke would make .1543 lb. in the cylinder at cut-off, and if it 
were all steam its volume would be 1581 cu.ins., denoted by point E 
on diagram. The ratio of AC to AE gives the amount of actual 
steam present in the cylinder at cut-off, to the amount of steam and 
water. The saturation curves CF and EG are drawn through C and 
E from tabular values and represent in the case of CF the volumes 
which would have been present in the cylinder at any point of stroke had the 
steam and water originally present expanded in such a way as to keep the 
ratio or dryness constant, and in case of EG, volumes at any point of the stroke 
if all the steam and water originally present had been in form of steam 
and had remained so throughout the stroke. Just as the ratio of A C to AE 
shows per cent of steam present at cut-off, so does the ratio of distances of any 
points Y and Z, from the volume axis denote the per cent of steam present at 
that particular point of the stroke. By taking a series of points along the 
expansion curve it is possible to tell whether evaporation or condensation is 
occurring during expansion, In this case the ratio, 



= = .795, and£T = .86. 
AE XZ 

Hence, it is evident that evaporation is occurring since the percentage of steam 
is greater in the second case. 



HEAT AND MATTER 479 

For some classes of problems it is desirable that the external mechanical 
work be separated from the latent heat, and for this reason latent heat is 
given in three ways : 

(a) External latent heat, 

(b) Internal latent heat, 

(c) Latent heat total. 

The external latent heat in foot-pounds is the product of pressure and volume 
change, or expressing pressures in pounds per square inch, 

144 
External latent heat = - T p(V v -V L ) (679) 

This is sometimes reduced by neglecting V L as insignificantly small as it really 
is for most problems which are limited to temperatures below 400° for saturated 
vapor, in which case, 

144 
External latent heat = -y-pV v (680) 

In all cases 

Internal latent heat = L— (Ext. Lat. Ht.) (a) 



= L-~~p(V v -V L )(b) 



or 

T 144 



rVv (C) 



(681) 



Fusion and freezing are quite similar to vaporization and condensation 
in that they are constant temperature processes with proportionality between the 
amount of substance changing state and the amount of heat exchanged. They 
are different in as much as little or no volume change occurs. As there is so little 
external work done it may be expected that there is little change in their latent 
heats with temperature and pressure, but as a matter of fact it makes very 
little difference in most engineering work just how this may be, because prac- 
tically all freezing and melting takes place under atmospheric pressure. There 
does not appear to be any relation established between heats of fusion like 
those for vaporization that permit of estimates of value from other constants, 
so direct experimental data must be available and some such are given for a 
few substances at the end of this Chapter in Table XL VI. As a matter of fact 
such laws would be of little use, and this is probably reason enough for their 
non-discovery. 

Example 1. Pigs of iron having a total weight of 5 tons and a temperature of 
2000° F. are cooled by immersing them in open water at a temperature of 60° F. If 
one-half of the water is evaporated by boiling, how much must there have been originally? 

The iron must have been cooled to the final temperature of the water, which 
must have been 212° F. Also the heat given up by the iron will be the 



480 ENGINEERING THERMODYNAMICS 

product of its weight, specific heat and temperature difference, or, considering the 
mean specific heat to be .15, 

10,000 X (2000 -212) X .15 =2,682,000 B.T.U. 

The heat absorbed by the water in being heated, considering its specific heat as unity 
will be its weight times its temperature change and, since one-half evaporates, the 
heat absorbed in evaporating it will be half its weight times the latent heat, or 

W [(212 -60) +.J X970] = 637^ B.T.U. 

These expressions for heat must be equal, hence 

^ 2 -M2=42101b, 

637 

Example 2. A tank of pure water holding 1000 gallons is to be frozen by means 
of evaporating ammonia. The water is originally at a temperature of 60° F. and the 
ice is finally at a temperature of 20° F. The ammonia evaporates at a pressure 
of one atmosphere and the vapor leaves the coils in a saturated condition. How 
many pounds of ammonia liquid will be needed, how many cubic feet of dry saturated 
vapor will be formed, and how much work will be done in forming the vapor? 

The heat to be removed is the sum of that to cool the water, the latent heat 
of fusion of ice, and that to cool the ice, or for this case 

[(60-32) +144+.5(32 -20)3x8333, 

8333 being the weight of 1000 gallons of water. Hence the B.T.U. abstracted 
amount to 1,466,608. 

Each pound of ammonia in evaporating at atmospheric pressure absorbs 594 B.T.U.'s 
as latent heat and, therefore, 2470 lbs. are needed. At this pressure each pound of 
vapor occupies 17.5 cu.ft., hence there will be 43,200 cuit. of vapor. At this same 
pressure the volume of a pound of liquid is .024 cu.ft., so that the work done per pound 
in evaporating the ammonia is 37,000 ft.-lbs. and the total work is 915 XlO 6 ft.-lbs. 

Prob. 1. How much ice would be melted at 32° F. with the heat necessary to 
boil away 5 lbs. of water at atmospheric pressure, the water being initially at the 
temperature corresponding to the boiling-point at this pressure? 

Prob. 2. What is the work done during the vaporization of 1 lb. of liquid anhydrous 
ammonia at the pressure of the atmosphere? 

Prob. 3. From the tables of properties of anhydrous ammonia check the value of 
the constant in Trouton's law given as 23.6 by Young. 

Prob. 4. As steam travels through a pipe some of it is condensed on account of 
the radiation of heat from the pipe. If 5 per cent of the steam condenses how much 
heat per hour will be given off by the pipe when 30,000 lbs. of steam per hour at a 
pressure of 150 lbs. per square inch absolute is passing through it? 

Prob. 5. Brine having the specific heat of .8 is cooled by the evaporation of ammonia 
in coils. If the brine is lowered 5° F. by ammonia evaporating at a pressure of 20 



HEAT AND MATTER 481 

lbs. per square inch gage, the vapor escaping at brine temperature, how many pounds 
of brine could be cooled per pound of ammonia? 

Prob. 6. Steam from an engine is condensed and the water cooled down to a 
temperature of 80° F. in a condenser in which the vacuum is 28 ins. of Hg. How many 
pounds of cooling water will be required per pound of steam if the steam be initially 
10 per cent wet? 

Prob. 7. A pound of water at a temperature of 60° F. is made into steam at 100 
lbs. per square inch gage pressure. How much heat will be required for this, and what 
will be the volumes at (a) original condition; (b) just before any steam is made; (c) after 
all the water has been changed to steam? 

Prob. 8. A sand mold weighs 1000 lbs. and 100 lbs. of melted cast iron are poured 
into it. Neglecting any radiation losses and assuming the iron to be practically at 
its freezing temperature how much of the iron will solidify before the mold becomes 
of the same temperature as the iron? 

Prob. 9. How many pounds of ice could be melted by heat given up by freezing 
50 lbs. of lead? 

11. Gas and Vapor Mixtures. Partial and Total Gas and Vapor Pressures. 
Volume, Weight, and Gas Constant Relations. Saturated Mixtures. Humidity. 

One of the characteristic properties of gases distinguishing them from liquids, 
and which also extends to vapors with certain limitations is that of infinite 
expansion, according to which no matter how the containing envelope or volume 
of the expansive fluid may vary, the space will be filled with it at some pres- 
sure and the weight remain unchanged except when a vapor is brought to 
condensation conditions, or the pressure lowered on the surface of a liquid 
which will, of course, make more vapor. A given weight of gas or vapor (within 
limits) will fill any volume at some pressure peculiar to itself, and two gases, 
two vapors, or a vapor and gas, existing together in a given volume, will fill it 
at some new pressure which is the sum of the pressures each would exert sepa- 
rately at the same temperature (if non-miscible) . This fact, sometimes des- 
ignated as Dalton's Law, permits of the derivation of equations for the rela- 
tion of any one pressure, partial or total, to any other total or partial, in 
terms of the weights of gas or vapor present, and the gas constants R. It 
also leads to equations for the various constituent and total weights in 
terms of partial and total pressures and gas constants. Such equations sup- 
ply a basis for the solution of problems in humidification and drying of air, in 
carburetion of air for gasolene and alcohol engines, or of water gas for illumina- 
tion, and are likewise useful as check relations in certain cases of gas mixtures 
such as the atmospheric mixture of nitrogen and oxygen, producer gas or gase- 
ous combustibles in general. 

Let w\, W2 and w x be the respective weights of the constituents of a mixture; 

w m = Xw be the weight of the mixture; 

Pi, P2, P x be the respective partial pressures of the constituents; 

P m = 2P be the pressure of the mixture; 

Pi, P2, R x be the respective gas constants; 

R m be the gas constant for the mixture. 



482 ENGINEERING THERMODYNAMICS 

Then if w\ lbs. of one, and w^ lbs. of another gas or vapor at temperature 
T m occupy the volume V m cubic feet together, 

V m Pi = wiRiT m , ....... (a)] 

and (682) 

V m P2 = w 2 R2T m , (6) J 

whence 

V m (Pi+P2) = (w 1 Ri+w 2 R2)T m , . (683) 

or, in general, 

2P=2(wR)T m (684) 

Or putting 

2P = P m , (685) 

and 

X{wR)=R m w m , or R m = ^ U ^-, . . . (686) 

w m 

then 

P m V m = w m R m T m , . . (687) 

so that the mixture will behave thermally quite the same as any one gas with 
such exceptions as may be due to a different gas constant R m . 
Dividing Eq. (682a) by Eq. (683) or (684) gives 

Pi = wiRi wiRj 

Pm w 1 R 1 +w 2 R 2 2(wR)' ■ ' * ^** } 

which gives the ratio of any partial pressure to that for the mixture in terms 
of the individual weights and gas constants. Hence 

Pi w^i t N 

P^ = ^ (689) 

which gives the ratio of any partial pressure to that for the mixture in terms of 
its own weight and gas constant and those for the mixture. 

It is possible to express the ratio of weights as a function of gas constants 
alone which will permit of a third expression for the partial pressures in terms 
of gas constants without involving any weights. For two gases 

W\=W m — W2. 

Whence 

W\ . W2 

^ =1 -^' • • • ( 69 °) 



HEAT AND MATTER 483 



But from Eq. (686) 

W1R1 + W2R2 = WmRmj 



or 



so that 



or 



w 2= 1 / p w x 

W m R 2 \ W m 



W m W m R2\ W m 



W\(^ R\\ R n 



>m\ R2) 



and 



W m \ R2J R2 

wi = R 2 — R r 

W m R2 — R] 



(691) 



which is the ratio of partial to total weights in terms of gas constants. On sub- 
stitution in Eq. (689), 

P\ R\ ( R2 — R 



1 m Km \*t 



^§f), (692) 



which gives the ratio of partial pressures of two gases or vapors to that for the 
mixture in terms of the individual gas constants and that for the mixture, and 
a similar expression can be found for more than two gases. The ratio of any 
one partial, to the total weight can also be found from Eq. (689) in terms of its 
gas constant and partial pressure, and the mixture gas constant and pressure, 
from Eq. (691) in terms of the gas constants for the constituents and for 
the mixture. This ratio of partial to total mixture weight gives the fractional 
composition by weight. 

It is sometimes necessary to know the volume relations in a mixture of two 
gases existing at the same pressure or two vapors or a vapor and gas, such, for 
example, as air and water vapor. In this case two different volumes existing 
together at a common temperature and pressure together make up a mixture 
volume equal to their sum. Using similar symbols 

........ . (693 

P m V 2 = w 2 R2T m \ 

where Vi and V2 are the volumes occupied by the two constituents respectively 
when at a mixture pressure P m and temperature T m , whence for the mixture 

P m (Vi + V2) = (w 1 R 1 +w 2 R2)T m (694) 

or 

P m 2(V) = i:(wR)T m (695) 



484 ENGINEERING THERMODYNAMICS 

These Eqs. (694) and (695) are identical in form with (683) and (684) except that 
V replaces P, and V, P, so that all equations just derived also apply to volumes 
as the volume proportion will be identical with pressure proportions. For 
convenience of reference these may be set down. 
From Eq. (688), 

vT?wR> (696) 

which gives the ratio of any partial volume, to that for the mixture in terms of 
the individual weights and gas constants. 



From Eq. (689) 

Vi wiRi 



V m W m R % 



(697) 



which gives the ratio of any partial volume to that of the mixture in terms of its 
own weight and gas constant and those for the mixture. 



From Eq. (692) 

Vi == Ri (R2-B m ) 

V m Rm (R2 — R1) ' 



(698) 



which gives the ratio of any partial volume to that of the mixture in terms of the 
individual gas constants and that for the mixture. 

The volumetric composition of air is given by Eq. (697) or its equal 
numerically Eq. (692), and since the partial pressure of oxygen and nitrogen 
in air are 78.69 per cent and 21.31 per cent, these are its volumetric per cents. 

When one of the constituents is a vapor, all the preceding applies, provided 
the condition of the vapor is such that at the temperatures assumed it is not 
near the condition of condensation, but then the relations become more definite 
since the partial pressure of the vapor is fixed by the temperature. In practical 
work with gas and vapor mixtures the failure^of the perfect gas laws near the 
condensation condition is ignored and they are assumed to be true for the very 
good reason that there is no other way as good, to get numerical results. 

All liquids, and many, if not all solids will, if placed in a vacuum chamber, 
evaporate until the pressure has reached a certain value depending on the tem- 
perature, at which time the liquid and its vapor are in equilibrium, and evapo- 
ration may be said either to cease or proceed at a rate exactly equal to the rate at 
which vapor condenses, or more precisely, at equilibrium the weight of vapor 
in the vapor form remains constant. The weight of vapor that will rise over a liquid 
in a given space depends on the temperature and pressure which are related 



HEAT AND MATTER 



485 








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Fig. 148. — Vapor Pressure of Hydrocarbons and Light Petroleum Distillates of the Gasolene 

Class. 






486 



ENGINEEEING THERMODYNAMICS 



in the so-called vapor tension or vapor pressure tables and curves, such as 
shown in Figs. 148, 149 and 150, for some liquid fuels or as given in the pre- 
vious section for water. At any fixed temperature the vapor will con- 
tinue to rise until it exerts its own vapor pressure for the temperature, 
the process being often described as evaporation without ebullition. If the 
liquid or solid be introduced into a chamber containing dry gas the evapora- 
tion will proceed precisely the same as in the vacuum until the pressure has 



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Fig. 149.— Vapor Pressure of Heavy Petroleum Distillates of the Kerosene Class. 

risen by an amount corresponding to the vapor pressure for the temperature, 
because each substance exerts the pressure it would if alone occupying the 
volume; when they both occupy the same volume the pressure will be their 
sum and equal to the pressure of the gas alone added to the vapor pressure for 
the same temperature. There is one important practical condition, and that 
is, time enough for the completion of the process of evaporation which proceeds 
very slowly toward the end. If time enough is allowed the vapor pressure will 
establish itself and the gas is said to be saturated, and this is an important 



HEAT AND MATTER 



487 



special case of gas-vapor mixtures. It is the condition in which the gas may 
be said to carry the maximum weight of vapor possible for the total pressure 





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Temperature Degrees, Fahrenheit 
Fig. 150. — Vapor Pressure of the Alcohols. 



and temperature. The gas in contact with the liquid may carry less vapor 
if it has not been in contact long enough at the given temperature, and a gas 



488 ENGINEEEING THERMODYNAMICS 

no longer in contact with the liquid may carry less, because, (a) of insufficient 
time of original contact; (b) of condensation of some it originally carried; (c) 
of a rise of temperature after leaving the liquid. To all such general cases the 
equations above apply without change, but for the special case of saturated 
mixtures they have a simpler form. 

Let P v be the vapor pressure of the liquid for temperature T, which is the 
partial pressure of the vapor in a saturated vapor gas mixture ; 
" P g be the partial pressure of the gas at same temperature. 
Then for a gas saturated with vapor at temperature T, Eq. (689), 

Weight vapor X __ / Vapor pressu reX ( R for mixture X . . 

Weight mixt./ \Mixture press./ \ R for vapor / 

But according to Eq. (646) and Eq. (651), 



R for vaporX _ /Density of mixtureX _ /Mol. wt. of mixtA 
R for mixt./ \ Density of vapor . / \Mol. wt. of vapor/ 



(700) 



whence 

w v Pv m v 



Also 



(701) 

IMm -Lin Wlm 



^-=^^, (702) 

t 

and 

**!*- . (703) 

W ff PglUg 

The presence of water vapor in the atmosphere, and problems connected with 
it, constitute a specific case of vapor-gas mixture, subject to the foregoing laws. 
This subject has been given most attention by the United States Weather 
Bureau; tables have been prepared for ready computation and for certain 
problems for which only experimental data or empiric formulas afford solution. 

Air is said to be " saturated with moisture " when it contains the saturated 
vapor of water. It might be better to say that the space is saturated since the 
presence or absence of the air has no effect upon the water vapor other than im- 
posing its temperature or imparting heat to the water vapor, and also that the 
air retards the diffusion of water particles. The weight of saturated aqueous 
vapor per cubic foot depends only on the temperature, and not on the presence 
of air. 

If the space contains only a certain fraction of the weight of aqueous vapor 
corresponding to saturation, that fraction is called the " relative humidity " 
or degree of saturation, and the corresponding percentage, the per cent of 
saturation. If air containing saturated water vapor be cooled ever so little, 



HEAT AND MATTER 489 

some of the vapor will be condensed and appear in the liquid form. If air is 
cooled at constant pressure, from a given initial condition, the degree 
of saturation approaches unity, and finally reaches that value at a temperature 
called the " dew point " corresponding to the initial condition. At this tem- 
perature the condition of saturation has been reached and any further cooling 
will cause the precipitation of liquid water, as occurs in the formation of dew, 
rainclouds or fog. 

A space or body of air carrying water vapor in smaller quantity than that 
of saturation, in reality contains superheated steam. If the vapor density 
and the temperature of the mixture be known, the degree of superheat may be 
ascertained from the temperature of the mixture, and the temperature corre- 
sponding to saturated water vapor having a pressure equal to the partial pres- 
sure of the vapor in the mixture. 

Humidity of atmospheric air is ordinarily determined by an instrument 
called the psychrometer, which consists merely of two thermometers, one with 
a bulb exposed directly to the air and the other covered with a piece of wick 
which is kept moist with water. The two are mounted together so that they 
can be whirled or swung about in the air until a stable condition has been 
reached. The dry-bulb thermometer should record the temperature of the air. 
The wet-bulb thermometer will record something lower than the air tempera- 
ture, dependent upon the rate at which evaporation takes place, since the process 
of evaporation abstracts heat. Were there no other influence, this process of 
evaporation would continue till the temperature of the wet bulb became that 
of the dew point. The temperature of the wet-bulb thermometer never falls 
to the dew point, however, because of conduction of heat between the cold 
bulb and the warmer surrounding air. From extensive experiments conducted 
by the U. S. Weather Bureau, Professor Ferrel has devised the following formula 
for the vapor pressure, h in ins. of mercury corresponding to given readings 
of the wet- and dry-bulb thermometers, U and t w degrees F. 

h = h , -mOZQ7h b (t d -t w )(l-j^\ (704) 

where h b is barometric height in inches, after applying all corrections, and h! is 
pressure of saturated water vapor, in inches of mercury, corresponding to the 
temperature t w . 

The relations shown by this formula are expressed graphically in much 
more convenient form in the curves of Figs. 151 and 152, devised by 
Prof. H. L. Parr. The use of the curves is best illustrated by an example: 
if the dry-bulb reading is 75° F. and the wet-bulb 65° F., find the dew point. 
The difference of wet- and dry-bulb temperatures is 10°. From 10° at the top of 
the diagram (B) Fig. 151 project downward, and from 75° air temperature at the 
left of diagram project to the right to the intersection, where the dew point is 
read by interpolation between the contour curves at (C) to be 59.5° F. These 
curves are drawn for a barometric pressure of 29.92 ins. (standard) and will 



490 



ENGINEERING THERMODYNAMICS 



Difference in Temperature: Wet and Dry Bulbs: Degrees Fahrenheit 
.0 2 4 6 8 10 12 14 1.6 18 20 




12 3 4 5 6 7 8 9 10 11 

Difference in Temperature:Wet and Dry Bulbs: Degrees Centigrade 



Fig. 151. — Relation between Wet and Dry Bulb Psychrometer Readings and Dew Point for 

Air and Water Vapor. 



HEAT AND MATTER 



491 



100° 




10 20 30 40 50 60 70 80 90 K 

Degree of Humidity, Per Cent 



Fig. 152. — Relation between Humidity and Weight of Moisture per Cubic Foot of 

Saturated Air. 



492 ENGINEERING THERMODYNAMICS 

not apply correctly, when the barometer is not equal to this, though with fair 
approximation, so long as the difference in barometer is not great. Where there 
is much departure the original the formula must be used. Fig. 152 gives 
weight of aqueous vapor per cubic foot of mixture, in grains (7055 lb.) and 
also the degree of humidity. The temperature of the dew point 59.6° F, is 
located at (C) on the right-hand side of Fig. 152. Interpolation between the 
ends of the contours for weight, gives 5.6 grains per cubic foot. On the same 
scale the temperature of the air, F., is represented at point (A) 75°, project- 
ing to the intersecting point D and down to the bottom of the diagram gives 
on the scale for degree of humidity, 60 per cent. 

Example 1. By means of the relation of gas constants find the proportion of nitrogen 
and oxygen in the air. 

R for nitrogen is 54.92 and for oxygen 48.25 and for air 53.35. From Eq. (698) 

Vo _ Ro (Rn —Rm) 

V m Rm (Rn—RoY 

which, on substituting the above values for R N , Ro and Rm gives 

Vo 48.25(54.92-53.35) 



V m 53.35(54.92-48.25) 



.213, 



or air is 21.3 per cent oxygen by volume. 

Example 2. At what temperature will air containing ^ lb. of water per pound of 
dry air at atmospheric pressure be saturated? 

If the vapor pressure be known, the temperature may be found from tables. 
From Eq. (701) 

Wv _ P v m v 
w a P a ma 



or 



P g mgW t and ,=760mm. ofHg. 

mvWg 



For air m g = 28.88 and for water ra p = 18, 

hence substituting those values 

(7 60-P„)(28.88x.5) 
18x1 
or 

D 750X7.22 OOP7 „ 

Pp = -1^22~ =337mm - Hg ' 

which corresponds to a temperature of 172° F. 



HEAT AND MATTER 493 

Example 3. A pound of alcohol requires 9.06 lbs. of air for a proper combustible 
mixture for gas engines. At what temperature will these proportions contitute a 
saturated mixture? 

From Eq. (701) 

P ^%^' 
m v Wg 

For alcohol m„-46, for air m g =28.88, and P v +P g =760 mm. of Hg for atmospheric 
pressure. Substituting these values in the above equation 

(760-P,)28.88 

P„ = — =49 mm. Hg. 

46x9.06 6 

From the curve of vapor tension of alcohol, the temperature corresponding to 49 mm. 
of Hg is about 72° F. 

Prob. 1. Air at 80 per cent lumidity, atmospheric pressure and 70° F, is cooled 
to 40° F. How much water will be thrown down per 1000 cu.ft. of moist air. 

Prob. 2. The same air is compressed adiabatically to five atmospheres, and again 
cooled to 40° F. at this pressure. How much moisture per 1000 cu.ft. of moist air 
will be separated out when the tenperature becomes 70° F, and bee n much at 
40° F. 

Prob. 3. What will be the weight of water in a pound of air and water vapor if 
the value for R for the mixture is taken as 55.25, for air as 53.35 and for water vapor 
as 91? 

Prob. 4. At what temperature will air containing its own weight of water vapor 
be saturated at atmospheric pressure? 

Prob. 5. An internal combustion engine uses a saturated mixture of air and gasolene 
vapor in which ratio of air to gasolene is 15.3. Considering the gasolene to be hex- 
ane, at what temperature will the mixture be? 

Prob. 6. Should kerosene regarded as decane, C 10 H 22 , be substituted for gaso- 
lene, in the above problem what would be the change in temperature of mixture, 
assuming it still to be saturated? 

Prob. 7. Air containing moisture equal to one per cent of the weight of the air 
alone is at a temperature of 150° F. How much is the water vapor superheated? 
What is the humidity? 

Prob. 8. The reading of a dry bulb of a psychrometer is 90° F. and of the wet 
bulb 70° F. By means of curves of Figs. 151 and 152, find the dew point, relative humid- 
ity, and grains of water per cubic foot of air. 

12. Absorption of Gases by Liquids and Adsorption or Occlusion by Solids. 
Relative Volumes and Weights with Pressure and Temperature. Heats of 
Absorption and of Dilution. Properties of Aqua Ammonia. In the attainment 
of high vacuua in steam condensers, separate removal of considerable quantities 
of non-condensible gases is found necessary by means of dry vacuum pumps, 
a fact that proves in a practical way the freedom with which the boiler water 
had absorbed gases. These gases for otherwise pure water are carbon dioxide 
and air, probably mainly air, but may include many others, notably the 



494 ENGINEERING THERMODYNAMICS 

products of organic decomposition especially when condensing water is taken 
from the neighborhood of sewers, as is generally the case when power plants 
are located on city water fronts. To a very much greater extent, however, 
is ammonia soluble in water, the latter being capable of taking up about 1000 
volumes of ammonia at 0° and one atmosphere, against about 30 volumes of air, 
and one-fiftieth of a volume of hydrogen. It is the freedom of solution of am- 
monia in water that makes the process useful as a means of removing 
anhydrous ammonia from the cooling coils in mechanical refrigerating plants, 
as a substitute for the mechanical removal by piston compressors. 

In all cases the solubility of gases in liquids decreases with rise of tem- 
perature, a fact associated with the separation of gases from boiler feed- 
water during their heating in feed-water heaters, economizers and the boiler 
itself. This property is also depended upon to free the aqua ammonia that 
has absorbed its ammonia charge from the cooling coils, of the amount so 
taken up, by heating the solution in a separate chamber from which the am- 
monia vapor escapes to the ammonia condenser to become liquid anhydrous 
while the weak liquor returns to the absorber for a new charge. To permit of 
the calculation of the quantity of liquor to be circulated, in order that a given 
amount of anhydrous ammonia may be absorbed from the cooling coils and 
delivered later by heating to high temperature to a condenser, requires accu- 
rate data on the maximum possible ammonia content of solutions at various 
temperatures and pressures. Rise of temperature always will reduce the 
gas content of the solution if originally saturated, but the volume dissolved 
is independent of pressure for slightly soluble gases like nitrogen or hydro- 
gen, the weight dissolved, of course, being greater and directly proportional to 
pressure at a given temperature by reason of the increased density. 

This law of independence of volume and pressure or proportionality 
of weight to pressure, is known as Henry's law, and is hardly true for gases 
as freely soluble as ammonia, probably due to some action between water and 
the gas, equivalent to that studied by Thomsen for solids, which tend to form 
hydrates of various kinds. For such gases as are soluble by weight in propor- 
tion to pressure, it is not the total pressure of the solution that is significant, 
but the partial pressure of the gas alone, so that the amount of mixed gases 
like air dissolved in water will depend, for the oxygen part, on the specific 
solubility of oxygen and its partial pressure in the air, which is approximately 
one-fifth that of the air, and the same is true for the nitrogen. Thus, in examin- 
ing the solubility conditions for ammonia in water, while in practice the total 
pressure only is known, it is to the separate pressure of the ammonia that the 
amount dissolved must be referred in any attempt to establish a law of 
relation. 

Just as gases dissolve in liquids, so are they absorbed by solids, though in this 
case the process is described as one of adsorption or occlusion. This phenomenon 
is now being studied in connection with coal, which it is found more or less 
freely absorbs air, the oxygen of which under comparatively low temperatures 
unites with the coal causing spontaneous combustion if the heat is conserved 



HEAT AND MATTER 



495 



as in a pile or its liberation in any way accelerated by heating or otherwise. 
Most investigations of adsorption or occlusion have been made with charcoals, 




110 130 150 170 190 

Temperature in Degress Fahrenheit 



210 



230 



250 



Fig. 



153. — Ammonia- water Solutions, Relation between Total Pressure and Temperature 
(Dotted Lines Mollier Data). 

the more dense varieties of which have greater adsorptive power than others. 
The quantity of different gases adsorbed is believed by Dewar to be the same 
in volume per unit of charcoal when each is held at its own condensation tern- 



496 



ENGINEERING THERMODYNAMICS 



perature. The quantity increases with rise of pressure but not in proportion, 
and decreases rapidly with rise of temperature and a curve showing the tem- 



120 



110 



100 



3 90 

o 

B 

< 

§80 

M 

si 
I 70 



T3 

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o 

Ph 



3 50 



40 



30 



20 



10 







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5 10 H.8 15 20 23.3925 30 33.73 35 4041.55 45 50 

Per Cent by Weight of Ammonia in Solution. 

Fig. 154. — Ammonia-water Solutions, Relation between Total Pressure and Per Cent NH 3 

in Solution. 

perature and pressure at which equal volumes are adsorbed is similar in form 
to a vapor tension curve. 

In the establishment of the properties of aqueous solutions of ammonia 



HEAT AND MATTER 



497 



progress was for many years very slow and the early experimental data of 
Abegg and Rosenfeld, 1903, for weak solutions and low pressures, that of Roscoe 
and Dettman, 1859, for highly concentrated solutions under low pressures at 



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J50.36 



10 11-8 15 .20 23.39 25 30 33.7335 ^O^-SS 45 

Percent by Weight of Ammonia in Solution 



50 



55 



Fig. 155. — Ammonia-water Solutions, Relation between Temperature and Per Cent NH 3 

in Solution. 

32° F., together with that of Sims, 1862, for four temperatures from 32° F. 
to 212° F., failing to check and having many gaps, has been more recently supple- 
mented by Perman, 1903, and Mollier, 1908. With these new data it has been 
possible to graphically fill in the data of the unexplored region with reasonable, 



498 ENGINEERING THERMODYNAMICS 

though not yet satisfactory accuracy. The results are given in three diagrams, 
grouping the three variables of pressure, temperature and concentration in 
pairs, Fig. 153, giving pressure-temperature, Fig. 154, pressure-concentration 
and Fig. 155, temperature-concentration, from which can be read off with reason- 
able accuracy any quantity needed in calculations and from which Table 
LI, has been prepared. In this table the lower numbers are new, and the 
upper those as given by Starr several years ago and since used by engineers 
engaged in refrigeration, as standards. 

These data refer to the equilibrium conditions of the solution, and in using 
them for practical problems care must be taken to avoid applying them to other 
conditions, for example to solutions that are not homogeneous, or in which 
there has not been sufficient time for the establishment of equilibrium. Over 
the surface of such solutions there exists a mixture of water vapor and 
ammonia, each exerting its own partial pressure and the sum of the partial 
pressures making up the pressure of the solution; it must not be assumed, 
however, that the partial pressures are those of the pure substances, as this is 
a true solution and not a simple mixture. Moreover, there is no certainty that 
it is always a solution of just ammonia in water, as there is a reasonably good 
possibility that hydrates of ammonia may form, which would further complicate 
the relative pressures of the two constituents. 

It is from Perman that the most accurate data on the composition of the 
equilibrium vapor mixture have come, and he calculated the partial pressures 
from the composition of the mixture determined by analysis and using Dalton's 
law, 

Partial pressure of ammonia _ Volume of ammonia 
Total pressure Total vapor volume' 

No success has yet been met with, in attempting to express a general relation 
between partial pressures and the two variables, pressure and temperatures, 
so Perman's values are given as found in Table LII, and for the unex- 
plored region it is not possible to do better than make a guess at a needed value. 
As a check on the Perman values and to assist in estimating needed values for 
other ranges the sum of the two partial pressures is given and compared with 
the accepted values for the total pressure, and it should be noted that as the 
partial pressure relations give the volumetric composition of the vapor mix- 
ture, it is unfortunate that the data are not extended to at least 35 per cent 
concentration to cover the solutions in the generator of the absorption refriger- 
ating machine, from which both ammonia and water-vapor are discharged 
in as yet unknown relative amounts and which must be separated as completely 
as possible before condensation. 

Any change in the ammonia content of a solution is thermal in character 
and is, therefore, accompanied by heat changes. When water absorbs ammonia 
heat is liberated, as is also the case when ammonia in solution is diluted with 
more water, the latter being really a further absorption in the fresh water of 
the ammonia already contained in its solution being diluted by it. Likewise, 



HEAT AND MATTER 499 

when ammonia is absorbed by an ammonia solution heat is also liberated, 
but heat is absorbed by a solution from which ammonia is escaping, as in 
evaporation of liquids. There are three cases of the exothermic process, each 
with an endothermic inverse and these are: 

(a) Absorption of ammonia by pure water. 

(b) Absorption of ammonia by ammonia solution. 

(c) Dilution by water of an ammonia solution. 

Data on the amounts of heat liberated in these cases are not sufficient 
to establish firmly any general law of change, but are sufficient to give an 
approximation. The first important fact in this connection is that the heat 
liberated per pound of ammonia when pure water absorbs ammonia depends on 
the amount of water. One pound of ammonia absorbed in a little water gives 
out a little heat, more is liberated when more water is present, but when the 
amount of water is large, put at 200 times the weight of ammonia by Thomsen, 
the heat of absorption is constant and does not increase beyond this point. 
It may easily be, however, that the point is reached with fifty water weights, or 
that some heat continues to be generated for any amount of water to infinity, 
but so small in quantity as to be impossible to measure in the great weight 
of liquid present. For example, if one B.T.U. were liberated in 100 lbs. 
of solution, the temperature rise would be somewhere near to 1 /ioo° F., and to 
detect this in the presence of radiation and conduction influences and make 
allowance for the heat of stirring would be difficult. 

For engineering purposes, however, it may be accepted that the heat of absorp- 
tion of a pound of ammonia is constant if the weight of water is large, and its value 
was fixed at 926 B.T.U. per lb. by Favre and Silberman and later redetermined 
by Thomsen at 8430 calories per gram molecule of NH3, which is equivalent to 

( - X — — ) =893 B.T.U. per pound. This value is accepted and defined as the 
\5 17 / 

heat of complete absorption for the want of a better term, and in view of the 
desirability of distinguishing it from the heat of absorption in small amounts 
of water or in solutions already containing appreciable amounts of ammonia, 
which will be designated as heat of partial absorption. 

Experiments have further established another important relation between 
the heats of dilution of solutions and their original strength. According to 
this, solutions behave like ammonia itself with respect to pure water and lib- 
erate a little heat when a little water is added, more with larger amounts, 
attaining a constancy for very large amounts of water. Thus a solution of a 
given ammonia strength, it may be assumed, will always liberate the same 
amount of heat when diluted with water, if the total amount of water after 
dilution exceeds two hundred times the weight of ammonia present or there- 
abouts. This heat per pound of NH3 contained will be designated as the 
heat of complete dilution, and defined as the heat liberated when a solution 
containing 1 lb. of ammonia in a given amount of water is completely 
diluted with water, or brought to the condition of 200 lbs. or more of water per 
pound of ammonia contained. 



500 



ENGINEERING THERMODYNAMICS 



There is a rational relation between these three heats, that of complete 
absorption, which is a constant, that of complete dilution, which depends only 
on the original ratio of ammonia to water, and that of partial absorption, 
which is a function of the character of solution receiving it, or if pure water 
the amount. This relation is 

[ Heat of 1 J (Heat of partial absorption) 1 „ m TT „_„'_' 

complete f = ]) , _ + * w ..*. ' =893 B.T.U. per lb.NH 3 . 705) 
absorption J I (+ Heat of complete dilution) j 

Numerical values for heats of partial absorption are entirely lacking, but 
Berthelot has given ten values for the heats of complete dilution for solutions 
containing from 1 lb. of water per pound of ammonia, to a little over one 
hundred, but at only one temperature, 57° F. Up to Thomsen's time these 
figures seem to have been the sole dependence for engineering calculation; 
he, however, added three more figures for more concentrated solutions, giving 
not the heats of complete dilution, but those of partial dilution, that is, the 
heat liberated when the water content is increased from one original value to 
three different final values not corresponding to the state of complete dilu- 
tion. However, these may be used to check and correct the Berthelot figures 
and are especially useful because they cover the doubtful range of his deter- 
mination and the range of ammonia strengths. 

Berthelot's results are given in the following Table XXV and plotted in 
Fig. 156, from which he derives a general law of relation given by the follow- 
ing formula Eq. (706), which is also plotted to show its agreement with the 
points, and which is an equilateral hyperbola asymptotic to axes of H and w. 

Table XXV 



BERTHELOT'S DATA ON HEATS OF COMPLETE DILUTION OF AMMONIA 

SOLUTIONS 



Original Solu- 
tion, 1 Lb. Am- 
monia in (w) 
Lbs. Water. 


When Completely Diluted gives 


Original Solu- 
tion, 1 Lb. Am- 
monia in (w) 
Lbs. Water. 


When Completely Diluted gives 


1.04 
1.06 
1.13 
1.98 
3.18 


136^ 

134 

124 

51 

41 J 


B.T.U. per lb. ammonia 


3.76 

6.11 

10.06 

57.39 

116.47 


34^ 
22 

2 



0- 


■ B.T.U. per lb. ammonia 



142 ^ 
Heat of complete dilution, B.T.U. per lb. NH 3 = -(Berthelot) . . (706) 



The agreement, it will be noted, is not very good for larger values of w than 
4 or 5, which is unfortunate, as commercial ammonia lies between one part 
ammonia to nine parts water, or w = 9, and one part ammonia to 39 



HEAT AND MATTER 



501 



parts water, or w = 39. Nevertheless engineers have been using this formula 
in these doubtful ranges for some years. 

By means of the few but probably accurate figures given by Thomsen and 
experimentally determined by him it is possible to check this practice. He 
measured not the heat of complete dilution as did Berthelot, but the heats of 
partial dilution, and the manner in which his data merge into those of Berthelot 
make the combined results of doubly great value because of the difference in 
method. Thomsen added a limited amount of water to a solution of ammonia 
containing 3.39 lbs. water per pound of ammonia and measured the heat, 
which, of course, was not the heat of complete dilution, with the following 
results; 



THOMSEN'S DATA ON HEATS OF PARTIAL DILUTION OF AMMONIA 

SOLUTIONS 



Original Solution. 
1 Lb. Ammonia in (w) Lbs. Water. 


Final Solution. 
1 Lb. Ammonia in (w) Lbs. Water. 


B.T.U. per Lb. Ammonia Absorbed 
by the Ammonia Solution. 


w= ■ 


f3.39 
3.39 
3.39 


W = ■ 


f 19.27 
29.86 
56.33 


+34 
+37 
+40 



These results have been fitted into those of complete dilution by the relations 
of Eqs. (707) and (708), and by the introduction of one assumption. 



Heat of complete dilu- 
tion of original am- 
monia solution per 
lb. NH 3 



Heat of partial dilution 
from original to some 
greater water content 
per lb. NH 3 . 



I + 



Heat of complete 
dilution of the 
new solution per 
lb. NH 3 . 



(707) 



or 



{Heat of complete dilu- 
tion of new solution 
per lb. NH 3 J 



{Heat of complete dilu-1 
tionof original solution [ 
per lb. NH 3 . J 



{Heat of partial dilution] 
from original to final f (708) 
solution per lb. NH 3 . J 



If the heat of complete dilution of the original solution containing 3.39 lbs. 
water per pound ammonia be taken from the Berthelot equation its value is 
42, so assuming this to be correct it may be introduced in Eq. (708) as a constant, 
giving with the Thomsen figures the following: 



Heat of complete dilution 

of new solution, per lb. r =42 — 
NH 3 . 



Heat of partial dilution 
from original to final 
solution per lb. NH 3 . 



Heat of complete dilution per pound NH3 with 

(w = 19.27) =42-34 (Thomsen) =8; 
(w = 29.86) =42-37 (Thomsen) = 5 ; 
(w = 56.33) =42-40 (Thomsen) = 2. 



502 



ENGINEERING THERMODYNAMICS 



These three new points are also plotted and agree well with the Berthelot 
equation, better even than the original Berthelot points themselves, so that 
Thomsen's partial dilution figures seem to confirm Berthelot's complete dilution 
data and the curve of Fig. 156 and Eq. (706) may be taken as truly represen- 




5 10 15 20 25 

Pounds of Water Per Pound of Ammonia 



30 



Fig. 156. — Heat of Complete Dilution of Ammonia— Water Solutions, by Excess Water. 



tative of the heat of complete dilution of ammonia solutions and indirectly, 
of course, heats of partial dilutions as well. 

Heats of absorption are more often needed in practical problems than heats 
of complete or partial dilution, but these heats follow on the assumption that the 
heat of complete absorption, must be equal to the sum of the heat of partial absorp- 
tion, and the heat of complete dilution of the solution so formed. 



HEAT AND MATTER 



503 



Hence 



| Heat of partial absorption 
in w lbs. water B.T.U. 
per lb. NH :t . 



Heat of complete 
absorption in ex- 
cess water B.T.U. 
per lb. NH,. 



' Heat of complete dilution by 
excess water of solution 
with w lbs. water per lb. 
NH 3 , B.T.U. per lb. NH 3 . 



or 



( Heat of partial absorption in water 
I B.T.U. per lb. NH 3 absorbed 



893 



142.5 



w 



(709) 




30 25 30 

Lbs. Water per lb. NH 3 

Fig. 157. — Ammonia-water Solutions, Relation between Heats 
r Partial Absorption | 
of \ Complete Absorption Y Shown Graphically. 
I Complete Dilution J 

It is interesting to note that the relation between these three quantities, heat 
of partial absorption in limited amounts of water, heat of complete absorption 
in excess water and heat of complete dilution of the solution in excess water, 
can be shown graphically in Fig. 157, where the designations are self- 
explanatory. 

Most important are the heats liberated when ammonia is absorbed not 
by water but by weak ammonia solutions themselves, and these heats of partial 
solution of ammonia in ammonia solutions are obtainable from the data already 
established by a comparatively simple relation. In this case there are two 
different solutions in question, an original one which becomes the second one 
on receiving more ammonia. If the water received all the ammonia contained 
in the second solution a certain quantity of heat would be liberated and it must 
be equal to the total amount liberated by absorption of the first ammonia in 
the water, and by the absorption of the second ammonia in the resulting solu- 



504 



ENGINEERING THERMODYNAMICS 






tion, whence this last quantity is obtainable by differences between the heats 
of partial solution of ammonia in water alone. 



Let w=lbs. water per lb. ammonia in original solution which, therefore, consists of 

w + 1 lbs. solution, 1 lb. of ammonia and w lbs. of water. 

" A =lbs. ammonia added per lb. ammonia already present, making new solution 

w 
containing w lbs. of water and A+l lbs. of ammonia or — — - lbs. water 

A +1 

per lb. ammonia in w+A +1 lbs. of solution. 



Then 



Heat of partial absorption of 

original 1 lb. ammonia in w [ = 893- 
lbs. water, B.T.TJ. 



142.5 



w 



fHeat of partial absorption of] , 
-j all (A +1) lbs. of ammonia I = /893 - 
[ in w lbs. of water, B,T.U. J 



142.4 N 

w 

I+T, 



,(A+i), 



Whence 



Heat of partial 
absorption of 
A lbs. NH 3 by 
solution con- 
taining 1 lb. 
NH 3 in w lbs. 
water,B,T.U, 



= /89'3- 



142 - 5 V +1) Jm- 14 



W 



H 



893 -=-^(A +2) (710) 

w | 



A+l/ 



Therefore 



Heat of partial absorption of A lbs. 
NH 3 by solution containing 1 lb. 
NH 3 in wlbs. water B .T.U. per lb. 
NH 3 



= 893- l -^ {A +2) 



(711) 



As ammonia solution strengths are often given in terms of per cent of 
ammonia present by weight and the heat of absorption in terms of changes in 
the per cent strength, the following conversion factors are useful : 



Per cent ammonia in original solution = Si = 
Per cent ammonia in final solution = $2 = 



100 
w+1 9 . 

100(A + 1) 

w+A + 1 ' 



Whence 



w 



100 -Si 



(a) 



100/ £2 -S i 

il ~s7uoo-&K6) 



(712) 



HEAT AND MATTER 505 

These on substitution in Eq. (711) give the heat of absorption per pound 
of ammonia absorbed to change the per cent NH3 from Si to $2. 



jat of partial absorption of 1 lb. ) /Si $2 \ 

of NH3 in a solution containing /- =893 — 142.51 ^^---^-+77^ — ^r). • (713) 

ilk T*rw„ ; n „,iK m »+^ rttt \100-^i 100-^2/ 



When, however, the absorbed ammonia is to be driven off from the solution by 
heating it, the discharge will consist partly of ammonia vapor and partly water 
vapor, so the heat of liberation of a given amount of ammonia from solution will 
be in practice that for the ammonia and equal to what would be liberated by its 
absorption, but also in addition the heat of vaporization of the water vapor. 
When, as in absorption refrigerating machine generators, the discharged 
vapors meet incoming solution and are thereby partly condensed, prac- 
tically all except perhaps 2 per cent of the heat of vaporization of water vapor is 
returned and the net heat of ammonia liberation is not materially different 
from the value for absorption. If this is not done a correction must be intro- 
duced for the water vapor. 

To assist in the solution of problems on the amount of non-condensible 
gases to be handled by dry vacuum pumps serving steam condensers, and on 
the change in the composition of gases when scrubbed by water in the course 
of cooling and cleaning after manufacture, a table of solubilities of various gases 
in water is added at the end of this Chapter in Tables LIII and LIV, as compiled 
from various sources and reported in the Smithsonian physical tables. The 
numbers in the tables are volumes of standard gas, that is, gas measured at 
32° F. and 1 atm. pressure, per unit volume of water, though they are at a 
different volume as absorbed or when absorbed at the temperature given. 
The pressure of the solution is in every case 29.92 ins. Hg., absolute and this 
is the combined pressure of both the gas and the water vapor. 

Example. In the absorber of an ice machine of the absorption type, a weak solu- 
tion of ammonia in water takes up the ammonia vapor coming from the refrigerating 
coils, the heat found being removed by water. The weak liquor, as it is called, is con- 
tinuously supplied and the rich liquor continually pumped away to the generator 
where, by heating, some of the ammonia vapor is driven off to the condenser. As- 
suming the action in the absorber to be merely one of ammonia dissolving in a weak 
solution and that no water vapor leaves the generator, what will be the heat produced 
in the absorber and needed in the generator per pound of ammonia for the fol- 
lowing assumed conditions: 

Weak solution, 15 per cent NH3; strong solution, 30 per cent NH3; 
temperature in absorber, 80° F. 

From Eq. (713), the heat of absorption per lb. of ammonia absorbed will be 



506 ENGINEERING THERMODYNAMICS 

where S\ and £2 are the per cents of ammonia in weak and rich solutions? respec- 
tively, or 

Q =893-142.51^+-^) =807. B.T.U. per lb. NH 3 absorbed 
\85 70/ 

Prob. 1. Ammonia is being absorbed by water at a temperature of 100° F. The 
solution contains 10 per cent of ammonia. If the total pressure is 15 ins. of Hg, what 
part of this is due to foreign gases, what part to ammonia and what part to water 
vapor? 

|Prob. 2. How many cubic feet and how many pounds of the following gases can 
be separately dissolved in 1000 gallons of pure water at atmospheric pressure and a 
temperature of 50° F.? Air, carbon dioxide, and hydrogen. How would the results 
be changed if the pressure were doubled? If the temperature were doubled? 

Prob. 3. When either water or ammonia is added to an ammonia solution, heat 
is evolved. Explain why this is so. 

Prob. 4. Ammonia is being absorbed by a stream of running water, there being 
5 lbs. of water per pound of ammonia. What will be the heat developed per pound 
of ammonia liquor formed? 

Prob. 5. Ten pounds of the above solution receives an additional pound of am- 
monia. How much heat is generated by this action? 

Prob. 6. A solution containing 10 per cent of ammonia receives an addition of 
another 10 per cent. What was the amount of heat developed per pound of ammonia 
and per pound of solution when the second portion of the ammonia was added? 

Prob. 7. The pressure in the absorber of an ammonia absorption machine is one at- 
mosphere and the temperature is 80° F. What is the maximum per cent of am- 
monia which can be absorbed by the water? 

Prob. 8. The generator is working under a pressure of 125 lbs. per square inch 
gage and the heat is supplied by a steam coil in which the pressure is 30 lbs. per square 
inch gage. What per cent of ammonia will be left in solution after passing through 
the generator and about how much steam must condense per lb. NH3 discharged? 

13. Combustion and Related Reactions. Relative Weights and Volumes 
of Substances and Elements, before and after Reaction. Not only may 
matter assume the three states of solid, liquid and vapor separately, in pairs 
simultaneously, or even all three together with various accompanying or causal 
temperature, pressure, or heat content, conditions, without chemical change 
of the matter itself, as already discussed, but matter itself may change in kind. 
As Mellor puts it — " matter appears to be endowed with properties in virtue 
of which two or more dissimilar substances, when brought into close contact, 
give rise to other forms of matter possessing properties quite distinct from the 
original substance." " The process of change is called a chemical reaction." 
Chemical changes are assumed to be characterized by molecular rearrangement 
according to which molecules of elements may divide into atoms and the 
separated atoms of one combine with those of another element, to form a molecule 
of a new substance to be called a compound. Similarly, the molecules of com- 
pounds may split and reassociate, part of one, with part of another, to make 
a new compound or a single compound may split up into its elements which 



HEAT AND MATTER 507 

remain separated, the last case being generally termed dissociation. All three 
classes of change of substance are classifiable as chemical reactions, and there is 
really no very rigid line to be drawn between the sub-classes of reaction known 
as combination and dissociation except when applied to the same substances, 
in which case one is the reverse of the other. The complete or partial destruc- 
tion of a substance as such is commonly termed decomposition as, for example, 
when hydrocarbon constituents of coal volatile, or liquid fuel, are changed by 
excessively high temperature into free carbon, and other hydrocarbons or 
even free hydrogen. 

Every chemical change whether of combination or dissociation is accompanied 
by a heat change of the system or group of substances. When the reaction 
is such that heat is set free tending to raise the temperature of the reacting 
mass unless it be carried away as liberated, the reaction is classed as exothermic. 
On the contrary, those reactions that are accompanied by heat absorption, 
tending to lower the temperature, unless heat be added to supply the absorp- 
tion, are classed as endothermic. 

It appears then that every reaction tends to change the temperature of the 
system, causing it to rise or fall according as the reaction is exothermic or 
endothermic, except in the one case where several reactions occur simulta- 
neously, in which all the exothermic, set free just enough heat to supply what 
is required for the endothermic. 

The most important reaction in engineering is combustion, defined as the 
chemical reaction between fuels and the oxygen of the air, which is exothermic 
or heat liberating, and the source of practically all the heat used in engines for 
conversion into work. Combustion is often classed as an oxidation process, and is 
thus distinguished from another important engineering class of related reactions, 
reduction or the reverse of oxidation, the most prominent case of which is 
the change from carbon dioxide to carbon monoxide in gas producers, and in 
a lesser degree in furnaces, according to which there is a reduction of oxygen 
content per molecule, and which process is endothermic. The formation of 
carbon monoxide directly from carbon by its oxidation, is sometimes defined 
as partial combustion of carbon or incomplete oxidation because the product, 
carbon monoxide, may further oxidize or burn to carbon dioxide by taking 
up more oxygen. The substances produced by the partial combustion or partial 
oxidation of one fuel element may also be considered as the result of the dis- 
sociation or reduction or deoxidation of the substances produced by its complete 
combustion. Ordinarily, partial combustion and reduction are considered 
as reverse processes producing the same substances by (a) exothermic reaction 
of the primary substance in partial combustion and (b) the endothermic deoxi- 
dizing reaction of the products of complete exothermic reactions of the same 
primary substances. It is also common to consider only the reaction of a sub- 
stance or so-called fuel element with oxygen, as combustion and other processes 
whether involving oxygen or not, as related reactions. Thus, carbon combining 
with oxygen to form carbonic acid is complete, and carbon combining with 
oxygen to form carbon monoxide incomplete combustion of carbon, while carbon 



508 ENGINEERING THERMODYNAMICS 

monoxide and steam reacting can hardly be considered as combustion, and is 
best classified as a related reaction. 

The number of elements entering into combustion and related reactions 
is small, but the number of possible substances and reactions between them is 
amazingly large, and the prediction of just what reactions will take place between 
various groupings or mixtures of these substances extremely difficult and in some 
cases quite impossible. The study of possible reactions has become the province 
of physical chemistry, especially when the conditions controlling the result 
are also subjects of study. These conditions include the temperature, pressure, 
electrical state and the mutual relation of the elementary compounds present, and 
the relation of these various conditions to the primary and resultant substances, 
and the intermediate, successive, simultaneous or parallel reactions constitute the 
subject matter of the study of chemical equilibrium. If chemical equilibrium 
were better understood than it is, it would be possible to predict the resultant 
from primary substances for specified conditions, but at the present time this 
is impossible even though some of the greatest scientists the world has ever 
known, have devoted their lives to the study and many volumes of specific 
results have been published. Even if the exact prediction of the direction in 
which reaction would proceed in an ordinary complex system and the extent 
to which it would go in that direction, were made possible by a more complete 
thermochemistry than now exists, it would not be of much use in engineering 
because it is seldom possible to define the conditions that are present or to be met. 
For example, in gas producers, solid coal, air and steam are the primary materials 
and the product or result of their mutual reaction is a combustible gas. Engi- 
neers would like very much to be able to predict and control the exact com- 
position of this gas, but this is not possible nor will it in all probability ever be 
possible, because it would first be necessary to fix the chemical and thermal 
character of the coal, air and steam, to fix their relative quantities, to maintain 
an absolutely uniform fuel bed as to size, porosity and, quality with some 
other conditions equally elusive. 

It must be understood, therefore, that while the possibility or even probability 
of certain reactions taking place may be known, it is quite impossible to 
predict just what will happen, or what products will result, when a given 
group of primary substances mutually, react so that many important problems 
of combustion in boiler furnaces and gas producers cannot be solved except by 
approximation. 

The approximation takes the form of a calculation which is exact, based on 
an hypothesis which does not represent the facts of the case. In other words, 
engineering calculations about combustion are always to be prefaced by a 
statement that certain substances are going to change completely or within 
a given degree to certain others, whether they will or not. Furthermore, the 
substances must be defined chemically by their symbols, and then will it be 
possible to calculate the relative weights and volumes of the various substances 
that can so react, the corresponding relative weights and volumes of the products, 
and the heat liberated or absorbed, but not otherwise. It is quite important 



HEAT AND MATTER 509 

that too much confidence be not put in these results, which are no more correct 
than is the assumption of what substances are to be formed. 

From the study of chemical equilibrium a few principles of guidance have 
been developed that help a little but not very much. For example, Van't 
Hoff's " law of movable equilibrium " says that when the temperature of a system 
is raised that reaction takes place which is accompanied by absorption of heat 
and conversely. Another similar law is to the effect that a rise of pressure in 
a system in equilibrium causes that reaction that is accompanied by a reduction 
of volume. There are more of this sort but they are entirely too general to make 
it possible to avoid the procedure adopted by engineers of assuming the kind 
of reaction, and then calculating quantities and heats that should and would 
accompany it, if it did take place and, however, crude this may look it is very 
useful, and in most cases good enough or rather as good as the knowledge of other 
conditions to be met. 

Carbon and hydrogen are the only chemical elements of fuel value, and all 
commercial fuels, including wood, peat, lignite, bituminous and anthracite 
coals, charcoal, coke, crude petroleum oils with their distillates, gasolene, 
kerosene, and their residues, tars, heavy oils, alcohols, benzole a bituminous 
coal product, natural gas as well as blast furnace, carburetted and uncar- 
buretted water gas, coal gas, producer gas, and oil gas, are compounds, 
mixtures and mixtures of compounds of these fuel elements, with oxygen 
in some cases. The one exception is sulphur, which exists in many solid, 
liquid and gaseous fuels as an impurity, but which also has some small 
fuel value. 

This being the case, the number of products to be formed by the complete 
combustion of fuel is also small, and includes only carbon dioxide and water 
vapor, with the nitrogen carried by the air, and a small amount of sulphur 
compounds, often ignored. 

The process of reaction, whether combustion or one of the related ones, 
is to be described by the usual chemical equation which has the additional 
significance of showing the relative weights involved directly, because, 

(a) The total number of atoms of each chemical element on each side of 
the equation must be the same. 

(6) The sum of the products of the atomic weight of each atom and the 
number present, must be the same on each side of the equation. 

This is the same as saying, (a) that the total weight of hydrogen in the prod- 
ucts must be the same as the total weight of hydrogen in the original mixture 
and so also for the other elements, and (6) the total weight of the original mixture 
of reacting substances must be the same as that of the products. 

Natural fuels while sometimes consisting of simple elements like carbon 
or hydrogen alone, or simple compounds like carbon monoxide or methane alone, 
are more often mixtures. Their reaction equations are then to be derived 
from combinations of the equations representing the reactions of elements and 
of compounds with oxygen or with each other. To facilitate this the following 
table of some characteristic reactions of simple substances is inserted: 



510 



ENGINEEEING THEKMODYNAMICS 



SOME COMBUSTION AND RELATED REACTIONS OF FUEL ELEMENTS AND 

COMPOUNDS 



Combustion of fuelel ements 



Product of 
partial car- 
bon combus- 
tion 



c+o 2 =co 2 . 

2H 2 + 2 =2H 2 



Combustion 

of 

Chemical 

Compounds 



2CO+0 2 =2C0 2 



Paraffin series CnH 2w + 2 : 

Methane CH 4 +20 2 = C0 2 +2H 2 . 

7 
Ethane C 2 H 6 + ~0 2 = 2C0 2 +3H 2 

/3n+l 



Hydrocarbons ' 



(1) 
(2) 



(3) 



(4) 

(5) 



General C»H*i+ 2 + (— — )o 2 = nC0 2 + (n+l)H 2 (6) 



Alcohols 



Olefines or Ethylene series C w H 2re : 

Ethylene C 2 H430 2 = 2C0 2 +2H 2 

9 
Propylene C 3 H 6 +~0 2 = 3C0 2 +3H 2 

A 

3n 
General C w H 2ra + — O, = rcC0 2 +nH 2 

2i 

f Monatomic alcohols C W H 2W +iOH : 

Methyl alcohol CH 3 OH +-0 2 = C0 2 +2H 2 . 

L 



(7) 
(8) 

(9) 

(10) 
(11) 



Ethyl alcohol C 2 H 5 OH+30 2 = 2C0 2 +3H 2 0. 
General C w H 2w +iOH+— 2 =nC0 2 +(n+l)H 2 (12) 



Related Reactions , Incomplete Combustion, Decom- . 
position, Dissociation. 



2C+0 2 

C0 2 +H 2 = 

CH4+CO+H 2; 

C+C0 2 = 

CO+H 2 = 

2C0 2 = 

2H 2 = 

CH 4 : 

C 2 H 4 = 

CeHu = 

2CH 4 = 
3C 2 H 2 : 



= 2CO . . . 
= CO+H 2 . 
= C 2 H 4 +H 2 0. 

2CO . . . 

C0 2 +H 2 . . 
:2CO+0 2 . 
= 2H 2 +0 2 . 
= C+2H 2 . 
= C+CH 4 . 
: C 2 H 4 +C 4 Hio 
: C 2 H 2 -(-3H 2 
= C 2 H 4 -J-2C 2 H 2 



(13) 

(14) 

(15) 

.(16) 

(17) 

•(18) 

.(19) 

.(20) 

.(21) 

.(22) 

.(23) 

• (24) 



All these and any other reactions are characterized by definite weight 
relations which are given directly by the reaction equation by introducing the 
weight of each element from a table of atomic weights, and for this purpose 
the nearest whole number is close enough. For example, the complete com- 
bustion of hexane, CeHi4, the main constituent of gasolene, to carbon dioxide 
and water, is given by, 



or 



2C 6 Hi4+1902 = 12CQ2+14H 2 0. 



C 6 H4+f02 = 6C02+7H 2 



HEAT AND MATTER 



511 



This may be interpreted as follows, taking the atomic weight of C = 12, of 
H = l, and of = 16. 

(6X12 = 72) lbs. carbon 
+ (6X2X16 = 192) lbs. oxygen 
+ (7X2X1 = 14) lbs. hydrogen 
+ (7X16 = 112) lbs. oxygen 



(6X12 = 72) lbs. carbon 
+ (14X1 = 14) lbs. hydrogen 
+ (9|X2X16 = 304) lbs. oxygen 



or 



(72+14 = 86) lbs. hexane 
+ (304) lbs. oxygen' 



malr* / (72 + 192 = 264 lbs. carbon dioxide 
maKe ^ + ( 14 + 112 = i26) lbs. water vapor 



This general weight relation may be simplified somewhat by considering 
the weight of each one of the substances as unity, thus yielding the following 
different, but equivalent interpretations: 

1 lb. hexane+— lbs. oxygen J make ( — lbs. car. diox.+— lbs. water.) 
1 lb. oxy.+^j lbs. hexane j make (^^ lbs. car. diox.+^^T^ lbs. water. 



304 



V304 



304 



v result from com- 

1 lb. carb. diox.+^ lbs.water ) plete combus-/^ i Ds . hex.+^| lbs. oxy. 

/ tionof \264 264 J 



264 



result from com- 



(l lb. water +^| lbs. car. diox.) plete combus-/^ lbs. hex.+~ lbs. oxy 
\ 126 / tionof \126 126 

It might also be said from the same data that, 

1 lb. hexane completely burned with oxygen yields 390 lbs. products. 

1 lb. oxygen will completely burn 304 lbs. hexane and yield 390 lbs. products. 

This particular example can be analyzed in still another way, yielding a general 
expression for the combustion of an analyzed fuel. 

Fuel analyses are reported in two ways : (a) Proximate, giving per cent of 
each independent compound, or separately determined constituent substance; 
(b) ultimate, giving the per cent of each chemical element. Applying this dis- 
tinction to the original mixture of hexane and oxygen, and its products the 
proximate analyses are: 



Original mixture - 



Hexane 

Oxygen 
Total 



86. 
390 

304 
390 



= 22.1% by weight. 
= 77.9 % by weight. 



100.00% 



512 



ENGINEERING THERMODYNAMICS 



Products 



Carbon dioxide 



Water vapor or liquid = 
Total 



264 
390 

126 
390 



= 67.6 % by weight. 
= 23.4 % by weight. 



100.00% 



Using the ultimate analysis method of designation for mixtures and products 



Original mixture ■ 



Products 



Carbon (in hexane) 

Hydro, (in hexane) 

Oxygen (free) 
Total 

Carbon (in CO2) 

Hydrogen (water) 

Oxygen (in CO2) 

Oxygen (water) 

Total 



72_ 
390 

ii 
390 

304 
390 



390 

II 
390 

192 
390 



= 18.4% by weight. 
= 3.6 % by weight. 
= 78.0 % by weight. 



100.00% 
= 18.4 % by weight. 
= 3.6 % by weight. 



304 



112i=390 =78 -° %byWeight - 
390 J 



100.00% 



Neither combustion or its related reactions take place with oxygen alone, 

but with air containing oxygen, 23.2 per cent, and nitrogen, 76.8 per cent 

76 8 
by weight, each pound of oxygen carrying with it —- = 3.31 lbs. of nitrogen 

Zo.Z 

or existing in 4.31 lbs. of air. The nitrogen is generally considered neutral, 

though it may form compounds with hydrogen, such as ammonia directly, 

or with oxygen, such as nitrous oxide if conditions are right. If neutral, it 

has the effect of changing the weight of the mixture by an amount depending 

on the proportion of oxygen that came from air. 

f (Weight of mixture with oxygen alone) 1 _ , . h+ , . . .., {) 

I +3.31 (wt. oxygen present) / " ^eignt ot mixture witn air;, 

or 

f (weight of nitrogen to be added when 1 _ o 01 /_* of OYV ™ nrP . e nti 
I air is used instead of oxygen) J 6 ' 61 (wt ' ot oxygen P resent ^ 



HEAT AND MATTER 



513 



This will add in the present case 304X3.31 = 1006 lbs. of nitrogen in the 
combustion of one molecule or 86 lbs. of hexane, the sum of 304 lbs. of oxygen 
and 1006 lbs. of nitrogen, giving 1310 lbs., the weight of air required, and rais- 
ing the total weight of the mixture to 1396 lbs. Of course, the per cent of the 
various constituents of the mixture and products is now changed, but the 
amount of change is quickly computed. All these relative numbers can be 
conveniently given in tables of conversion factors, such a table for hexane 
being given below, as a type form, useful in every-day work when it is nec- 
essary to make repeated calculations with some one fuel. 

CONVERSION FACTORS FOR WEIGHTS IN THE COMPLETE COMBUSTION OF 
HEXANE WITH AIR TO WATER AND CARBON DIOXIDE 



Original 


Final 


Hexane i 


Oxygen 


Nitrogen 




Carbon Di- 


Water 


Mixture. 


Mixture. 


CeHu. 


o 2 . 


N, 


Air. 


oxide CO2. 


H2O. 


1396. 


1396. 


86. 


304. 


1006. 


1310. 


264. 


126. 


1. 


1. 


• .0616 


.218 


.720 


.938 


.189 


.091 


16.22 


16.22 


1. 


3.54 


11.68 


15.22 


3.07 


1.46 


4.59 


4.59 


.283 


1. 


3.31 


4.31 


.87 


.41 


1.387 


1.387 


.085 


.302 


1. 


1.302 


.262 


.125 


1.066 


1.066 


.066 


.232 


.768 


1. 


.202 


.096 


5.29 


5.29 


.326 


1.15 


3.81 


4.96 


1. 


.48 


11.08 


11.08 


.683 


2.42 


7.98 


10.40 


2.10 


1. 



The average analysis of pure air is given by Parkes as being 02 = 20.96 per cent; 
CO2 = .04 per cent; N2 = 76.00 per cent by volume, giving 02 = 23.20 per cent; 
CO2 = .06 per cent; N2 = 76.74 per cent by weight. Regarding the CO2 as 
being negligible, the relation may be used for the purpose of computations of 
this sort, 02 = 23.2 per cent and N2 = 76.8 per cent. On the assumption that 
any nitrogen present came from air and not from any other compound, such 
as ammonia, it must have been associated with oxygen in the proportion 3.31 
to 1 of O2. The weight of oxygen on both sides must be equal and the weight 
of air on one side must be 4.31 times the weight of nitrogen. Of the nitrogen 

that is present - — X(the weight of nitrogen present) must have come from air 
o.ol 

and the rest from the other substances. 

A much-used weight relation is, for the weight of air and products per pound 

of fuel, and on the assumption that 

H = part of hydrogen by weight per pound fuel, 
C=part of carbon by weight per pound fuel, 



(Weightofairperlb.offuel)=CXY2X4.31+HX8X4.31 = 11.49C+34.46H,(71 



514 ENGINEERING THERMODYNAMICS 

Weight of products per lb. fuel = 3|C lbs. C0 2 +9H lbs. H 2 0+^| 

X3.31C lbs. N+8X3.31H lbs. N 
= 3.67C in form of C0 2 +9H in form of H 2 0+8.82 C in form of N 
= 12.49C+35.46H, . . . (715) 

for complete combustion in air with no more air supplied than enters into the 
reaction 

Most of the practical problems concerning the relative amounts of substances 
involved in combustion and reduction processes are concerned with gases, at 
least on one side of the equation and sometimes on both. Direct combustion 
and the gasification of fuels in producers and complete combustion in boiler 
furnaces always yields gas mixtures, the composition of which is always volu- 
metrically determined by analysis, while the explosive mixtures or primary 
working substances of gas engines are gaseous initially and remain so after 
combustion. It is quite necessary, therefore, to transform the weight relations 
of the reaction equation into another form yielding volumes. There are three 
ways of doing this, all equivalent and all yielding the same result if the constants 
used are consistent, and if the gases and vapors follow the Avagadro law, on 
which the most useful of the methods depends. 

1. The volume at standard conditions of any substance can be found from 
the weight present by multiplying that weight by the specific volume of the 
substances ; in cubic feet per pound, at standard condition of 1 atm. at 32° F. 

2. The molecular weight in pounds of any gaseous or vapor substance 
assumed to follow Avagadro's law occupies 358 cu.ft. 

3. The volumetric relations are given directly by the coefficients of the sub- 
stance in the chemical equation when the substances are gaseous and all enter 
into the reaction, and when each is expressed in terms of molecules present. 

The first method needs no explanation or further development and the second 
and third are really one. If in the reaction equations there be written for 
any substance its molecular weightX358, the product will be the volume of 
that substance in the reaction in cubic feet at 32° F. and standard atmospheric 
pressure of 29.92 ins. Hg, if the substance could so exist and even if not, the 
relation between the volumes will hold at any other simultaneous pressure and 
temperature for all. When the substances all are gaseous the coefficient of the 
symbol, representing the number of molecules, indicates the relative volumes 
at once. . Thus, for complete combustion of carbon with oxygen alone 

c+o 2 =co 2 

or (one molecule of carbon) + (one molecule) of oxygen makes (one molecule of 
C0 2 ) or l2 lbs. carbon +368 cu.ft. oxygen makes 358 cu.ft. of C0 2 . One pound 
of carbon requires 29.8 cu.ft. of oxygen and will make 29.8 cu.ft. of C0 2 at 32° 
F. and 29.92 ins. Hg. 
Also for hydrogen 

2H 2 +0 2 = 2H 2 0, 



HEAT AND MATTER 



515 



or (two molecules of hydrogen) -f (one molecule of oxygen) makes (two molecules 
of water); or 2 cu.ft. of hydrogen+1 cu.ft. oxygen makes 2 cu.ft. gaseous water; 
or 4 lbs. hydrogen requires 358 cu.ft. of oxygen, hence 1 lb. of hydrogen re- 
quires 89.5 cu.ft. of oxygen. 
For hexane there was written 

C 6 Hi4+9i02 = 7H 2 0+6C02, 

or 1 cu.ft. of hexane requires 9§ cu.ft. of oxygen and makes 7 cu.ft. of gaseous 
water and 6 cu.ft. of carbon dioxide. It thus appears that the volumetric 
analysis of gaseous mixtures or products is given simply by the ratio of the num- 
ber of molecules of each substance to the total number of molecules present. 
When the oxygen is to be derived from the air each cubic foot of oxygen 

79.04 
will carry with it ' =3.771 cu.ft. of nitrogen which is to be added, or is to 
20.96 

be found in 4.771 cu.ft. of air. 

The air requirements for various fuels chemically are given in Table 

XXVI below in pounds and cubic feet of air per pound of fuel and per cubic 

foot of fuel (standard), for the exact combining proportions: 

Table XXVI 

AIR REQUIRED FOR COMBUSTION FOR VARIOUS SUBSTANCES 

(Combustion complete in every case except for C burning to CO) 



Substance 



Carbon, C to CO : 

Carbon, C to CO 

Hydrogen, H2 

Carbon monoxide, CO 

Sulphur, S 

Methane, CH 4 . . . . 

Ethane, C 2 H 6 . . . 

Ethylene, C 2 H 4 . . . 

Acetylene, C 2 H 2 . . . 

Propane, C 3 H 8 . . . 

Propylene, C 3 H 6 . . . 

Allylene, C 3 H 4 . . . 

Butane, C4H10 . . . 

Butylene, C 4 H 8 . . . 

Pentylene, C 5 Hi . . . 

Hexane, CeHu. . . 

Benzole, C 6 H 6 . . . 

Heptane, C 7 Hi 6 . . . 
Methyl alcohol, CH 3 OH. 
Ethyl alcohol, C 2 H 5 OH 



1 Lb. of Substance 
Requires Air 



Lbs. 



11.55 

5.77 

34.64 

2.47 

4.32 

17.32 

16.16 

14.85 

13.32 

15.75 

14.85 

13.86 

15.53 

14.85 

14.85 

3.53 

13.32 

15.24 

6.49 

9.04 



Cu.Ft. 
Standard 



143.10 

71.55 

429.19 

30.6 

53.52 

214.59 

200.22 

183 . 99 

165.07 

195.14 

183.99 

172.73 

192.42 

183.99 

183.99 

43.79 

165.07 

188.85 

80.47 

111.96 



1 Cu. Ft. of Substance 

(Standard) 

Requires Air 



Lbs. 



.193 
.193 

.774 
1.354 
1.157 

.964 
1.929 
1.736 



.543 
.508 
315 
890 
66 



243 
.58 
1.17 



Cu.Ft. 
Standard. 



2.39 
2.39 

9.59 
16.73 
14.34 
11.95 
23.90 
21.51 
19.12 
31.07 
28.68 
35.85 
45.45 
35.84 
52.58 

7.17 
14.34 



516 ENGINEERING THERMODYNAMICS 

Prob. 1. Complete the following equations for combustion with oxygen and with 
air: 

CH 4 +2H 2 + -0 2 = 

C6H6-I 2 = 

C+3H 2 +4S+-0 2 = 

Prob. 2. What will be the weight of the air needed in each of the cases of Prob. 1 
and what will be the weight of the products, the volumetric composition? 

Prob. 3. When 1 lb. of propylene is burned how much air is needed and what 
is formed? What is the proportion of each product by weight? What would be the 
approximate volume and volumetric composition of the products at a temperature of 
60° F.? 

Prob. 4. After combustion the products of a combustible mixture were found 
to be 17 cu.ft. of nitrogen, 3 cu.ft of carbon dioxide and water, equivalent to 6 cu.ft. 
of water vapor. What was the composition prior to combustion and what was the 
volume at 60° F.? 

Prob. 5. 5 lbs. of carbon dioxide, 4 lbs. of oxygen, and 2 lbs. of water would result 
from the combustion of how much carbon and hydrogen in what amount of oxygen, 
and what would be the complete weight of the products if the combustion had taken 
place in air? 

Prob. 6. When steam is passed through a bed of hot coke the coke takes 
the oxygen from the water to form CO and the resulting gas is called water gas. For" 
10 lbs. of carbon how much gas by weight could be made and how much steam would 
be required? 

Prob. 7. A sample of producer gas gave upon analysis the following per cubic foot. 

H 2 =.18 cu.ft. 

CO =.25 cu.ft. 

CH 4 = .03 cu.ft. 

C0 2 = .07 cu.ft. 

N 2 = .47 cu.ft. , • 

What will be the cubic feet of air needed to burn a cubic foot and what will be the 
composition of the products? 

14. Heats of Reaction. Calorific Power of Combustible Elements and 
Simple Chemical Compounds. B.T.U. Per Pound and Per Cubic Foot. Com- 
bustion of the so-called fuels is always exothermic and the heat set free 
per pound of fuel is its calorific power. This heat of reaction may also be 
expressed per unit weight or per unit volume of any other element or substance 
entering into the reaction, but all are derivable from the calorific power by the 
weight and volume relations. Each is more directly useful in some particular 
calculation, for example, the B.T.U. per cubic foot of combustible gaseous 
mixture is an important factor in the mean effective pressure of gas engines, 
and to the B.T.U. per pound of products is the temperature rise during com- 
bustion proportional. To facilitate boiler calculations and reduce the large 
number of B.T.U. per pound of fuel, its equivalent evaporative power from 
water at 212° F. to dry saturated steam at the same temperature is often given 
and is the B.T.U. per pound divided by 970.4. This value is described as 



HEAT AND MATTEE 517 

equivalent evaporative power, the number of evaporation units, or as evap- 
orative power from and at 212° F. 

All calorific powers can be determined exactly only, by direct experimental 
observations of burning the fuel in water cooled calorimeters, and the heat of 
combustion is directly given by the pound-degrees of the water with a correc- 
tion for the metal and other parts simultaneously heated, called the water 
equivalent of the instrument, and also for radiation during test. It is assumed 
in calorimeter tests that all the substances are brought to their original tem- 
perature after combustion, if not, another correction must be made for the 
residual heat remaining in the hot products, ash or instrument parts, and which 
has not been imparted to the calorimeter. 

It should be noted that whenever hydrogen burns to superheated water 
vapor in a calorimeter, the products in cooling will somewhere reach the satura- 
tion temperature for the pressure existing. When this point is reached water 
vapor will condense and the calorimeter water will receive the latent heat 
of condensation of the water produced by combustion. If, for example, the 
combustion proceeded at atmospheric pressure as in all gas calorimeters, this 
point would be 212° F. and as the condensation proceeds at this temperature 
there may be two calorific powers reported for products of combustion cooled 
to 212° F., the first for products condensed and the second for the products 
not condensed. The former will be greater than the latter by 970.4 B.T.U. 
per pound of water condensed and not per pound formed because all will not 
condense at any temperature. Of course, this same difference exists when 
the products are cooled below 212°, but there is in this case another difference 
inasmuch as the substances giving up heat to the calorimeter water are not 
all gaseous, but a liquid, some fixed gases and some water vapor, a little of which 
continues to condense for each degree drop, and this is in contrast to the con- 
ditions above 212®, where nothing but gases and superheated vapor, which 
behaves the same as gases, are being cooled. 

It is thus apparent that not only may there be two calorific powers for fuels 
containing hydrogen, one high and the other low, but that the exact measure 
of the difference between them per pound of hydrogen burned or per pound of 
water formed is a question of how far or to what temperature the cooling be 
carried. In view of this possibility of _ more than one interpretation, it is 
sufficiently close for engineering work of the usual routine to accept as the dif- 
ference, 970.4 X weight of water vapor formed and this has the advantage of 
being definite and possible to calculate, even though it may not be exactly 
in accordance with actual conditions. Since 1 lb. of hydrogen makes 9 lbs. 
of water in combustion, the weight of water is nine times that of the hydro- 
gen, therefore, it should be accepted that 

[.Difference between high] rV7n/l v. ft vw > u , e u -, , - ,s 

\ , , , . c \ =970.4 X9X (weight of hydrogen per pound fuel). 

[ and low calorific power J 

= 8734x( (weight0f ^ r °rl • • • • (716) 
I per pound fuel) J 7 



518 ENGINEERING THERMODYNAMICS 



Engineers have been forced to take this difference into consideration more 
particularly since the greatly increased use of hydrocarbon gaseous and liquid 
fuels and the alcohols, the former for burning both under boilers and in internal 
combustion engines and the latter in gas engines alone. In both these classes 
of apparatus the products of combustion leave quite hot, always above 212° 
F., so that it may be said with some propriety that they did not receive the 
full heat of combustion as reported by the calorimeter for products reduced 
to initial temperature. On the other hand, it may with equal propriety be 
contended that they have received even less than the low value because the 
products are discharged hotter than 212° F., but that they should receive no 
credit for the difference which is to be charged against them because they are 
unable to utilize low temperature heat. In this book the most commonly 
accepted practice will be adopted and this is to charge against boilers and all 
open or atmospheric pressure fires or furnaces the high calorific power, and 
against explosive combustion or that taking place in closed chambers at 
pressures considerably in excess of atmosphere, the low value as previously 
defined. To do otherwise is to enter a controversy already ten years old and 
still without end. 

Unless otherwise defined calorific power shall be understood to mean the 
heat liberated by combustion obtained by burning 1 lb. of fuel at constant 
pressure in free oxygen, both fuel and oxygen being at equal temperature 
before, and products reduced to that same temperature after combustion. 
This is the high value, but not so named unless for special reasons. Of course, 
constant volume combustion would yield a different value which some authori- 
ties insist on applying to gas engine explosions, but this is in the nature of 
hair-splitting in engineering, where conditions of service are always somewhat 
undefined and practical factors of performance introduced as an allowance* 

Since the mean specific heat of the original fuel and oxygen may not be the 
same as that of the products, it will make some difference what the base tem- 
perature of the experiment is. This effect will be greatest in the case of hydro- 
gen burning to water liquid or vapor, and the fact is illustrated best by a specific 
computation. Julius Thomsen gives the experimental heat of combustion 
of hydrogen as being 61,200 B.T.U. per pound hydrogen, when the original 
hydrogen and oxygen are at a temperature of 18° C. or 64° F., and the result- 
ing water brought to the same temperature. Since 9 lbs. of water are formed 
per pound hydrogen the heat obtained from the calorimeter, in case the prod- 
uct, water, were finally at 212° would be less than the above by the dif- 
ference between heat of the liquid at 212° and at 64° for the 9 lbs. Then per 
pound of H2 (H2 and O2 at 64° F. burning to liquid water at 64° F.) gives 
61,200 B.T.U. ; (H 2 and 2 at 64° F. burning to liquid water at 212° F.) gives 
61,200-148X9 = 59,868 B.T.U. If, however, the original substances H 2 and 
O2 are at 212° instead of 64°, more heat would be removed from the calo- 
rimeter, by the amount of heat necessary to raise the temperature of 1 lb. H 2 
and 8 lbs. 2 from 64° to 212° F. Or (H 2 and 2 at 212° F. burning to 
liquid water at 212° F.) gives 59,868+148(1 X3.4+8X. 217) =60,626 B.T.U. 






HEAT AND MATTER 519 

per lb. H2. If the final products of combustion be vapor of water at 212° 
instead of liquid, the (H2 and O2 at 212° F. burning to water vapor at 212° 
F.) gives 60,626-9X970.4 = 51,892 B.T.U. 

The temperature 212° F. has no significance when the pressure is other than 
atmospheric, and in the products of combustion there are nearly always 
other substances than water vapor present, so that very rarely is the partial 
pressure of the resulting vapor equal to standard atmosphere. Nevertheless, 
to conform to the practice of regarding the difference between high and low 
value as being equal to the product of the weight of water vapor formed by the 
latent heat at 212°, and at the same time to be consistent with scientific informa- 
tion available, it will be regarded that the expression " high value " for heat 
of combustion of hydrogen refers to the figure 60,626 B.T.U. per pound hydro- 
gen, as derived above, while the term " low value " refers to the figure 51,892 
B.T.U. per pound hydrogen, both being referred to an arbitrary base temperature 
of 212° F. 

Investigations of the thermo-chemists have shown that the heat of com- 
bustion is not only determined by the chemical elements that burn, in kind 
and amount as might be expected, but also on their molecular structure, whether 
a single chemical element is involved or a chemical compound. In other 
words, assuming carbon and hydrogen to be the fuel elements in a fuel, its 
calorific power is not equal to the sum of the products of these two calorific 
powers into their respective partial weights. There is a difference and 
the difference is not the same for different compounds but follows a simple 
law for the compounds of one series, like the hydrocarbons of the paraffine 
series, for example. Furthermore, for carbon alone there are several different 
calorific powers, that for soft, porous carbon being at one end and that for 
crystalline diamond at the other. Fortunately, these differences are not to 
be considered in engineering as two successive samples of the same natural 
fuel will differ from each other more than would warrant fine corrections like 
that for the carbon. The molecular structure as a cause of difference is, 
however, useful in explaining the lack of agreement between a calorimeter test 
which is always conclusive and a calculation based on chemical composition 
and calorific power of the elements. 

As combination and dissociation are inverse processes the heat liberated 
by the combustion of hydrogen to form water which may be regarded as the 
heat of water formation must be equal to the heat necessarily absorbed in 
separating the water into hydrogen and oxygen, whence the heat of dissocia- 
tion of a compound formed by combustion is equal to the heat of combustion 
of its fuel element. This is important in high temperature work in which 
steam is often dissociated as hydrogen is burned and where in successive 
steps steam dissociation and hydrogen combustion may follow each other. 
In high temperature combustion as, for example, gas engine explosions, it is 
quite possible that before the fuel is burned entirely the temperature has 
risen to the value at which dissociation of products takes place and, there- 
fore, combustion or combination will be retarded until heat be carried away 



520 ENGINEERING THERMODYNAMICS 

When this occurs, all the heat of combustion does not enter into the raising 
of temperature indefinitely but only partly so, the rest acting as liberated to 
maintain a high temperature as heat is being abstracted or otherwise being 
disposed of. 

As the calorific power of a commercial fuel is a function of the calorific power 
of the carbon, hydrogen and sulphur it contains, it is important that their cal- 
orific powers be firmly established, and some of the more authoritative values 
are given in the Table LV at the end of this chapter. In the case of car- 
bon, which may burn to carbon monoxide or to carbon dioxide, it is import- 
ant that the corresponding reaction heats be consistent. This will be the case 
when the heat of combustion of 1 lb. of carbon burning to CO, added to the 
heat of combustion of the amount of CO containing a pound of carbon, be 
together equal to the heat of combustion of carbon to CO2 directly. One lb. 

28 . 

of C is contained in — = 2.33 lbs. of carbon monoxide according to atomic weights 

compared to = 16, or — ' QQ — = 2.324, according to atomic weights com- 
pared to H=l. 

Therefore, for atomic weight = 16. 

{B.T.U. per lb. C } , j 2.333 XB.T.U. per lb.] = {B.T.U. per lb. C) 
I burning to CO J "^ I CO burning to C0 2 J I burning to CO J 

or for atomic weights H = l. 



(B.T.U. per lb. C] = / 2.324 XB.T.U. per lb.] (B.T.U per lb. C 

l 



' burning to COJ [ CO burning to CO2 j | burning to CO2 



This is but a special form of the general thermo-chemical law. The heat of 
combustion of a compound (CO) together with the heat of formation of that com- 
pound from its elements (C and O) must be equal to the heat of combustion of 
the elements (C and O) direct to the final products CO2 or the heat of formation 
of products directly from the elements. 

Taking the Favre and Silberman value for the complete combustion of 
carbon to carbon dioxide, which is accepted by Julius Thomsen one of the 
greatest of the thermochemists as 14,544 B.T.U. per pound C, and Thomsen's 
own value for the heat of combustion of CO to C0 2 as 4369 B.T.U. per pound 
CO, the heat. of combustion of the compound must be (using atomic weights 
C = 12 and O = 16 because Thomsen did so in reporting his value) 

j (B.T.U. perpound C burning to CO2) ] 
B.T.U. per pound C burning to CO = ! - 2.333 X (B.T.U. per pound CO burn- ( 

[ ingtoC0 2 ), ' 

= 14,544-10,193 = 4351. 

From the above it is clear that the heat of combustion of CO burning 
to CO2 is 10,193 B.T.U. per pound of carbon. 



HEAT AND MATTER 



521 



It is evident from the general law that the heat of combustion of com- 
pounds like the hydrocarb6ns and alcohols cannot be computed from the heat 
of combustion of the carbon and hydrogen they contain, without correcting for 
the heats of formation of these compounds from their elements. These heats 
are, however, pretty well determined for some of the definite compounds that 
are constituents of natural fuels ; and among these the most important are the 
hydrocarbons. For these Thomsen has determined many of the heats of for- 
mation of the compound and finds some positive and others negative, so that 
in the former cases the heat of combustion of the compound would be that 
calculated for its hydrogen and carbonless the heat of formation of the com- 
pound, and in the latter cases plus the heat of formation. These heats of 
formation are generally small as indicated by the following table: 

HEAT OF FORMATION IN TERMS OF HEAT CIRCULATION 



Methane CH 4 


+ 10.2% of its heat of combi 


Ethane C 2 H 6 


+ 7.6% 




Ethylene C 2 H 4 


- -8% 




Acetylene C2H2 


-15.4% 




Propane C3H8 


+ 6.5% 




Benzene CeH6 


- 1-6% 




Trimethyl methane CH :(CH 3 ) 3 = C 4 Hi 6 


+ 6.2% 




Tetramethyl methane C :(CH 3 ) 4 = C5H12 


+ 5.6% 




Propylene CH 2 :CH :CH 3 = C 3 H 6 


- .7% 




Trimethylene C 3 H6 


- -7% 




Isobutylene CH 2 :C:(CH 3 )2 = C 4 H 8 


+ 1.6% 




Isoamylene C2H 4 +C:(CH 3 )2 = C5Hio 


+ 2.2% 




Diallyl C 3 H5:C 3 H 5 = C6Hio 


- .1% 




Allylene CH :C :CH 2 = C 3 H 3 


- 8.6% 


1 1 1 



Such figures as these are not of very much use in the practical work of dealing 
with the volatile of coals or liquid fuels because it is not possible to tell precisely 
what hydrocarbons are present though some of the above may be and others 
for which these heats are unknown. For all such practical work approximate 
methods are used and some of them are surprisingly accurate, as will be seen 
from further discussion. Fig. 158 shows curves plotted for several series of 
hydrocarbons, giving the heat of formation of the products of combustion. This 
is the heat which would be obtained by the burning of that amount of hy- 
drogen and carbon in the free state which exists in the combined state in the 
hydrocarbon. For instance, if the hydrocarbon is represented by the symbol 
CxHy, the heat of formation of the products is (high value) 

1 2X14,544x+60,626t/ 
•12x+j/ 

which is not the heat of combustion of the hydrocarbon, since such combustion 
requires the supply of heat of formation of the hydrocarbon itself, either posi- 



522 



ENGINEERING THERMODYNAMICS 













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HEAT AND MATTER 523 

tive or negative. The experimental values of heat of combustion of a small 
number of these compounds are indicated by points, .and the difference between 
point and the curve, shown by brackets, is this heat of formation of the 
hydrocarbon. These curves bring out the remarkable fact that however com- 
plex the hydrocarbons are as to molecular constitution, the heats of combustion 
of the majority of them do not vary very much and the ethylenes have a constant 
heat except as the heats of formation vary a little. 

Calculations dealing with the combustion of gases and vapors are often more 
conveniently carried through, on a volume than on a weight basis. In doing 
this, it is convenient to have available the volumetric calorific powers of the stand- 
ard gases and vapors reduced to 32° F., and 29.92 ins. Hg absolute pressure 
even if the particular compound is a liquid under these conditions, for the pro- 
portionality will hold for higher temperatures as if it were a vapor or gas. Gases 
reported for these conditions of pressure and temperature are termed standard 
and the reduction to this standard from any other pressure and temperature 
or to any desired pressure and temperature is to be made on the assumption 
of perfect gas behavior. 

Let P29 -92 = absolute pressure in inches of mercury =(29.92), of standard 

gas or vapor; 
" 1^32/29.92 = volume in cubic feet per pound of standard gas; 
" ^32/29.92 = density in pounds per cubic foot of standard gas; 

" (B.T.U. per cubic foot) 3 2/29.92= ' ' : = B.T.U. per cubic foot 

^32/29.92 

standard gas; 
" P = any other pressure in inches of mercury at which the gas may be 

under study; 
" £ = any other Fahrenheit temperature; 
" V and d = corresponding specific volume and density for P and t; 

(B.T.U. per cubic foot) = corresponding volumetric heating power for 

P and t. 



Then 



PV P29.92 X F32/29.92 29.92 X F32/29.92 



(£+460) 292 492 

Whence 

7 "0^)nr) Fw/29 - M ' 



or 



d =(^60A2£92r 32/29 - 92 ' 



524 ENGINEEEING THERMODYNAMICS 

and 



,-dttt n, B.T.U. per lb. 

(B.T.U. per cu.ft.) = 

\ ACkO I \ D / "32/29 



(B.T.U. per cu.ft.)32/29.92 
/H-460X /29._r ^ 
V 492 )[• P 



.U. per CU.K.; 3 2/29.92 , 717 x 

(t+m\ /29.92\ ' " " * K } 



The values for the (B.T.U.per cubic foot)s2/29.92, that is, of standard gases given 
in the table at the end of the Chapter, are derived from the B.T.U. values per 
pound reported, both high and low values, and from the standard volumes of 
the pound taken from table at the end of the Chapter, the direct experimental 
determination being used wherever it is available otherwise, as calculated from 
the molecular weights. 

A consistency is desirable whenever not seriously in conflict with facts, 
and the molecular relations of density are necessary in dealing with liquid fuels 
and the gasification of solid ones ; the values per cubic foot for hydrogen and 
carbon monoxide are often derived in this way from the pound value instead of 
making use of direct calorimetric values given in the table. This is always 
done for the vapors. 

According to the formulas 

2H 2 +0 2 = 2H 2 
and 

2CO+0 2 = 2C0 2 , 

it appears that the calorific power of hydrogen per cubic foot standard must be 



(718) 



B.T.U. per cu.ft. hydrogen = ~^— = 339 high (a) 

2 X51892 
= —353 — = 290 low ( b ) 

Then if a molecule of CO has the same volume as a molecule of H 2 

B.T.U. per cu.ft. carbon monoxide = — ^ — = 342. . . (719) 

For the hydrocarbons the same procedure can be put in a general form, neglect- 
ing the heats of formation of the compounds. 



HEAT AND MATTER 525 



Thus, for the paraffines C„H 2n +2 taking the molecular weight of C = 12, 

. (720) 



Wt. C in compound = — — r-= — rH== — r? P er 1°- f ue * ( a ) 

12n+2w+2 7n+l 



Wt. H in compound = ir> ", — r-= = = — — per lb. fuel (6) 
12n+2n+2 7n+l 

__ TT u , . 14544 X6n+60626(rc+l) 147890n+ 60626 .. . x 
B.T.U. per 11). fuel=~ — " 7n+l = 7n+l _(hlgh) (a) 

_14544X6n+51892(n+l) 139156n+51892 



7n+l 7w+l 



14n+2 

139731n+ 51872 

7n+l 139159/1+5246. x ,. N 
358 = 179 (l0W) ib) 



14n+2 
For the ethylene series C n H2 n 

Weight C in compound - == = .857 per lb. fuel (a) 

Weight H in compound = ~ = = = . 143 per lb. fuel (b) 

Hence 

B.T.U. per lb. = .857X14544+. 143X60626 = 21134 (high) (a) 
= .857X14544+. 143X51892 = 19885 (low) {b) 



Therefore 



DfrTT ,. ,. As .857X14544+. 143X60626 oof7 

B.T.U. per cu.ft. vapor (std.) = ^ = 827n(high)(a) 

Tin 

.857 X14544+.143X51892 „ . 

= 3^ = 778n(low) (6) 



(721) 



Also cu.ft. vapor (st'd) per ^ = l2n ^ n+2 = T^2' ' ' (722) 

Therefore 

147890n+60626 

T}TTT ,. 7n+l 14789071+60626,,. . , s 
B.T.U. per cu.ft. vapor = g— = — (high) (a) 



(723) 



(724) 



(725) 



or o 

Cu.ft. vapor (std) per lb. =—- (726) 



(727) 



526 



ENGINEERING THERMODYNAMICS 



These results are plotted to a base of n in Fig. 159, and illustrate beautifully 
how the calorific power varies with the position of the hydrocarbon in the series 
with the paraffine series at the top as the highest of all. These values are 
the basis of work on coal and oil gas and liquid fuels to be taken up later. To 
the calculated values on the curve of Fig. 159 may be added the available direct 
determination for comparison from Fig. 158 and the Slaby empiric formula 

(B.T.U. per cubic foot) = 112+ 18880 X (pounds per cubic foot.) . (728) 

The calorific power of a power of a commercial fuel or lighting gas per cubic 
foot is the sum of the volumetric per cents of its constituents into the calorific 



Heptane, 



C,H 



ye 









500 



Hexane, CoH 



* 



<> 



> 



s& 



7. 

Pentylene, 



/ 






^Benzene, C C H 



5300 

Pi 



200 



Butane, C^Hip,*- r 

S V^Butylene, GMg 



7^ 



Propane, 



C,H 



i& 



~Z> 



y?i 



/ 



Propylene, O3H 



Ethane, C 2 H 



/ 



/• 



'~7 
/ 

2CEthyl 



Z* 



jne, C2H4 



100 



k/ B 

< Methane, CH 4 



3 4 5 6 

Value of n in Chemical Formula 



Fig. 159. — Heat of Formation of Products for Combustion of Hydrocarbons, in B.T.U., 

per cu.ft. at Standard Conditions. 

power of those constituent gases which include carbon monoxide, hydrogen, 
methane, ethylene, ethane and sometimes some higher hydrocarbons, the nature 
of which is assumed, like benzene. Coals and oils cannot be so treated, but the 
methods used here with varying degrees of approximation will be taken up 
later. Of course, in all cases an estimate may be made from the combustible 
chemical elements if they are known in amount, by guessing at their molecular 
condition. 



Example 1. A certain fuel consists of 5 per cent hydrogen, 8 per cent oxygen, 
1 per cent sulphur, and 87 per cent carbon, by weight. Find the heat of formation 



HEAT AND MATTER 527 

of the products of complete combustion, if all of the oxygen in the fuel is originally 

combined with hydrogen in form of moisture. 

OS 
.08 lb. oxygen with '-— = .01 lb. hydrogen constitute .09 lb. water originally in 

8 

fuel, leaving .05 -.01 =.04 lb. hydrogen to be burned. 

.04x60626= 2425B.T.U. 
.01 X 3998= 40B.T.U. 
.87X14544 = 12653 B.T.U. 



Heat of formation of products = 15118 B.T.U. high value 

Prob. 1. A certain fuel contains 20 per cent H 2 and 80 per cent C by weight. If 
the heat of formation of the compound was 8 per cent of the heat of formation of the 
products, what are the high and low values of the heat of combustion? 

Prob. 2. What would be the heat of combustion for the reactions given in Problem 
1, Section 13? 

Prob. 3. It is known that a cubic foot of natural gas under standard conditions 
contains 980 B.T.U.'s. What would be its calorific power at 70° F. and at an altitude 
of 8000 ft.? 

Prob. 4. A sample of producer gas gave the following analysis per cubic foot. 
What would be its high and low heating value per cubic foot? 

H 2 =.18; CO=.25; CH 4 = .03; CO 2 =.07; N 2 = .47. 

Prob. 5. What would be the cost per 100,000 cu.ft. of standard gas and perbbl. 
of oil to give the same cost per B.T.U. as coal at $4.00 per ton, when analyses are as 
follows? 

Coal. Oil. Gas. 

C=.75 C 5 H 12 =.10, N 2 =.04, 

CH 4 = .07, C 6 H 14 = .30, H 2 = .02, 

Moisture = .08, C 9 H 20 =.50, CH 4 = .94, 

Ash = .10, Ci 3 H 28 =.10. 

Analysis of coal and oil by weight, gas by volume. 

Prob. 6. An evaporation of 7.7 lbs. of water from and at 212° F. was obtained 
with the following coal. What per cent of the heat of formation of products was 
delivered to the steam? 

C=.758; H 2 =.049; O 2 =.085; S=.023; Ash = .085. 

Prob. 7. An internal combustion engine gives a horse-power hour with a 
fuel consumption of .5 lb. of hexane, while another engine uses 8 cu.ft. of methane 
during the same period and develops the same power. What are the respective 
efficiencies? 

Prob. 8. By burning a pound of the oil, the analysis for which by weight is given 
below, 500 lbs. of water were heated 40° F. in a calorimeter. What was the heat of 
formation of the compounds? 

Ci H 22 = 10 per cenl; C 8 H 18 =80 per cent; C c Hi 4 = 10 per cent. 



528 ENGINEERING THERMODYNAMICS 

15. Heat Transmission Processes. Factors of Internal Conduction, Surface 
Resistance, Radiation and Convection. Coefficients of Heat Transmission. 

When air supports combustion, the gaseous products may be said to be heated 
directly, and to define the process, the term " internal combustion heating " 
is generally applied, and likewise when a liquid cools by evaporation from its 
surface due to a lesser surface vapor pressure than corresponds to the tempera- 
ture, it might be said to be directly or internally cooled. With such exceptions 
as these it is universally necessary in dealing with heat in commercial apparatus 
to transmit or transfer it from one place or one substance to another place or 
substance. Thus, the heat for boiling water in steam boilers must come through 
tubes and plates from the fire or hot products of combustion, and a similar 
transfer takes place in closed feed-water heaters in economizers, steam super- 
heaters, in all heated surfaces of boiler, pipe and engine parts, in surface con- 
densers for steam, ammonia or carbon dioxide, in steam and hot-water radiators, 
through the walls of buildings outward when heated during the winter and inward 
when cooled for cold storage below the atmosphere, likewise in the brine coolers, 
ice cans and other parts of mechanical refrigerating systems, in apparatus for 
concentrating solutions in chemical manufacture or in distillation, in the cooling 
of gas engine and compressor cylinders by water jackets and cooling air between 
compressor cylinders, or drying and superheating steam between cylinders of mul- 
tiple expansion engines, or preheating of compressed air for air engines. Thus, the 
practical application of heat for the doing of a thing desired, or the abstraction of 
imdesired heat, generally though not always involves transmission, and most 
often through solid containers of great variety ranging from the thin brass tubes 
of surface condensers to the iron or steel pipe of boiler walls or ammonia 
evaporating coils, to steam pipe covering or the brick wall of a dwelling house. 
That the movement of heat from one place or substance to another of lower 
temperature follows some law of flow was long ago recognized, even before any- 
thing was known of the transformation of heat into work. Old as is the realiza- 
tion of the existence of heat flow laws, there is, however, no general acceptance 
of what are those laws except in a few special cases, and the reason for this 
situation, which seems somewhat queer after all these years of experimental 
study and formulation of theories, becomes clear after even a brief analysis 
of what must really happen in the course of an actual case of transmission. 
In spite, however, of the absence of such laws as will permit of reasonably 
exact calculation of the relations between quantity of flow, the dimensions and 
kind of surfaces and quantities and kinds of material exchanging heat, with 
their corresponding changes of temperature or state, engineers must design 
apparatus that is needed, as best they can, and for this purpose there has been 
developed a method of approximation with which is used a series of coefficients 
developed by experience for the more common and recurrent cases. The treat- 
ment of this subject will, therefore, be divisible into two parts, first the examina- 
tion of such laws as are fairly well fixed, and second, the development of an 
approximate method of treatment for the practical everyday problems that the 
fundamental laws do not solve. 






HEAT AND MATTER 529 

Heat may pass from one substance or place to another in three essentially 
different ways and every practical case includes one or all of these and no others. 

1. Heat may flow along a bar or wire or along any line in a solid body from 
a high to a low temperature point, by simple molecular communication. Each 
molecule heats the next without any discontinuity and with a regular con- 
tinuous fall of temperature from the high to the low temperature point. This 
is internal conduction of heat, and the conducting capacity of nearly all the com- 
mon substances is pretty accurately determined and represented by coefficients 
of thermal conductivity, some selected values for which are given at the end of 
this Chapter in Table LVL Of course, by internal conductivity only, heat 
could not flow beyond the limits of one piece of substance, but at the boundary 
it may be communicated to the next body and so on pass any number of dividing 
surfaces or through any number of bodies. This is a most common mode of 
conduction about which there is no exact information comparable with that 
for internal conduction. 

2. Heat may flow from a hot body through space to a distant colder body 
entirely without reference to the separating medium, and the best example is 
the passage of the sun's heat to the earth. This mode of transmission is defined 
as radiation and heat so received by the cold, or discharged by the hot body, as 
radiant heat. The capacity for radiation of heat for different substances is 
practically unknown, though there is fair though not good agreement as to 
the relation of heat quantity to the two controlling temperatures. There is 
practically nothing known of the radiating capacity of any but originally black 
bodies brought to high temperature conditions or incandescence. 

3. Heat may get from one place or substance to another by being carried 
as a charge on a particle that is moved bodily between the two places. Thus, 
a particle of water heated at the bottom of a vessel becomes less dense and floating 
in the colder or more dense, actually carries its heat to the top as is also the case 
in chimneys where the flotation tendency produces the necessary draft. This 
is convection communication of heat and takes place only in liquids and gases, 
where in action it is generally associated most intimately with radiation and 
conduction. There may be a movement of particles of liquid or gas not caused 
by differences in density characteristic of convection, but by pumps or fans, 
and which is quite effective in carrying or transmitting heat, but this is an 
externally forced circulation of heated particles rather than a self-caused heat 
transfer. 

Transmission of heat by internal conduction takes place according to a 
law of proportionality to temperature difference, but as the conductivity of sub- 
stances depends on the temperature itself there must be involved a temperature 
term. This gives for solids a comparatively simple law which is apparently 
also applicable to liquids and gases, but for gases variations enter that have 
been the subject of lengthy study by such eminent physicists as Maxwell, 
Clausius, Kelvin, Tyndall, and some hundreds or so less famous but good men 
without acceptable solution. 



530 ENGINEERING THERMODYNAMICS 

For solids and liquids the conductivity is given by the following Eq. (729) : 

K t = K 32 [l+x(t-S2)] (729) 

in which 

X* = B.T.U. per hour per square foot of cross section per degree F. 
difference in temperature between the points under considera- 
tion, per inch of path, when the body has the temperature 
«°F.; 
i?32 = B.T.U per hour per square foot per degree F. per inch of thickness 
at32°F.; 
£ = temperature degrees F.; 
x = & constant. 

This conductivity at any temperature is not directly useful except in compar- 
ing the conductivity of substances, but in computing the amount of heat flow- 
ing between a high temperature and a low temperature point the mean con- 
ductivity for this range must be determined. As conductivity is a straight 
line function of temperature the mean conductivity between the temperature 
is the arithmetical mean of the conductivities at these temperatures, hence 

Let tn = high temperature degrees F.; 
U = \ow temperature degrees F.; 
K^ t h-u) =mean conductivity between t h and ft. 



Then 



_ ^_ K 32 [l+x(tn-32)]-{-Ks2[l+x(t l -S2 )] 
-tv (th—n) — q ~ ~~ ' — 

= K 32 (l+|fe-32) + tt-32)]) (730) 

From this mean conductivity expression Eq. (730) the amount of heat that 
will flow through a body by internal conduction is given by Eq. (731) 

Q = Al(t h -ti)K(t h -t l ), 



l+|[(**-32)-(fc-32)] 



where 



Q = B.T.U. per hour; 

A = cross-section of conducting path in square feet; 
1 = length of conducting path in inches. 

Applying expressions like this to solids is a direct and safe proceeding if the high 
and low temperatures are taken within the conducting body itself and not in 



HEAT AND MATTER 531 

some external substance like a liquid or gas in contact. The same expression 
and limitation applies to liquids if there is no convection flow, in which case 
the heat flow will be the algebraic sum of the conduction and convection parts. 
For gases there enters likewise a radiation factor as a third disturber as well as a 
more general and important doubt as to the value of the expressions at all for 
gases except as a means comparing them with liquids and solids. In no case 
can these equations be applied to heat flow across a surface boundary or joint 
in a solid, or from solid to liquid to gas. 

Inspection of the table of conductivities reveals some most interesting 
and valuable relations — thus, silver is about the head of the list as a conductor 
and carbon dioxide at the foot with an enormous difference between them, 
silver being able to conduct about 35,000 times as much heat as carbon dioxide 
for equal temperature differences. In fact, the substances group themselves 
naturally into good and bad conductors, all the gases being in the 
latter class and many solids, like felt and wool. These solids are the natural 
heat insulators and their insulating value, seemingly dependent on their porosity, 
at least to some degree, has led to the theory that the gases confined between 
fibers or granules of solids are strong contributors to the insulating values 
of otherwise solid bodies. Liquids fall into an intermediate class, being 
universally better conductors than gases, not so good as the metals, which as 
a class are the best conductors, but of the same order as the group of solids 
generally considered as insulators. 

As the relative conductivity of solids, liquids and gases, or their reciprocals 
the resistances to heat, are most useful in explaining or even predicting the 
relative characteristics of the common heat transfer paths in commercial 
apparatus, it is important that a table be developed from existing data." These 
data are of two sorts, first, the absolute internal conductivities of the table and 
second, direct experimental measurement of relative conducting power, which is 
determined by different methods than the absolute value. These relative con- 
ductivities (Table LVII) are similarly useful in selecting from conflicting values 
like those for iron, that which is most probable, though, of course, commercial 
irons being alloys merging into the steels must have small differences depending 
on composition, but so small as to be negligible for most engineering purposes. 
Thus, comparing the common metals with silver, Wiedermann and Franz 
find the following numbers as representing the conductivity, silver being 100, 
lead 8.5, iron 11.9, steel 11.9, copper 73.6, zinc 28.1, tin 15.2, of course, all 
at the same temperature, which is most conveniently fixed at 0° C. = 32° F. 
Applying these ratios to the absolute conductivity of silver by Weber, 1.096, 
the most closely corresponding of the direct measurements when there are more 
than one is found and marked in the Table LVI by an arrow *— . For liquids, 
the value for water being 100, the following relative values are given: Benzole 
28.8, by Weber; ethyl alcohol absolute 30.9; 90 per cent, 32.05; 50 per cent,54.29; 
by Henneberg, and methyl alcohol 27.37 by De Heen. For gases with air taken 
as 100, hydrogen is fixed at 701, oxygen 102, carbon dioxide 62, methane 139, 
ethylene 74 by Stefan, hydrogen at 710, carbon dioxide 59, carbon monoxide 



532 ENGINEERING THERMODYNAMICS 

98, by Kundt and Warberg, ammonia 91.7 and illuminating gas 267 by Plank. 
Where these ratios do not check, adjustment has been made in the table 
of relative conductivities with carbon dioxide as unity for all and resistances 
with silver unity for all, so that all numbers, are whole numbers, and to avoid 
conflict of testimony only two significant numbers are retained for large dif- 
ferences and three for small. 

It is extremely likely that the conductivity of gases does not follow the 
linear law with respect to temperature that seems satisfactory for solids, and 
it is from Clark Maxwell that the best suggestion for a substitute has come. 
He predicted from the kinetic theory of gases that the conductivity is propor- 
tional to the product of the coefficient of gas viscosity and specific heat at 
constant volume or 

K = [lC v X (a constant) . 

Dalby from a study of values of [i finds it can be put proportional to the three- 
quarter power of the absolute temperature T so that, 

K = cC v (T)\ 



or 

K T 



( T° FA * 
V 492 / ' 



approximately if specific heats are assumed constant. This more than doubles 
the conductivity value for 32° F., at the temperature 930° F. This is an 
important addition because the gases of furnaces are very hot indeed while 
the tubes, plates and water are comparatively cool. 

Conductivity beyond the boundaries of a body is known to suffer a rapid 
decrease or heat flow to encounter a resistance through joints in metal bars 
or between plates at point of contact. Boiler plates offer appreciable resistance 
and boiler seams are forbidden in contact with fires by certain inspection laws, 
because experience shows the plates frequently burn, which they would not 
do were there not a large joint resistance. It is also known that several layers 
of thin boards make better insulation than equal thickness of single boards 
for refrigerated boxes and rooms. Whether this joint resistance is due to a 
thin layer of poor conducting air or is a strictly separate phenomenon is not 
known, but most carefully made accurately fitted joints in experimental bars 
in which there could not be over one ten-thousandth of an inch air layer gave 
measurable joint resistance, smaller, of course, than a bad joint. It is probably 
true that there is a real joint or surface resistance of unknown character at the 
boundary of every body and likewise a fluid film of some thickness as well, 
though probably not uniform. When heat passing from fluid to plate or 
plate to fluid, encounters a film of fluid adhering to the plate and for all practical 
purposes sticking to it, the heat resistance is materially increased. Inspection 



HEAT AND MATTER 533 

of the table indicates that, taking the relative conductivity of iron at about 
6000, a layer of carbon dioxide of good of an mcn thick will offer as much resist- 
ance to the passage of heat as a one-inch thickness of metal, and taking water 
as 40, relative conductivity, a layer of water goob = 150 of an inch thick would 
offer the same resistance, both in addition to the separate joint resistance. 

The existence of such heat-resisting films of gas and liquid in heat transfer 
plates is conclusively proved, so that the problem of estimating heat flow from 
a body on the other side involves an estimate of the thickness of the adhering 
fluid film, which must be the purest sort of guess. In addition to the film 
and joint resistance difficulty, there is another when dealing with boiler, con- 
denser, feed-water heater, economizer, superheater and similar tubes and plate 
surfaces, and that is the scale, grease, soot, rust or dirt layer that is always 
present in some condition, of density and thickness and also highly resistant 
as belonging in the heat insulator group. Furthermore, such a layer on both 
sides involves another joint resistance, so that the practical everyday problem 
of deciding on how much surface to allow for a given heat flow or how much heat 
can flow past a given surface, is not one that can be solved by the laws of con- 
duction even if data on all substances were available, and probably never 
will be, because it requires first an hypothesis fixing the thickness of fluid film, 
the joint resistance, the kind and thickness of fouling or layer of dirt. Problems 
of this character are to be solved in another way by means of coefficients of 
transmission found by experiment and applied to the same kind of apparatus 
as that on which the experiment was made and covering all resistances together, 
though each cannot be separated from the rest. 

There is no doubt now as to the existence and supreme importance of the 
surface films, especially on the gas side of plates receiving heat from or delivering 
it to gases or involving the presence of gases such as deposit on steam condenser 
tubes or intermittent layers of vapor on boiler surfaces when ebullition is taking 
place, nor can there be longer any doubt about the reduction of film thickness 
and resistence by fluid agitation. This agitation generally is the result of 
velocity increase of the fluid over the surface, but such data as exist on this 
effect do not permit of a determination of actual film thickness or resistance, 
but rather of overall coefficients of transmission to be considered later. 

If a series of resistances to heat flow as reciprocals of conductivity are known, 
the whole heat flow temperature relations, can be set down algebraically; the 
real practical difficulty arises from their numerical evaluation. 

Let lij h, etc. = thickness of each portion of plate or film along the path 

of heat flow, in inches; 
" K\j K2, etc. = mean or constant coefficients of conductivity of each material; 

pi, p2, etc. = — , — , etc., thermal resistances of each material; 

ill A2 

cji, <J2, etc. = surface resistances; 

tn and U = high and low temperatures at the limits of the path. 



534 ENGINEERING THERMODYNAMICS 

Then 

^ffe§ • • < 732 > 

For the whole complex heat flow path there can be introduced a general overall 
coefficient U inclusive of all internal and boundary resistances such that 

Q = AU(t h -t l ), . (733) 

where 

u =^rh;- •••••••• (734) 

Where C/ = B.T.U. per hour per square foot, per degree difference of temperature 
between any two points usually in practical problems taken in 
two separated fluids, one receiving heat from the other. 

Direct evaluation of this coefficient of transmission U is the usual practical 
procedure, inclusive of all separate conductivities or thermal resistances together 
with radiation and convection influences without separately evaluating each. 

Heat flow by radiation alone has been experimentally studied and reduced 
to laws by a number of skillful physicists, and while they do not agree, the so- 
called Stefan and Bolzmann law is now quite generally accepted. According 
to this law heat will be radiated from an incandescent " black " body according 
to Eq. (735). 

B.T.U. per hour radiated per square foot = 16Xl0- 10 (7 7 i 4 -7 7 2 4 ), (735) 

where 

T\ = absolute temperature F of the " black " radiating body, 
and 

T2 = absolute temperature F of the receiving body. 

A black body is defined as one capable of absorbing all heat rays received, neither 
reflecting or transmitting any, and for practical purposes may be assumed to 
be any porous carbon, that is all carbon except the crystalline. Bodies not 
" black," like hot gases or fire brick or boiler plates, do reflect some heat 
received, and fail to radiate to the same extent as does a black body. For 
them absorption and radiation increase with darkness (when cold) and rough- 
ness, which is one reason for nickel plating and polishing surfaces from which 
radiation is to be resisted. 



HEAT AND MATTER 



535 



For various substances the following figures, Table XXVII, give numerical 
values on relative radiation or absorption and reflecting power the sources of 
which are uncertain, but which have been for a number of years quoted by 
handbooks. These figures are probably wrong, but are given in the absence 
of any better, which even if available would be of but little real use. 

Table XXVII 
RADIATION COEFFICIENTS 





Radiating and Ab- 
sorbing Powers. 


Reflecting Power. 


Porous carbon (black body).. . 

Glass 

Ice 

Polished cast iron 

Wrought iron polished 

Steel polished . 


1.00 
.90 
.85 
.25 
.23 
.19 
.07 
.07 
.03 


0.00 
.10 
.15 
.75 

.77 
.81 
.93 
.93 
.97 


Brass polished 


Copper hammered 

Silver polished 





For the black body itself the following curves, Fig. 160, are calculated from the 
Stefan and Bolzmann law for various temperatures of the radiator up to 6000° 
absolute F. to a heat absorber at 500, 600, 700, 800, 900 and 1000° absolute. 
The very great quantities of heat that may thus pass, independent entirely of 
conduction, convection and mechanical carrying by gases and equally independ- 
ent of dead gas films from an incandescent coal fire to boiler plates or furnace 
interiors is at first surprising. It must not be assumed, however, that all this 
will be taken up directly by the plates of a firebox, for example, as they will 
reflect possibly a quarter to a half of what is so received, sending it out again 
to other walls, the reflection being especially severe when the heat ray strikes 
at a small angle, when nearly all the heat will be immediately reflected. How- 
ever, these quantities are significant of the great steam-forming capacity of 
enclosed fireboxes common to locomotives and to the Scotch marine boiler. 
Dalby estimates that assuming the hydrocarbon volatile gases which on com- 
bustion separate out solid carbon are themselves equal to an incandescent 
black body and filling the firebox so that area of radiating surface equals that of 
absorption, each square foot of firebox can evaporate from this source alone 
about 134 evaporation units, equivalent roughly to 4 J boiler horse-power. 
Experiments all prove the firebox capacity for evaporating greater than all the 
rest of the boiler as ordinarily made. 

To compare the Stephan and Bolzmann law of fourth power of temperature 
with other proposals, Prof. Callender has plotted all on one sheet, which is repro- 
duced in Fig. 161, with the scale changed to English units by Prof. Dalby. Ordi- 
narily the heat radiated is not separated from that otherwise transmitted, but 



536 



ENGINEEBING THERMODYNAMICS 




1500 2000 2500 3000 

Radiating* Temperature Eahr. 



3500 



1000 



Fig. 160. — Heat Radiation from " Black Body " According to Stefan and Bolzmann Law. 



HEAT AND MATTER 



537 



the whole transmission determined from general experimental coefficients where 
known. 



lM 



12 



CO 

Ph 



cs 8 



^ 



§ 4 





















































































/ t 
1 i 

' 1 


























\ 


/ / 


1 

i 

1 


























I 

• 

; i 


/ / 

r / 
/ / 


1 

i 

t 


























r / 


// 




























U/ 


7 




























1/ 


/ 
























































^M 






























//I 




























// 


Of- 


Newton T\ 

Hosetti T 3 , 

i 




















/ 

* 
/ / 


'/// 


* 


Stefan T 4 | 
-Petra'velT^ 








/ 


i 




/ ; 


i 














./ 


/ / 
/ / 


/'/ 


i 


Bottomley |T 5 - 7 








//& 




f rT^ 
















^ 


/ 


A 




/ 




.Webe 


r a T 












^-r: 


^-" 






,J' 





















400 



800 1200 1600 2000 

Radiating Temperature Fahr. (Absolute) 

Fig. 161. — Comparison of Heat Radiating Laws. 



2400 



Example. The wall of a house consists of 6 ins. of. brickwork lined with 1 in. of 
plaster. The junction of the plaster to the brick being equivalent to 10 ins. of plaster. 
The windows are of glass | in. thick and have a film of moisture on them .003 in. thick, 
the junction of the water and glass being equivalent to 3.5 ins. of glass. How does 
the coefficient of heat transfer for the walls compare with that of the windows? 



From Eq. (734), 



For the wall 



U 



U 



2pH-2a 



-|x6 + |xl0+l 



538 ENGINEERING THERMODYNAMICS 

For the window 



Hence 



p=5 for the glass, 
=3.5 for the water. 

U=— * = 1.38. 

7U +3 - 5 ) + 3T5 X - 003 • 

Prob. 1. A bar of copper and one of glass are heated by the same source of heat. 
How much longer will it take for the end of the glass one furthest from the source of 
heat to reach a uniform temperature than that of copper? 

Prob. 2. Neglecting joint effects, how much will the heat lost from a pipe \ in. 
thick carrying steam at 500° F. and running through a room at 50° F. be reduced by 
the application of a plaster of Paris coating 2 ins. thick? 

Prob. 3. A standard 2 in. boiler tube has a thickness of .095 in. On the inside 
of the tube is a layer of scale ^ in. thick which may be considered as plaster of Paris; 
on the outside of the tube is \ in. coating of soot having the same conductivity as wool. 
Neglecting the resistance due to the junction of these substances, what will be the 
difference in heat transmitted through the above tube and a new clean one? 

Prob. 4. The walls of a room are as follows: 6 ins. concrete, 6 ins. sawdust, 3 
ins. concrete. What will be the heat lost per square foot per hour for an inside 
temperature of 40° F. and an outside temperature of 90° F.? 

Prob. 5. The copper tubes of a condenser are .05 in. in thickness. On each side 
is a water film .01 in. thick. How will the heat transferred through this tube compare 
with that of a dry one in contact with air? 

Prob. 6. A tin tube .05 in. thick is silver plated, the plating being .01 in. thick 
and the joint resistance being equal to 1 in. of silver. How does the plating affect 
the value of U? 

Prob. 7. Considering steel to have a radiating power equal to .2 of carbon, what 
will be the value of the radiant heat of a steel sphere having 10 sq.ft. of surface at a 
the temperature of 1500° F. when the receiving body is at 50° F.? 

Prob. 8. A boiler has 1700 sq.ft. of heating surface of which 100 are exposed to 
direct radiant heat. The remainder is heated by hot gases of average temperature 
of 1000° F. The temperature of the radiating fire is 3000° F. and the temperature 
of the water in the boiler is 350° F. If the surface not subject to radiation consists 
of iron tubes .1 in. thick with .1 in. scale and the junction of scale to plate is equal 
to an inch of iron, what is the relative importance of each part of the boiler? 

16. Heat Transmission between Separated Fluids. Mean Temperature 
Differences, Coefficients of Transmission. Heat flow, in terms of B.T.U. 
per hour transmitted, is usually taken to be proportional to the temperature 
difference when one fluid is giving up heat to another separated by plates or 
tubes, as for internal conduction, though experimental determinations have 
indicated the existence of other relations. For example, Grashof, Rankine and 
others have announced the square of the mean temperature difference between 
the two fluids as the correct function, while Orrok, recently experimenting with 
surface condensers, reports a seven-eighths power instead of the first power. 
The temperatures that are measurable in engineering work are those in 
the body of fluid, gas or water entering or leaving a coil or pipe or the tern- 



HEAT AND MATTER 539 

perature in a condenser or boiling mass at some selected points. It is not pos- 
sible to determine outside of the laboratory the skin temperature of a boiler 
tube, the mean temperature of a kettle shell or even the temperature of the fluid 
next the cooling or heating surface. Exact scientific analysis of heat flow laws 
would involve those and many other quantities, hence the practical necessity 
for approximate calculations and the absence of any generally inclusive law or 
equation. Each class or case must be studied by itself though in the light of 
relations to others. 

Mathematical analysis and experimental observations both in the labora- 
tory on special apparatus and in the field on standard working equipment has 
led to the following generalizations beside the proportionality of flow to some 
power of the mean temperature difference between the giving and receiving 
fluid. The quantity of heat transmitted is found to be proportional to the 
velocity of the fluid to some power when one of the fluids is in motion and to 
some power of each not necessarily the same when both move. Thus, for 
the case of water in the tubes of surface condensers and feed-water heaters, 
the one-third power has been used by Joule and Ser, the one-half power by 
Hageman, Josse and Orrok, and the first power by Stanton. The heat flow has 
been related to the steam velocity at the condensing surface by Hausbrand and 
Ser as a function of its one-half power, but found to be independent for such con- 
ditions as exist in surface condensers and feed-water heaters by Orrok, which 
corresponds to the zero power. Air or other gas in motion giving up heat does 
so at a rate, proportional not to velocity according to Jordan and Reynolds, but 
to the weight passing per unit of cross-section of gas stream directly, a conclu- 
sion that may reconcile the discrepancies noted previously where the velocities 
used by different experimenters might be the same, whereas temperature, density 
and channel cross-section were in all cases different. When steam condenses, 
the heat flow is strongly controlled by the collection of gaseous matter on the 
surfaces, and data by Smith and others show that in ordinary apparatus this 
may affect the heat flow as much as 50 per cent. 

For the purposes of calculation the cases are grouped into types, for example, 
surface condensers and feed-water heaters are typical of heat transmission 
from condensing steam to water entering at a low and leaving at a higher tem- 
perature, but still lower than that of the steam. If the water temperatures 
are fixed and the amount of heat to be exchanged likewise, the problem becomes 
one of finding the necessary surface. On the other hand the problem may be 
set to find the rise of water-temperature for a given amount of surface. 
j ?~i A different situation exists, for example, in steam superheaters in which steam 
temperature rises by heat received from hot gases, the temperature of which 
is falling. In this case, given the surface and the initial temperatures of the 
gases and the steam the final temperature of both may be required for a given 
amount of surface or inversely the object sought may be the surface necessary 
to accomplish a given rise of steam or given fall of gas temperature. 

In all these practical cases of transfer what is sought is the relation 
between amount of surface and the corresponding heat flow or its equivalent 



ENGINEERING THERMODYNAMICS 

in quantity of substance heated or cooled, vapors condensed or liquids evap- 
orated. The temperatures of the substance are not always uniform throughout 
the mass or over the heating surface and ordinarily change as the substance 
passes over the surface. Therefore, while the quantity of heat passing is 
proportional to some power of the temperature difference at any instant, 
the instantaneous temperature differences being indeterminate, it is necessary 
to establish a mean temperature difference for the whole period of thermal 
contact of the substance in question. 

In addition to the fixing of a mean temperature difference for each typical 
kind of fluid flow to which heat flow is related, it is also necessary in the solu- 
tion of numerical problems to have the constant of proportionality for each class 
apparatus or typical transfer. This constant of proportionality is known 
&s the coefficient of heat transmission and is defined as the B.T.U. per hour, 
transmitted through one square foot of surface or cross-section of heat flow 
path, per degree F. difference in temperature between the two substances 
measured in the body of each and not at the surfaces of contact. Accordingly, 
Let Q = B.T.U. per hour transmitted; 
" U — coefficient of heat transmission = B.T.U. per hour per square foot 

per degree F. measured in the substances ; 
" Z w = mean temperature difference for whole period of transfer; 
" A = square feet of surface through which transfer is taking place or 
the cross-sectional at area of the heat flow path in square feet. 
Then 

Q = AUtm (736) 

There is a definite relation between the mean temperature difference and 
the initial and final temperatures of the fluids exchanging heat, and in establish- 
ing this relation five different cases must be recognized as follows : 

Case I. The substance giving up heat may be at a constant temperature 
and the substance receiving heat suffer a rise of temperature. 

Case II. The substance giving up heat may suffer a drop in temperature 
and the substance receiving heat remain at a constant tem- 
perature. 

Case III. Both substances may remain at a constant temperature. 

Case IV. Both substances may change temperature, that giving up heat 
falling and that taking the heat rising by parallel flow, the final 
temperatures of both tending to become equal. 

Case V. Both substances may change temperature by counter current flow, i.e., 
in opposite directions, the final temperature of one substance 
tending to become equal to the initial temperature of the other. 

For each of these cases there is a different relation between the initial and 
final temperatures of both substances and the mean temperature difference, 
and these algebraic relations will be derived after a further examination of the 
typical cases to show how these five are inclusive of all ordinary conditions, 



HEAT AND MATTER 



541 



Constant temperatures characterize the two cases of condensing vapors 
and evaporating liquids, though not strictly so, because it is well known 
that in a boiler or condenser the water on the vapor side is not quite constant 
in temperature, but for the purposes of such calculations as these it must be 
taken so, or no calculation at all would be possible. Substances changing 
temperature regularly and indefinitely may be liquid or gaseous. With these 
distinctions the following classification of heat transfer cases can be made 
with reference (a) to the kind of thermal change taking place in the substances ; 

CLASSIFICATION OF TYPICAL HEAT TRANSFER APPARATUS 



Thermal Action in Substances 


Class of Relation be- 
tween temperatures 
and mean tempera- 
ture differences. 


Examples of Standard Thermal 


Giving up Heat. 


Receiving Heat. 


Apparatus. 


Liquid Cooling 


Liquid warming 


Case IV or V 


Aqua ammonia exchanger. Water 
coils of ammonia absorber. Brine 
water cooler. 




Gas warming 


Case IV or V 


Hot-water house-heating radiator; 
cooling tower. Automobile radia- 
tor. 




Liquid boiling 


Case II 


Ammonia and carbon dioxide 
brine cooler. Evaporator with hot 
liquid coils or jacket. 


Gas cooling 


Liquid warming 


Case IV or V 


Brine coils in cold storage rooms. 
Economizers. Cylinder jackets. 
Forecooler ammonia condenser. 
Brine coil air dryer or cooler. Com- 
pressor intercooler. 




Gas warming 


Case IV or V 


Dense air machine coils in cold 
storage rooms. Steam superheaters. 
Compressed air engine preheater. 




Liquid boiling 


Case II 


Steam boiler. Ammonia or carbon 
dioxide direct expansion coils in 
cold storage rooms. 


Vapor condensing 


Liquid warming 


Case I 


Exhaust steam boiler feed-water 
heater. Steam, ammonia and carbon 
dioxide condenser. 




Gas warming 


Case I 


Steam radiator and tempering 
coils. Steam piping radiation. Mul- 
tiple expansion engine reheating 
receiver. Dry air condenser. 




Liquid boiling 


Case III 


Exhaust steam vac. evaporator. 
Multiple effect evaporator. Coil 
and jacketed evap. kettle. Aqua 
ammonia generator. Evaporative 
condenser. 



542 ENGINEERING THERMODYNAMICS 

(b) type of relation between initial, final temperatures and mean temperature 
difference, and (c) specific examples of apparatus in which each of the type 
actions take place which, therefore, are grouped as similar or dissimilar. | 

It might be noted in the case of ice cans submerged in brine that before 
freezing begins the case is one of heat of water cooling to brine warming, and 
after freezing begins, constant temperature latent heat of freezing to brine 
warming, but as the brine is not allowed to rise more than a few degrees the 
process is nearly one of constant temperature on both sides nd is not listed 
above because it is a peculiar case without parallel. A similar complexity 
exists with respect to reheating receivers of multiple expansion engines in which 
the heat of condensing high pressure steam is added to steam passing between 
cylinders and where if the working steam is wet it is first dried at constant 
temperature and afterward warmed or superheated to a temperature approach- 
ing that of the live steam. Still other cases of possible complexity are the 
cooling tower and evaporative condensers in which more than one action may 
take place and in which the specific construction exerts a strong or controlling 
influence. 

It is structure entirely that determines whether Case IV for parallel flow or 
Case V for counter-current flow shall apply to the transfer between liquids 
and gases, one to the other in which both substances suffer temperature change. 
In some constructions it is quite impossible to fix the flow relations and this 
is in general true when a large mass of substance is under treatment as in a 
tank, instead of in coils pipes or flowing between guiding partitions, and in 
these cases it is necessary to assume the nearest typical case as representative, 
which, of course, requires some good judgment. Derivations of mean differences 
will first be based on proportionality of heat transmitted to the first power of the 
temperature difference. 

Case I. Mean temperature difference, for constant high temperature source 
to rising temperature cold substance. In Fig. 51 the temperature of the two 
substances is plotted vertically, and the surfaces over which the flow causes 
temperature change, horizontally for the conditions specified as Case I. 

Let t h = constant temperature of the hot substance degrees F. ; 

t c = any momentary temperature of the cold substance degrees F.; 
i Cl = initial temperature of the cold substance degrees F. ; 
t C2 = final temperature of the cold substance degrees F. ; 
w c = pounds of cold substance flowing per hour over surface. 
C c = specific heat of cold substance; 
A = surface in square feet; 

Z m = mean temperature difference between hot and cold substance; 
U = coefficient of heat transfer = B.T.U. per hour per square feet per 
degree F. 

Then will the heat transmitted per hour through the elementary surface 
dA be given by U(t h — t c )dA, and as the temperature rise for this period will 



HEAT AND MATTER 



543 



bed/ c ,the heat transmitted per hour is also given by C c w c dt c which is the amount 
taken up by the cold substance. These two quantities must be equal, whencte. 



dA = 



C c w c dt c 



u h-t; 



If, as is here assumed to be the case, the hot substance be at a constant tem- 
perature, h is independent of t e and the coefficient of heat transfer U like- 



250 














































Ten 










^U^-*o 2 


































( 










200 


A 

i 


A 


1 






















\ 


.___ 


^== 






i 
i 
i 




!• 


i 






























150 


. — [.. 
1 


— \- 


i. 




7^ 


g 


[£*-* 


z^^- 


F 











' 


—IT 













ttffjt 


h ) i 


Jk 


$> 


s^^ 




























100 


1 

1 
i 


h 


r 






















*!»a 










I 


'i* 


































50 






































i 

t c 


! 





































1 

t 


t 
























* 











50 100 150 200 250 300 350 400 

Sq. Ft. Surface 

Fig. 162. — Heat Transfer from Constant Temperature Source to Rising Temperature Fluid, 

Case I. 



wise, then will the relation between temperature rise of the cold substance and 
the surface traversed be given by, 



_C C Wc f l c2 

~ U Jt 



dt, 



(t h -t c y 



or 



*-*$■**(££)> ■■■•■■ < 737 > 

For the whole period of transfer if U is independent of the temperature difference 

t m U A. = C c^c Kycz *cij ) 



or 



trn-^iU-Q . . (738) 



544 



ENGINEERING THERMODYNAMICS 



Substitution in Eq. (738) of the value of UA from Eq. (737) in terms of tempera- 
ture differences gives the mean temperature difference 



{t C2 tci) 



«*£=£ 



(739) 



This value is indicated by the distance from the high temperature line t h to 
the dotted line on the diagram Fig. 162. 

Case II. A constant temperature of the cold substance T c with a falling 
temperature of the hot one constitutes the second case and is represented in 
Fig. 163. 



3b 



37 



36 



,35 



2 34 



33 



i>32 



31 



30 





































1 


































1 








2^£ 


2*g, 
























— 1- 




~*T~ 








i^ 


■2°* 


fe* 


'(~T 


































^b) 














































ir 3 
i 




(t/lf 


t c ) 






























tm 


























i 
i 
i 
































Ctj 


'2~t c ) 
| 


































1 

1 




1 




Y 






Ten 


pera< 


ure < 


<f Cc 


Id F 


uid 


tc) 








1 







































20 30 40 50 60 

Sq. Feet of Surface Passed over 



TO 



Fig. 163. — Heat Transfer from a Falling Temperature Source to a Constant Low Tempera- 
ture Fluid, Case II. 



With the use of similar symbols and as indicated on the diagram, 

C h w h dt h 



dA 



U (tn-tcY 



6j 2 

t m UA = C h w h (t hl — t h2 ) , 
(thi — tfc) 



U 



log, 



f t hl -t c Y 

\th 2 — tc/ 



■ (741) 






HEAT AND MATTER 



545 



Case III. When both temperatures are constant the relations shown in Fig. 
164 are most simple since the mean temperature difference is constant and the 
heat transmitted per hour is directly proportional to the surface in action. 



*m — *h *c 



(742) 



250 


■ 




























































4 


A 
























200 


























































& 

1 
































( 


t h ~t e 


)=t m 






















s 

3 




hi 


























islOO 

S-i 

a 






























p 






T 
























H 
50 






I 


























i c 





























t 


1 

























10 



20 30 40 

Sq. Feet of Surface Passed Over 



50 



Fig. 164. — Heat Transfer between Two Fluids Each at a Constant Temperature, Case III. 

Case IV. Parallel flow conditions of two substances both changing temperature 
are indicated in Fig. 165 which, therefore, represents diagrammatically the 
fourth case. After passage over the elementary surface dA, the cold sub- 
stance will rise in temperature an amount dt c and the temperature of hot 
substance fall correspondingly dt h ; therefore, the heat transmitted per hour is 
given by 

U(tn — Q dA = C c w c dt c = CnWndtn . 

It appears, therefore, that the relation between surface and temperature change 
can be given two forms, as follows, 



. _ Cj»c_ ft<* dt c _ _ CnWn f hl dt h ■ 
U J tci (tn-Q- U L {tn-Q 



These are really identical because t h is a function of t c at any moment since 
the heat gained by the cold body must be equal to that lost by the hot one, 
which fact fixes the relation between temperatures. 



546 ENGINEERING THERMODYNAMICS 

Hence after any interval of time or surface traversed 

l C h w h (t hl — t h ) = CcW c (tc— Q. 



Therefore 



th ~ thA '^k {tc ~ tcl 



Subtracting from each side t c , 



C h w h 



(^ | CcW c \ t 



U;« 2 -tc 2 ) 




A = 



10 15 20 25 

Sq. Feet of Surface Passed over 

Fig. 165. — Heat Transfer between Fluids, Each Changing'Temperature, Parallel Flow. 

CASE^IV. 
Substituting this in the, differential equation" where the variable is t e , gives 

C C W C 

CcW c 



CcWc r c * dtc 



C c Wc 



u(i+$ 



log, 



C h wJ 



f i CcWc f 

^hi ~T~ n l ^\ 



H 



C c w c \ 



*»,+ 



C c w c 



tc, 



Therefore 



ChWh 

CcW c 



U 



1+% 



■Wc\ 

ChwJ 



,CcV>c\. t , 

log e (^Er),i fc (743) 



HEAT AND MATTER 547 

This can be put in a slightly different form by the relation 

1 I C C W C _ - . thi —fa* _ (tg ~ U\ ) H" (thi — th 2 ) 
C hlVfi la tc\ *ci tci 

which may be inserted in Eq. (743) where the problem indicates a necessity 
for it. 

For the whole transfer, 

t m UA = CcWc(t C2 —tci) = C h w h (t hl —th 2 ), 



whence 



CcW c (tc 2 — t Cl ) _ C c Wc(tc 2 — t Cl ) 



UA 



CrW c 

1 „ / Ml tci 

lo & ITTZT 

\ t7*2 °C 2 



(tc 2 ~ lei) + (thi "~" fo:) 



(tC2 hi) 



Therefore 



/ _ (^2 ~ hi) ~\- (thi — th 2 ) _ (thi ~ Q ~ (th 2 — ta) (744) 



\ M 2 *>Ct / \ t/»2 ^C2 / 



Case V. Counter-current flow conditions for this case are indicated on Fig- 
166, which shows a rising temperature of the cold body to a value approach, 
ing that of the initial temperature of the hot one, instead of its final temperature 
as in the case of parallel flow. Beginning with the entrance of the cold bod}' 
into the system its temperature will have risen dt c degrees from t Cl , after 
passage over the elementary surface dA, and in the same time the hot body 
will have fallen dt h degrees to the value % t . Therefore, as for parallel flow 

. _}C c w c C l c% dt c _ChW h C l hi dtn 

Elimination of one variable t h by introducing its relation to t c is done in a similar 
manner but by terms differently involved. After any appreciable interval 

ChWh (th — t h2 ) = CcW c (t c — t Cl ). 
Therefore 

th = th 2 -\--Ti (tc — ta)y 

thWh 



and 



th tc-t h2 S^+^-Y- 



548 ENGINEERING THEEMODYNAMICS 

Substitution in the differential equation where the variable is t c gives, 



A = 



\w c ft C2 dtc 

\CnWh 

























\ i 












i\ ! 












1 IV 












1 1 V 

1 1 

1 1 












1 J 
1 


\^ 










(**rO ! 


i 










i 

i i 
i i 


1 
I 


^.^ 






100 


i i 


i 

i 
i 




^A 






*fc! 


tm 




"^^^ 






> 


i i 


1 






t t 




"»^/ 


\ i 


80 


i r""- 


^! 






1 

1 1 


60 


— 1 K 

1 | 


— ^~^=^-^ 


___^ 


, 


(*fci7*oi) j 




1 1 
1 1 




^=^ 


^---JL^_ 


i ! 




i 

tc 2 








\ t 




1 

[ 








1 1 
1 




r i 
f f 








1 r 












1 





i ; 








1 1 



5 10 15 

Sq. Feet of Surface Passed Over 



Fig. 166. — Heat Transfer between Fluids, Each Changing Temperature, Counter-current 

Flow. Case V. 



Therefore 



C c Wc 



\C h w h 



log* 



/ thl — tc 2 \ 
\ th 2 — td I 



(745) 



HEAT AND MATTER 549 

The weight, specific heat ratio can be eliminated by the relation between it 
and temperatures which is given by 

C C W C _ -J _ thj — tht _ -J _ ( th! — tft 2 ) — (t C2 — Ul) 

C hWh let *ci lei *ci 

For the whole transfer 

t m UA = C c Wc(t C2 -t cl ) =C h w h (t hl -t h2 ). 



Whence 

t m = 



CcW c (t C2 — t Cl ) C c Wc (ta — t Cl ) 



UA 



r fa-y-fe-o ] loge t^zh) 

L Khz — tci) J \ t h2 — t Cl / 

(thi ~ ^2) ~ (la — tci) 
i / thi *cz \ 

\th 2 — tci / 

t„= fa-^)-^-^) (746) 

log. 7_ 7 

All of these equations for mean temperature difference are identical when 
properly interpreted, however different they may look in terms of the specific 
symbols. Inspection will show that the numerator is in every case the initial 
temperature difference less the final temperature difference, while the denom- 
inator is only the hyperbolic logarithm of the ratio of initial to final temperature 
differences. 

,.., ,.„. N (Initial temp, diff.) — (Final temp, diff.) /n A^\ 

(Mean temp, diff.) = /T .. . , / — ,.„ ^ . (747) 

, / Initial temp. dinerence \ 

\ Final temp, difference / 

By the adoption of coefficients of transfer XJ, or rather numerical values 
for U, for the common cases of practice, it becomes possible to calculate rela- 
tion between quantities of substance heated or quantity of heat transmitted 
and surface but only approximately. The coefficients used in most engineering 
work allow a good safe margin in the form of sufficient surface so the apparatus 
will always be big enough to perform the required service, but economy in 
the use of material demands care in design to bring to bear as large a coefficient 
as possible by whatever means are available and which means are confined to 
the use of high velocities of fluid or equivalent means of fluid film reduction 
and cleanliness. In Table XXVIII are given some average values of U that 
have been in common use in designing apparatus. These are presented not as 
accurately determined values but as average value illustrating the most 
important fact that in all cases where gases are involved the coefficient is ever 
so much less than when it is not. 



550 



ENGINEERING THERMODYNAMICS 

Table XXVIII 

COEFFICIENTS OF HEAT TRANSFER 
Average Practice 



Thermal Action in Substances. 


B.T.U. per Hour per 

Square Foot per 

Degree. 


Apparatus. 


Giving Up Heat. 


Receiving Heat. 




Liquid warming 


50-75 


Liquid heat exchangers, aqua 
ammonia water and beer 
coolers, ammonia absorber 
cooling coils 


Liquid cooling 


Gas warming 


2-6 


Hot-water radiators and cool- 
ing tower surfaces, depending 
on air velocity and character 
of water surface 




Liquid boiling 


100 
10-20 
30-50 


Shell brine coolers with circu- 
lator; tank brine coolers 
without circulator; double 
pipe brine coolers depending 
on velocity and hot liquid 
evaporators 




Liquid warming 


2-5 


Brine coolers in cold storage 
rooms depending on air circu- 
lation. Air coolers with 
water or brine coils; econo- 
mizers 




Gas warming 


2-4 


Steam superheaters 




Liquid boiling 


2-5 


Direct expansion ammonia coils 
in cold storage rooms de- 
pending on air circulation. 
Steam boilers 


Vapor condensing. . . . 


Liquid warming 


150-350 
1000 


Feed-water heaters and steam 
condensers depending on wa- 
ter velocity and removal of 
air on steam side. Experi- 
mental feed-water heater 
high velocity 


Gas warming 


2-4 


Steam radiators and pipes 




Liquid boiling 


400-600 


Vacuum evaporators with con- 
densing exhaust steam de- 
pending on viscosity of solu- 
tion 



HEAT AND MATTER 551 

Example 1. Case I, Fig. 162. A feed-water heater, with exhaust steam at 220° F. 
and a water supply at 60° F. and U =200, specific heat = l, water 300,000 lbs. per 
hour, which in round numbers corresponds roughly to the feed of a 100 H.P. 
boiler. Compute the surface to give the feed water a final temperature of 90°, 120°, 
150°, 180°, 210° and find t m . 



C c Wc , Ith — k\ 

. / 220-60 \ 

log < Uo=W sq ' ft - 



160 
= 150 log* — - = 150 loge 1.23 = 150 X .207 =31.05 sq.ft., 
loO 

160 
" ^ = 120° F., ,4=150 loge— -=150 loge 1.6 = 150X.47=70.5 sq.ft., 

" t C2 = lo0° F., A=150 loge ~= 150 X. 833 =124.95 sq.ft., 
► 70 

" * Cf = 180°F., A =150 loge --=150 XI. 386 =207.90 sq.ft., 

1 (\f\ 

" ^ = 210° F., A =150 loge— =150X2.77 =415.5 sq.ft., 

t^-tu 210-60 150 
™ = Z, — rv = TT^I — = W^ = 54 degrees mean temperature 

-(s^) >-© diff — 

Example 2. Case II, Fig. 163. An ammonia brine cooler where the evaporation 
takes place at 30° F. and the brine enters the coils at 40° and leaves at 35° F., a 
total fall in temperature of 5° F. Take the specific heat of the brine at .68 and 
assume 17,700 lbs. per hour, which about corresponds to one ton of ice per 24 hours. 
Take 17 = 100. Find the surface corresponding to 39° F., 38°, 37°, 36° and 35° F. 

. ChWh , (thi — t c ) 

.68X17700, (40-30) 
loge 



= 120 log 



100 ° D (fo a -30) 

10 



it* -soy 



For t h2 =39° F., A = 120 log e ^ = 12.6 sq.ft., 
" ^ 2 =38°F., A =120 log c ^=26.8 sq.ft. 



552 



ENGINEERING THERMODYNAMICS 



For t ht = 37° F., A = 120 loge -=- =42.9 sq.ft., 



36° F., A =120 log e ^ =61.4 sq.ft. 
o 



10 



fo t =35°F., A =120 loge — =83.2 sq.ft., 
5 



tjn — " 



hi — h 



l0g e 



\tfl2 — t C 



40-35 
log, (f 



= 7.21 degrees mean temperature difference. 



Example 3. Case III? Fig 164. Assume that exhaust steam at a temperature of 
235° F. is used to supply heat to an evaporator in which alcohol is being evapor- 
ated at a temperature of 70° F. So long as the pressures and the quality of the alco- 
hol remain constant, these temperatures will remain the same. What will be the 
surface required if 3,600,000 B.T.U. are to be supplied to the evaporating fluid per 
hour? Assume {7=400. 

(h—tc) =t m =235—70 = 165 degree temperature difference. 

Heat transfer per square foot = 165 X400 =60,000 B.T.U. per hour. 

3600000 
Required surface = = 60 sq.ft. 

oOOOO 



Example 4. Case IV, Fig 165. Water leaving a still at a temperature of 180° F. 
passes through a cooler before entering ice tanks in which it is to be made into ice. 
Let this cooler be assumed to have a parallel flow of distilled water and of cooling 
water. The cooling water enters at a temperature of 50° F. The cooler is to be 
capable of cooling 10 tons of water per day from a temperature of 180° F. to 90° F. 
using for cooling 2| lbs. of cooling water per pound of distilled water. Take U = 60. 

Per hour, 1^=834 lbs. per hour; w c =2080 lbs. 

C h = l; C c = l. 

Find the required cooling surface for each 15° fall of temperature of the hot water. 
Heat necessary to cool 20,000 lbs. through 15° will heat 50,000 lbs. of water | Xl5° =6°, 
which is accordingly the corresponding increment for t c as shown in tabular form. 



th 


tc 


th~ tc 


180 


50 


130 


165 


56 


109 


150 


62 


88 


135 


68 


67 


120 


74 


46 


105 


80 


25 


90 


86 


4 



HEAT AND MATTER 553 

Substitution in Eq. (743) gives the following expression for area corresponding 
to a given temperature difference, th 2 —U 2 



A = 



1X2080 



/ 1X2080\ 



. /l80-50\ ftA . /130\ 



130 
For t h2 = 165° F.,A =9.9 log*— =1.74 sq.ft., 

ISO 
For t h2 = 150° F., A =9.9 log,—- =3.88 sq.ft., 



130 
For fo, = 135° F.,A =9.9 log e -^ =6.56 sq.ft., 

130 
For t h2 = 120° F.,4 =9.9 log c — = 10.30 sq.ft., 

4b 

130 
For t h2 = 105° F„A =9.9 log e — =16.32 sq.ft., 

zo 

130 
For t h2 = 90° F„A =9.9 log e -^ =34.5 sq.ft. 



The mean temperature difference is 



_ (fc-U-fe-U 130^4 .126 , 

l m— /! ork \ -« i^-w uegrees. 



-tea -ra 



Example 5. Case V. Fig. 166. Assume a problem similar to that for Case IV except 
that the flow is counter current instead of parallel, i.e., into a counter-current cooler 
flows 834 lbs. per hour of distilled water at a temperature of 180° F., which is to be 
cooled to 90° upon leaving. Cooling water is supplied at 50° F. at the rate of 2080 
lbs. per hour. How much cooling surface is required to cool by intervals of 15°? 
What is the mean temperature difference when cooling to 90°? 

Ca-1; & = 1; ^=834; w c =2080. Assume £/=60; ^=50°. Since for any 
given interval, with counter flow 

ChWhithl ~th 2 ) =C C Wc(t C2 -t C y), 

C h w h 



tc 2 =ta-\-(tj n ~-th 2 ) 



= 50 + (180 -90) 



w c C c 
834X1 



2080X1 

=86° F. as shown in the table. 

834 
Next, let *a 2 = 165° F., and t ci =86-(180-165) ^=80° F. 



554 



ENGINEERING THERMODYNAMICS 



For each successive fall for fa of 15°, there is a fall of 6° for fa since the cold fluid is 
flowing in the opposite direction and the change of temperature of the one is due to 
the change of temperature of the other. 



fa 


tc 


(fa-fa) 


180 


86 


94° 


165 


80 


85 


150 


74 


76 


135 


68 


67 


120 


62 


58 


105 


56 


49 


90 


50 


40 



From Eq. (745) the area from the point of entrance of the hot liquid to the point 
at which temperature has fallen 15° is given below. 



For fo 2 = 165°F.,A 



1 X2080 / !80-86 \ 

_ /2080 \ Oge U65-80/ 
3 °V834~7 



94 

= 2.31 Xloge — =2.33 sq.ft. 

o5 

94 
" ^ 2 = 150 o F., A =23.1 loge ^=4.94 sq.ft., 



94 
" fc t = 135° F., A =23.1 log e ^r=7.82 sq.ft., 

o7 



94 
" ^ 2 = 120 o F., A =23.1 loge — = 11.15 sq.ft. 

05 



94 
^ 2 = 105 o F., A =23.1 log e 7^ = 15.1 sq.ft., 



94 
t hi = 90° F., A =23.1 loge 77j = 19-8 sq.ft. 



The mean temperature difference is 



C h w h {t hl -t h2 ) 834X90 ■ _ 
t m= ^rz — ~ = „„. ..„ „ =o3.1 degrees, 



UA 



60X19.8 



or by Eq. (746) 



94-40 54 on - . , 

= Au\~ = Ts^ degrees, as previously. 

io& y 



HEAT AND MATTER 555 

Prob. 1. In the manufacture of condensed milk, the water is evaporated at a low 
pressure by the use of steam. If the pressure in the vacuum kettle is such that the 
temperature is 90° F., the latent heat per pound will be about 1100 B.T.U.'s. On the 
assumption that this condition remains the same during the entire process, that U = 300, 
and that the heat is derived from steam at a temperature of 215° F., how much 
surface will be required to evaporate 2 tons of water per hour? 

Prob. 2. The rate of heat transfer in a surface condenser is 400. The vacuum 
desired calls for a temperature in the condenser of 100° F. The cooling water amounts 
to 1,000,000 lbs. per hour, enters at' 50° F. and leaves at 95° F. How much 
surface will be needed if the heater is a five-pass heater, and what will be the 
temperature at the end of each pass. 

Prob. 3. The boiler for a steam-heating plant runs at a pressure of 5 lbs. per 
square inch gage, and evaporates a thousand pounds of steam per hour. The products 
of combustion have an initial temperature of 1000° F., a final temperature of 400° F. 
and a specific heat of .25. Assuming 1000 B.T.U.'s needed per pound of steam and that 
U is 2.5, how much surface will be needed? 

Prob. 4. The oil in an electric transformer is cooled by running it through annular 
tubes through the inside of which is circulated cold water in a counter-current direction. 
1000 lbs. of oil per hour must be cooled from 150° F. to 80° F. while there is available 
5000 lbs. of water per hour at a temperature of 50° F. The specific heat of the oil may 
be taken as .4, and U as 50. At what temperature will the water leave at the end of 
the cooler and how much surface will be needed for every 10° of cooling? If it were 
desired to cool oil down to 70° F. how much additional surface would be required? 

Prob. 5. To reduce the quantity of moisture in air it is passed over coils con- 
taining cold brine and thereby lowered below the dew point. Disregarding the heat 
to be removed from the moisture what must be the area of coils to lower 500,000 
cubic feet of air per hour from 70° F. to 20° F.? The air and brine pipes enter 
the cooling duct at the same end. Using the following data, what will be the re- 
quired surface? Specific heat of air .25, of brine .7, brine enters at —20° F. and 
leaves at 15° F. What will be the mean temperature difference? 

Prob. 6. Solve Prob. 5 for counter-current flow. 

17. Variation in Coefficient of Heat Transmission Due to Kind of Substance, 
Character of Separating Wall and Conditions of Flow. Nothing could well 
be more striking than the fluctuations in the accepted workable values of U 
for different conditions of surface, kinds of substance and rate of flow, rang- 
ing as they do from about U = 2, to somewhere near [7=1000. It is quite 
natural in view of the uncertainty that.] must accompany the selection of the 
correct value for a given practical problem that many investigators have sought 
to explain by experiment and analysis the variations that exist with a view to 
associating all values by a general law. This ultimate aim appears, how r ever, 
to be quite hopeless of attainment, and even if it were not without hope it 
would probably be useless because, any general law must include terms to 
account for the conditions of surface, the gas content of liquids or vapors, 
something equivalent to the thickness or resistance of fluid films and other things 
equally intermediate by a computer engaged in predicting >what will happen in 
an apparatus for which these things must first be evaluated but cannot be. 
About all that one could reasonably expect in this connection is a relation of 



556 ENGINEERING THERMODYNAMICS 

U to the definable variables in each characteristic class of heat transfer cases 
with perhaps a few general principles, and some of these principles are 
fairly well established, though none quite beyond the limits of controversy. 

Among the general principles, one already cited appears most important 
of them all, that when a gas is either giving or receiving heat the coefficient 
of heat transfer is very much lower than when a liquid is involved as such, 
or as changing state with its vapor. So very high is the resistance and so 
very low the value of U when a gas is involved, that the main resistance to 
flow is always on the gas side unless most extraordinary means are used to 
change it. As transfer always proceeds from one substance to another and 
through a third generally, it may be considered as a three-stage operation, of 
which the first step is the giving up of heat by the hot one, the second its trans- 
mission proper and the third its absorption by the cooler body. The rate 
with which it will pass while a function of all the heat resistances encountered 
is practically controlled by the one single highest resistance when that one is 
much larger than the others. Thus, in a given case the rate of heat exchange 
may be limited by the ability of the fluid on one side or the other to give or 
to take and in the great majority of cases this is so, for a steam boiler the water 
can take up heat faster than the hot gases can give it, except possibly for 
surfaces receiving radiant heat, while in a steam condenser the limit is imposed 
by the ability of the circulating water to take the heat, and the latter 
is also the case for steam feed-water heaters as well. Therefore, even 
if means were available to increase the heat-absorbing capacity of water next 
the heating surface of boilers it would be of no value until the gases had been 
first caused to give up heat at a very much faster rate than they do. These 
cited, are plain cases, but there are others in which conditions are not so clear 
such, for example, as water heat exchangers and steam superheaters where 
in each case the same fluid and fluid conditions exist on each side and where 
it is impossible without further information to state just where the principal 
resistance is, that is on which side. 

The usual plan of experimenters seeking to relate the coefficient of heat 
transfer or the hourly transfer of heat to some prime variable has been to vary 
every condition that might produce a change, one at a time, plot the results 
to coordinates and seek an equation for the curve. Practically all that is known 
of the conditions that control the value of U has been derived in this manner 
with the exception of one result announced by Osborne Reynolds and derived 
by mathematical analysis based on the kinetic theory of gases and on certain 
hypotheses with regard to the condition of affairs at the metal surface. 

Some of these results will be reviewed to illustrate the complexity of the 
relations involved and the almost complete hopelessness of any attempt to 
generalize much beyond the following principles in addition to those of a single 
controlling resistance and the almost universally higher value of gas resistance 
over liquid. 

The transmission per hour is said to be proportional to the temperature 
difference of the fluids directly according to most experimenters, so that 



HEAT AND MATTER 557 

the coefficient in B.T.U. per hour per square foot per degree is independent 
of temperature difference, however much it may vary with other things. How- 
ever, Werner, Grashof and Weiss give the hourly rate as proportional to the 
square of the temperature difference or the value of U as proportional to tem- 
perature difference to the first power, while Orrok, experimentally studying 
heat transfer in steam condensers, reports the hourly rate as proportional 
to the seven-eighth power of the temperature difference and U inversely as 
its one-eighth power. 

When the heat transmitted per hour is directly proportional to temperature 
difference, the value of U, being the B.T.U. per square foot per degree mean 
temperature difference, is independent of the temperature difference and th 8 
mean temperature difference t m is a function of the initial and final as deter- 
mined in the last section. Should the heat flow per hour be proportional 
to any but the first power of the temperature difference then the value of U 
will be itself a function of temperature difference, and the mean temperature 
difference a different function of the actual temperatures than those previously 
derived. This can be shown by analysis for one case as follows, a similar 
method being applicable to other kinds of flow and cases of transfer though not 
worked out here because of the uncertainty of the value of the exponent. 

Consider the case of flow of heat from condensing vapor at a constant 
temperature to a liquid or gas with rising temperature which, of course, includes 
surface condensers and feed- water heaters. Then, an element of surface dA 
there will pass U(t h — t c )dA, B.T.U. per hour, which is equal to the product 
of weight per hour, specific heat and temperature rise of the substance 
receiving the heat, whence, 

7 . Cw dt c 
dA 



U k-t c 



If now the heat per hour be proportional to the nth power of the temperature 
difference, then U is proportional to the (n— 1) power of the temperature dif- 
ference or, 

u=K(tn-t c y-\ 

Therefore 



A= ^i C2ft "^"^ = Z(l-n) 1(4 " 4)1 " n " (feHc2)1 " W]> ' (748) 



which is the relation between the surface and the temperatures resulting from 
the passage over that surface. The mean temperature difference is given 
by the relation 

t m U m A = Cw(t C2 — t Cl ), 
in which U m is the average value of U for the whole heat exchange. 



558 ENGINEERING THERMODYNAMICS 

Whence 

= Cw(t Ct — t Cl ) = Cw(t C2 -t Cl ) 

UmA u m ^^[{k-t c y^-{k-t c .y--] 

K(l-n) {tc-Q 



u m (t h -t cl y- n -(t h -t C2 y- n 



(749) 



It appears from this that the mean temperature difference is a function of 
the temperature rise of the water or gas being heated, of the (1—ri) power 
of the initial and final temperature differences and of the mean value ^of the 
coefficient of heat transfer which is itself now a variable. This mean value 
of U can be found by integration, but is not, because, first the resulting expres- 
sion is very complicated for practical use, and second the completed expression 
is hardly worth while in spite of its apparent accuracy, because of the uncer- 
tainty in the value of n and the real value of U, so that the real result is not 
and cannot be accurate. 

Both hourly rate and the value of U assumed independent of temperature 
difference, that is, forn = l, are found to be more clearly a function of flow 
conditions which are defined by different experimenters in different ways, none 
of which seem to quite reach the heart of the matter. For example, all observers 
agree in assigning higher values to U when the flow is vigorous than when 
it is sluggish, that is, U is found to increase in some manner with rate of flow 
in all cases but one, and that is when the increased rate of flow takes place in 
the fluid on that side of the surface where the resistance is but a small frac- 
tion of the total. Thus, for hot gas warming water, increase in flow of the water 
does not sensibly increase the value of U, but increase in the rate of gas flow 
will do so. This is one source of discrepancy in the reports of experiments 
not conducted in such a way as to control both rates when seeking to relate 
heat flow to one of them, many experimenters reporting results for variations 
in one rate of fluid flow without making sure that the rate of flow of the other 
fluid was either constant or of negligible resistance. This point is clearly 
demonstrated by the following experimental data, comparing one set of experi- 
ments with another. 

The first series of experiments was conducted by Mr. W. D. Monks 
in the mechanical laboratories of Columbia University, under the writer's 
direction, on the transfer of heat from hot gases to water under varying 
conditions of water flow with the gas flow intentionally neglected, except 
as was necessary to control gas temperature. The metal walls were those 
of a 2-in. tube 88f ins. long, set vertically in a brick flue and heated out- 
side by the products from a gas fire. The water flow was controlled partly 
by varying velocity in the tube as set and partly by varying the area of the 
water passage in the tube, while surface conditions of the tube on the gas side 
were also varied by applying in one case cast-iron rings of the Foster steam 
superheater elements. The first series was run with a perfectly plain tube 



HEAT AND MATTER 



559 



supplying varying quantities of water while holding the gas temperature 
constant, then the process was repeated with a higher gas temperature, later 
a plugged 1-in. standard pipe was inserted as a core and the same repeated, and 
finally a helical strip of metal substituted to give the water a rotary movement. 
Some of the results are plotted in Fig. 167, showing the relation of U to water 
velocity up to 160 ft. per minute, and these are representative. They show 
a small increase in U with water velocity as might be expected, as there is surely 




80 100 

Velocity of Water Ft. Per Min. 



120 



^M40 



Fig. 167. — Variation of Heat Transfer Coefficient with Velocity of Water when Water Receives 
Heat from a Gas, Illustrating Substantial Constancy and Proving the Controlling Resist- 
ance to be on the Gas Side. 



some water resistance but hardly enough to be worth while, for example, the 
increase along any one of the curves is less than 1 B.T.U. per hour per square 
foot per degree, and all values for whatever water velocity or tube condition 
lie between U = 2, and £7 = 13. There are two influences affecting U more than 
water velocity, and these are gas temperatures and the outside condition of 
the tube. For the plain tube the use of gas temperatures from 1000° F. to 
1700° F. had the effect of approximately doubling the value of U, but it is 
not possible to say whether this effect is due inherently to temperature or to 



560 



ENGINEERING THERMODYNAMICS 



increased gas velocity which accompanies it, or to some other related quantity 
such as perhaps gas density. In every case the jacketed tube gave much 
lower values than the un jacketed, calculating surface as that exposed to the 
water and not that exposed to the gas, or about the same when calculating 
surface as that exposed to the gas, illustrating well the constancy of gas film 
resistance for equal gas flow and temperature conditions. The whole series 
demonstrates absolute^ the controlling character of the resistance on the gas 
side and leads to the conclusion that whatever variations in U may be found 
are due to varying film resistance, and may be related to velocity of fluid 
directly or indirectly, and the situation appears to be somewhat as illustrated 
in Fig. 168. 

Temperature Degrees F. 

§ g 

J I : J L 




Fig. 168. — Probable Relations between the Heat Resistances or Drops in Temperature when 
Heat Passes from Hot Gas to Water. 

Another series of experiments carried out in great detail by H. P. Jordan, 
three years ago, on the transmission of heat from hot air to water confirms 
the above conclusions but extends them, as the amount and velocity of the air 
were accurately determined and the transmission phenomenon divided into 
into two parts, first from air to metal and second from metal to water. The 
superior resistance on the gas side is shown by the curve of Fig. 169, plotted 
from one experiment, by the temperature drops, which on the air side was 246° 
F. with only 6.7° F. on the water side, that through the metal itself being 
negligible. In every case the rate of transmission from the hot gas to metal 
is found to increase with increase of flow, and here a most striking relation 
is demonstrated by an appropriate selection of the prime variable represent- 
ing rate of flow. When this variable is taken as pounds of air per squre foot 
of area of cross-section of air passage per second, the B.T.U. per hour per square 
foot per degree difference between air and metal relation, is linear and 
the curve a straight line as shown in Fig. 169, for one series. These lines all 
have the equation 

w 
U = A+B- (750) 



where U = B.T.U. per square feet per degree difference between gas and metal; 
w = pounds air per second; 
a = square foot cross-section area of air pasage. 



HEAT AND MATTER 



561 



In the above equation A and B are factors not the same numerically for 
different circumstances and found by Jordan to depend on the following 
conditions: Thus A, which is the zero flow value of U appears to be indepen- 
dent of both cross-section of channel and temperature, but probably fixed by 
the condition of the surface as to cleanliness or smoothness, while B, which 
measurjs the rate of increase in U with flow, seems to be dependent on both 
the air temperature and on the dimensions of the air passage and given by 



















5 


= C 


+c' 


i+> 


Z tm 


j 


















80 








































(2) 




O 
0) 
























• 












3* 


























J&s 


/ 






















B 
o 
Q20 

3 
























,y 
































tf 


^ 


/ 


























d-15 
u 














/ 






































y 


/ 








































,} 


^ 






vir - 








































































H 5 


"^y 


f 








































/ 










































ii 













































3 4 5 6 7 

Lbs. Air Per Sq.Ft. Passage Per Second 



10 



Fig. 169. — Variation of Heat Transfer Coefficient with Rate of Flow per sq.ft. of Stream 
Cross-section of the Gas (Air), when Water Receives Heat from a Gas, Illustrating Linear 
Relation between U and Gas Mass Flow. 

where c, c' and c" are constant, t m the arithmetical mean of the air and 
metal temperature, and 

, -, ,. , ,, . . , area of flow of channel. 

<? = mean hydraulic depth in inches =■ . (751) 

perimeter of cooling surface 

Substituting the values of the constants and factors he found that 

U w 

— - = .0015+[.000506 -.00045?+. 00000165k, ]-. 

oOUU • d 

This gives the coefficient of heat transfer in terms of all variables investigated, 
including weight of flow of gas in pounds per second, cross-section of channel, 
shape of channel or mean hydraulic depth and temperature of air and metal. 
The laws are summarized by Jordan as follows: 

(a) For a constant mass flow f - = c) the rate of transfer is proportional 
to temperature difference directly; 



562 ENGINEERING THERMODYNAMICS 

(fe) For a given temperature difference (t m = c) the rate of transfer increases 
with speed by a linear law; 

(c) For a given rate of flow and temperature difference the rate of transfer 

increases with the value of the temperature ; 

(d) The rate of transfer depends on the condition of surface; 

(e) The rate of transfer depends on the size of channel, and the smaller 

area 

the ratio, ; = q, the greater the rate of transfer. 

perimeter 

These experiments, as well as another series by Dr. T. E. Stanton, fourteen 
years ago, on the transfer from water to water through metal, may be regarded 
as confirming the theoretical predictions on the laws of transfer as announced 
by Prof. Osborne Reynolds nearly forty years ago. He gave on purely theoretic 
grounds a linear law of relation between rate of transfer per hour per degree and 
the product of density and velocity of fluid, but this product is equal to the weight 
per second per square foot of cross-section or the mass flow as it may be called. 

The conclusion that the coefficient of heat transmission when a gas is present 
on one side or the other increases in a linear law manner with mass flow is a 
most important one, because the greatest practical use of this sort of trans- 
mission is made in the steam boiler, the laws for which have for over a century 
been a subject of controversy and doubt, from which some order and agree- 
ment is beginning to appear. In the case of the boiler the temperatures on 
the two sides of the surface are unknown, greatest variations and uncertainty 
are found on the gas or flame side. This being the case it is quite impossible 
to find, or if found for one case, to use in engineering practice, any particular 
value of the coefficient per hour per degree per square foot. It is possible, 
comparatively easy, and entirely practicable, to discuss and use the rate of trans- 
mission per hour per square foot, and this evidently in any case will also bear 
a linear relation to the quantity of gas per hour being discharged through the flues 
or tubes. This is proved by all accurate tests on boilers in existence and which 
will be taken up in the next Chapter, as the subject is important enough to 
warrant a more detailed treatment. 

The next important class of heat transfer cases in view of practical applica- 
tions in standard engineering apparatus is the transfer from condensing steam 
to moving water, typical of exhaust steam feed-water heaters, surface con- 
densers, steam-heated kettles and similar equipment. This has long been a 
subject of experimental investigation and there are available many test results, 
all of which do not, however, agree in detail, though all do agree in assigning 
increasing values to U with water velocity, usually making it proportional 
to some power of the water velocity, Ser and Joule the one-third power, Hege- 
man the one-half power and Stanton the first power. In some of the work 
attempts have been made to relate U to the steam velocity or mass flow of steam 
approaching the condensing surface as well, but these are in the main unsuc- 
cessful. One great source of variation in U noticed by some and ignored 
by others as peculiar to this class, is the collection of non-condensible gases in 
bubbles and films on the condensing surface, preventing actual vapor-metal 






HEAT AND MATTER 



563 



contact and interposing great heat resistance. In general it may be said that 
for this class the principal resistance is on the water side and the steam cai? 
condense as fast as the water is able to carry heat away so that increase of 
water velocity always increases U by decreasing the film thickness and cor- 
responding heat resistance. The value of U is always higher than for cases 
where gases are present, because liquid film resistances are always less than those 




3 4 5 

Velocity of circulating water-ft. persec.=V«, 



Fig. 170. 



-Variation in Heat Transfer Coefficient with Water Velocity, when Water Receives 
Heat from Condensing Steam (Orrok). 

for gases of equal thickness. When gases collect on the condensing side the 
principal resistance may no longer rest in the water film, but may be trans- 
ferred to the vapor side, in which case increase of water velocity results in 
little or no gain. 

In no series of experiments has more care been exercised to get accurate 
data with a full understanding beforehand of the difficulties and interferences 



564 



ENGINEEEING THERMODYMAMICS 



to be overcome than those of Orrok made a year or so ago. In this series 
extraordinary precautions were taken to exclude air and other gases, and steam 
generated from a constant mass of water in a closed system was condensed 
under vacuum on tubes of different compositions and surface conditions sup- 
plied with water of varying velocity. The results reported are consistent and 
reliable as no others have been, probably for this reason, but are not applicable 

80,000 [-1600-1 1 -i 1 1 1 1 1 1 1 1-33- 



70,000 



60,000 



.50,000 



£.40,000 



30,000 



20,000 



10,000 




4 5 6 7 8 9 10 11 

Velocity of circulating water-f t. per sec.=V» 



Fig. 171.— Heat Transfer Coefficient as a Function of Water Velocity for Steam Condensers, 
Showing Curve of Accepted Law and Experimental Points by Orrok. 

to commercial steam generated by fresh supplies of water carrying gases in 
solution, as all water supplies do, without correction or allowance for the increase 
of heat resistance due to collection of gas films on condensing surfaces. 

The curves of Fig. 170, reproduced from Orrok's paper, give a graphic sum- 
mary of the value of U in terms of water velocity as determined by various 
experimenters and includes his own result. These crossing lines of different 
curvature and slope prove conclusively the complex nature of the case under 



HEAT AND MATTER 



565 



examination and the fact that U cannot be regarded as a function of water velo- 
city alone. More in detail Orrok's results for the two vacua 15 and 27 ins. Hg 
are given in Fig. 171, curves (.4) and (B), being the values of U and (C), that of 
B.T.U. per square foot per hour or UXt m and it should be noted how widely 
distributed are the points about the curve in spite of great care in testing. 
All these curves, as located in the bands of points, follow the square root of 
velocity law having the equations, 



U = S0SVV W for 27 ins. vacuum. 
U = 224:VVw for 15 ins- vacuum. 



(752) 




12 3 4 5 6 7 

Water velocity. -.ft. per sec. 

Fig. 172. — Comparison of Values of U for Surface Condensers. 



These are compared with the work of the more reliable experimenters whose 
numbers are available for recalculation of results in Fig. 172. The effect of 
steam velocity, amount of air present in the steam as represented by the partial 
pressure of the air and of the nature of the tube were also investigated, the 
results being summed up in the following equation, 



U=K 



C'pV 



tj 



(753) 



[L is material coefficient 



566 ENGINEERING THERMODYNAMICS 

Where C is a cleanliness coefficient varying from 1.00 to .5; 

1.00 for copper; 

.98 for admiralty metal; 

.97 for admiralty aluminum lined; 

.92 for admiralty black oxidized; 

.87 for admiralty aluminium bronze; 

.80 for cuprous nickel; 

.79 for tin or admiralty lead lined; 

.75 for zinc; 

.74 for monel metal; 

.63 for Shelby steel; 

.55 for admiralty badly corroded; 

.47 for admiralty inside vulcanized; 

.25 for glass; 

.17 for admiralty vulcanized both sides; 

u x- £ Partial pr. steam , . , _ _ 

p = ratio of — ^ , . — and varying from 1.00 to 0. 

Total pressure 

" V d = water velocity feet per second; 

" t m = mean temperature difference; i 

K = a constant = 630 approximately. 

For high vacuum condensers the standard vacuum is 28 ins. and guarantees 
are usually based on this and on a circulating water inlet temperature of 70° F. 
with 20° F. allowable rise. For these conditions 



C/ = 435C 



m 



The effect of air or other gases deposited on the condensing surface on the 
rate of heat transmission was investigated most thoroughly by J. A. Smith, 
about six years ago, and his results are the best available. He finds that enough 
gases to be equivalent to a partial pressure of 20 m - Hg will at 90° F. decrease 
the transmission 25 per cent, and ^ in. Hg, 50 per cent. 

Probably the most complex of the heat transmission cases so far as con- 
cerns the variability of U is that in which the two fluids are the same in kind 
and both suffer a change of temperature, and these include heat passage from 
liquid to liquid, or gas to gas. For all of these it is quite impossible to fix 
off-hand the location of the controlling resistance, as it may be on either side, 
and with variations in flow may pass alternately from one side to the other, as 
in liquid heat exchangers, steam superheaters and similar apparatus. As 
there are no available data covering a sufficient range of conditions, about the 
best that can be done is to analyze such cases by trial with assumed metal 
temperature, taking Jordan's data for air relating U to mass flow and Orrok's 
for water relating U to the square root of the water velocity. These cases are 
not so common or so important in practice, which probably accounts for lack 
of data, but they are quite important enough tq warrant some trouble in 
establishing experimentally the needed laws. 



HEAT AND MATTER 567 



18. GENERAL PROBLEMS ON CHAPTER IV 

Prob. 1. What will be the final temperature of all the substances if 10 lbs. of iron 
at a temperature of 800° F. and 25 lbs. of stone at a temperature of 1000° F. are thrown 
into a tank containing 1000 lbs. of water at 40° F. and 10 lbs. of ice? 

Prob. 2. A room has a volume of 60,000 cu.ft. and the air in it is changed ten 
times per hour. The entering air is warmed by steam coils from a temperature 
of 20° F. to 60° F. The steam enters the coils with a pressure of 3 lbs. per square inch 
gage and containing 10 per cent of moisture. The condensed steam leaves at a tem- 
perature of 100° F. How many pounds of steam per hour will be required to heat the 
room? 

Prob. 3. How many cubic feet will a pound of saturated ammonia vapor contain 
at —15° C. and what will be the pressure at this temperature? What will be the 
absolute F. and C. temperature corresponding to a pressure of 100 lbs. per square inch 
gage pressure for steam, ammonia vapor, and carbon dioxide vapor? 

Prob. 4. If a glass jar held exactly T l of a gal. of water at 32° F. what would be 
its capacity at 100° F. Should it be full of water at 60° F. how much more could be 
added when the temperature was 39° F. and how much would have spilled over when 
the temperature had risen to 100° F.? 

Prob. 5. The unit elongation of an iron bar is 25Xl0 6 Xunit stress. What will 
be the force exerted by an iron beam of 7.5 sq.ins. cross-section when heated from 
0° F. to 100° F.? 

Prob. 6. A room full of air is heated from 20° F. to 100° F. How much air 
was forced out upon being heated? 

Prob. 7. An ammonia refrigerating machine has a condenser in which ammonia 
vapor at a pressure of 150 lbs. per square inch absolute and at a temperature of 225° 
F. is converted into liquid by means of water supplied at 70° F. passed over the tubes 
of the condenser. If each pound of water surfers a rise in temperature of 20° F. 
and the liquid ammonia is cooled to 60° F., how many pounds of water per pound 
of ammonia will be needed? 

Prob. 8. To produce one ton of refrigeration in twenty-four hours it is necessary 
to absorb 200 B.T.U. per minute. What will be the capacity of a machine in tons, 
in the coils of which 100 lbs. of liquid ammonia are evaporated per minute? The 
ammonia enters the coils at a temperature of 60° F., the pressure in the coils is 25 
lbs. per square inch absolute and the vapor leaves the coils at a temperature of 20° F. 

Prob. 9. A thermometer placed in a well in a pipe carrying saturated steam reads 
350° F. when half the scale (50° F. to 400° F.) is exposed. What will be the difference 
between the pressure due to reading as corrected for exposed stem as advised by the 
Bureau of Standards and that due to the reading as taken? 

Prob. 10. With a boiler efficiency of 70 per cent, 1 lb. of a certain grade of 
coal made 7.5 lbs. of steam at 100 lbs. per square inch gage from feed water at 60° F., 
the steam having 50° of superheat. If it were possible to convert 30 per cent of the 
heat of the steam directly into work, what would be the thermal efficiency referred to 
the coal and the water rate of the engine? 

Prob. 11. A volumetric analysis of boiler flue gases gave the following result: 
Carbon dioxide = 12 per cent; carbon monoxide =2 per cent; oxygen, 7 per cent; nitrogen 
79 per cent; what would be the error in the calculation of the sensible heat carried up 



568 ENGINEERING THERMODYNAMICS 

the stack on the assumption that the specific heat was the same as that of air rather 
than its proper value? What would be the total heat carried up the stack for 1000 lbs. 
of flue gases for a room temperature of 80° F. and a stack temperature of 480° F.? 

Prob. 12. The water level in a hot water tank is shown by a gage glass. The 
glass shows the level to be 6 ft., but the temperature of the water in glass is 80° F., 
while that in tank is 210° F. What is the true height of water in tank? 

Prob. 13. The mean temperature of the gases in a chimney is 350° F., while the 
outside air is 40° F. For a stack 150 ft. high what would be maximum draft possible? 
In another similar stack the gases are used to warm water in tubes at the base of 
the chimney and the average temperature is reduced to 280° F. What would be the 
loss of draft? 

Prob. 14. A thousand cubic feet of air are compressed from one to 10 atmospheres. 
If half is so compressed that the heat is removed sufficiently fast to prevent any rise 
in temperature and the other half in such a way that no heat is removed, what will 
be the sum of the volumes at end of process before and after mixing? 

Prob. 15. The barometer reading at a certain time is 29.5 ins. of Hg, the tem- 
perature is 60° F. and the air is saturated with water vapor. What is the pressure 
of air alone and the water content of the air? How much water will be lost by cooling 
to 20° F.? 

Prob. 16. Ammonia vapor leaves the coils of an ice plant at a temperature of 
10° F., at which temperature it is superheated 10°. What pressure was there in the 
coils and how much heat per pound NH3 was taken up if the liquid NH3 had orig- 
inally a temperature of 60° F. 

Prob. 17. Carbon dioxide gas is being compressed into a cylinder at a temperature 
of 60° F. At what pressure will condensation occur and how much heat must be 
removed per pound of stuff to cause this action? 

Prob. 18. Steam is being generated at a pressure of 100 lbs. per square inch gage 
and at the rate of 100 lbs. per minute. What horse-power is being developed in over- 
coming the external resistance to the formation of the steam? 

Prob. 19. The following is an analysis by volume of a sample of water gas. How 
much air would be needed to burn 1 cu.ft. of it and what would be the products 
resulting from such combustion by weight and by volume and the heat of combustion 
of the gas per pound and per cubic foot? 

CO =25.3; H 2 =9.2; CH 4 =3.1; C 2 H 4 = .8; C0 2 =3.4; N 2 =58. 

Prob. 20. Upon analysis a pound of Pocahontas coal yielded the following results 
by weight. C=84.9; H=4.2; 0=2.8. What would result upon burning a pound 
of this coal in heat developed and composition of products? 

Prob. 21. Why should the specific heat at constant volume differ from that at 
constant pressure and what is the relation between them for any assumed gas? 

Prob. 22. The walls of an ice storage room are composed of material through which 
the rate of heat transfer is .5 B.T.U. per square foot per degree difference per hour. 
If the room is 100x100x15 ft., what must be the pounds of ammonia evaporated 
at a pressure of 15 lbs. per square inch absolute to maintain a temperature of 25° F. 
within when the temperature without is 80° F.? The ammonia liquid enters the 
coils at 60° F. and the vapor leaves in a saturated condition. How much coil 
surface is needed? 

Prob. 23. A boiler is supplied with fuel, the ultimate analysis of which showed 
70 per cent C, 15 per cent H, and the remainder ash and moisture. A similar boiler 



HEAT AND MATTER 569 

is supplied with natural gas containing 99 per cent of CH 4 . Assuming that both have 
an efficiency of 70 per cent, and that the heats of combustion of the fuels equal the 
heats of formation of the products, how many pounds of steam at 100 lbs. per square 
inch absolute pressure could be made from feed water at 60° F. per pound of each fuel? 
Prob. 24. A boiler horse-power is given as the heat equivalent 34.5 lbs. of water 
evaporated from and at 212° F. To develop 1000 H.P. with a feed-water tempera- 
ture of 200° F. and a steam pressure of 200 lbs. per square inch gage would require how 
many feet of heating surface based on a transfer rate of 3 and a mean gas temperature 
of 800° F.? 

Prob. 25. At a given time the dry-bulb thermometer reads 75° F. and a wet-bulb 
thermometer reads 60° F. What is the relative humidity and how much would the 
temperature have to fall to produce rain? 

Prob. 26. An ammonia absorption machine is producing 10 tons of ice per day 
from water at 50° F., the ice being cooled to 25° F. The brine is cooled by evaporation 
of ammonia at a pressure of 10 lbs. per square inch gage. How much water will be 
required to remove the heat from the absorption if each pound is allowed to rise in 
temperature 20° F.? What will be the maximum per cent of ammonia in the 
solution in the absorber if the pressure be 10 lbs. gage and the temperature 
80° F.? 

Prob. 27. The following data were taken during a test of a feed-water heater; 
what was the rate of heat transfer per square foot of surface per degree mean temperature 
difference per hour? What was the boiler horse-power equivalent of the heat? Heat- 
ing surface 11 sq.ft. Steam pressure in heater 2 lbs. per square inch gage. Steam 
10% wet. Pounds of water heated per hour 3800. Temperature of inlet 
water 44° F. Temperature of water at end of successive passes, 78°, 106°, 133°, 154° 
170°, 180°, 202°. 

Prob. 28. Fuel containing 70 per cent C, 15 per cent H and 15 per cent ash is 
burned under a boiler. 20 per cent of the heat developed goes to the stack. If there 
is twice as much air supplied as is chemically needed, what will be the temperature 
of the flue gases leaving the boiler and its volumetric analysis cold? 

Prob. 29. The gases from above boiler pass through an economizer in which the 
flow of fluids may be considered as counter current. 3000 lbs. of water per hour are 
heated from a temperature of 50° F. Assuming 4 lbs. of gases per pound of water 
and the specific heat of the former to be .25, what will be the surface required for 
every 20° rise from inlet temperature to 230°? 
Note: Let C/=3.5. 

Prob. 30. 400 lbs. of coal are burned on a boiler grate per hour and during the 
same time 4000 lbs. of water are passed through an economizer. If the water is raised 
in temperature 200° and the gas is cooled 300°, what is the weight of flue gas per pound 
of coal burned, assuming the sp.ht. of the gases to be .25? 

Prob. 31. A pound of coal contains 14,000 heat-units. If 70 per cent of these are 
absorbed by the boiler, how many pounds of steam may be made per pound of coal 
at a pressure of 150 lbs. gage from feed-water at a temperature of 60° F.? 

Prob. 32. 60,000 cu.ft. of air are taken into a building per hour. Assuming that 
the air is at 40° F. and the humidity to be 7 per cent, how many pounds of moisture 
will the air contain? 

Prob. 33. An ammonia refrigerating machine is required to remove 1,000,000 B.T.U. 
per hour from a cold-storage room. The liquid ammonia coming to the expansion valve 
has a temperature of 60°. The pressure in the vaporizing coils is 10 lbs. per square 



570 ENGINEERING THERMODYNAMICS 

inch absolute and the temperature in the coils is 25° F. How many cubic feet of ammonia 
vapor per minute must the compressor handle? 

Prob. 34. The temperature of the outside air is 40° F. The average temperature 
of the gases in a chimney is 350° F. The gases may be considered as 80 per cent nitrogen, 
10 per cent oxygen, and 10 per cent carbon dioxide. How high must be the stack to 
give a hypothetical draught of 1 inch of water? 

Prob. 35. By the use of a feed-water heater the temperature of the water fed to a 
boiler is raised from 50° to 200°. The pressure in the boiler is 100 lbs. gage and the 
steam has 5 per cent moisture. What is the percentage of heat saved by the heater? 

Prob. 36. Steam is generated in a boiler at a pressure of 100 lbs. gage from feed- 
water having a temperature of 60° F. When the steam leaves the boiler it has 5 per 
cent moisture, and is carried to a separately fired superheater from which it leaves 
with 100° superheat. What per cent of the total heat in the superheated steam comes 
from the boiler and what per cent from the superheater? 

Prob. 37. A volume of 50 cu.ft. contains 5 lbs. of hydrogen and 10 lbs. of nitrogen 
at a temperature of 500° absolute. Find the pressure due to the hydrogen and to the 
nitrogen. 

Prob. 38. A mixture of air and saturated water-vapor has a temperature of 55° F. 
The total pressure is 1 atmosphere. What is the pressure due to the air and due to 
the moisture? 

Prob. 39. A wet bulb hygrometer gives readings of 70° and 48°. What is the 
humidity? 

Prob. 40. A certain illuminating gas is .8 as heavy as air. What will be its value 
of R? 

Prob. 41. The specific heat of a certain gas at constant pressure is 2.8. The value 
for R=7Q0. Find the value of its specific heat at constant volume. Also the value 
of y. 

Prob. 42. The temperature in a condenser is 120° F. The vacuum shows 23 ins. 
of Hg with the barometer standing at 30 ins. What is the pressure of the air in the 
condenser? 

Prob. 43. A pound of air is cooled from atmospheric pressure and 32° F. to —310° F. 
If the cooling is at constant volume, what will be the final pressure? 



TABLES 

Table XXIX 

TEMPERATURES, CENTIGRADE AND FAHRENHEIT 



571 



c. 


F. 


C. 


F. 


C. 


F. 


C. 


F. 


C. 


F. 


C. 


F. 


C. 


F. 


-40 


-40. 


26 


78.8 


92 


197.6 


158 


316.4 


224 


435.2 


290 


554 


950 


1742 


-39 


-38.2 


27 


80.6 


93 


199.4 


159 


318.2 


225 


437. 


300 


572 


960 


1760 


-38 


-36.4 


28 


82.4 


94 


201.2 


160 


320. 


226 


438.8 


310 


590 


970 


1778 


-37 


-34.6 


29 


84.2 


95 


203. 


161 


321.8 


227 


440.6 


320 


608 


980 


1796 


-36 


-32.8 


30 


86. 


96 


204.8 


162 


323.6 


228 


442.4 


330 


626 


990 


1814 


-35 


-31. 


31 


87.8 


97 


206.6 


163 


325.4 


229 


444.2 


340 


644 


1000 


1832 


-34 


-29.2 


32 


89.6 


98 


208.4 


164 


327.2 


230 


446. 


350 


662 


1010 


1850 


-33 


-27.4 


33 


91.4 


99 


210.2 


165 


329. 


231 


447.8 


360 


680 


1020 


1868 


-32 


-25.6 


34 


93.2 


100 


212. 


166 


330.8 


232 


449.6 


370 


698 


1030 


1886 


-31 


-23.8 


35 


95. 


101 


213.8 


167 


332.6 


233 


451.4 


380 


716 


1040 


1904 


-30 


-22. 


36 


96.8 


102 


215.6 


168 


334.4 


234 


453.2 


390 


734 


1050 


1922 


-29 


-20.2 


37 


98.6 


103 


217.4 


169 


336.2 


235 


455. 


400 


752 


1060 


1940 


-28 


-18.4 


38 


100.4 


104 


219.2 


170 


338. 


236 


456.8 


410 


770 


1070 


1958 


-27 


-16.6 


39 


102.2 


105 


221. 


171 


339.8 


237 


458.6 


420 


788 


1080 


1976 


-26 


-14.8 


40 


104. 


106 


222.8 


172 


341.6 


238 


460.4 


430 


806 


1090 


1994 


-25 


-13. 


41 


105.8 


107 


224.6 


173 


343.4 


239 


462.2 


440 


824 


1100 


2012 


-24 


-11.2 


42 ' 


107.6 


108 


226.4 


174 


345.2 


240 


464. 


450 


842 


1110 


2030 


-23 


- 9.4 


43 


109.4 


109 


228.2 


175 


347. 


241 


465.8 


460 


860 


1120 


2048 


-22 


- 7.6 


44 


111.2 


110 


230. 


176 


348.8 


242 


467.6 


470 


878 


1130 


2066 


-21 


- 5.8 


45 


113. 


111 


231.8 


177 


350.6 


243 


469.4 


480 


896 


1140 


2084 


-20 


- 4. 


46 


114.8 


112 


233.6 


178 


352.4 


244 


471.2 


490 


914 


1150 


2102 


-19 


- 2.2 


47 


116.6 


113 


235.4 


179 


354.2 


245 


473. 


500 


932 


1160 


2120 


-18 


- 0.4 


48 


118.4 


114 


237.2 


180 


356. 


246 


474.8 


510 


950 


1170 


2138 


-17 


+ 1.4 


49 


120.2 


115 


239. 


181 


357.8 


247 


476.6 


520 


. 968 


1180 


2156 


-16 


3.2 


50 


122. 


116 


240.8 


182 


359.6 


248 


478.4 


530 


986 


1190 


2174 


-15 


5. 


51 


123.8 


117 


242.6 


183 


361.4 


249 


480.2 


540 


1004 


1200 


2192 


-14 


6.8 


52 


125.6 


118 


244.4 


184 


363.2 


250 


482. 


550 


1022 


1210 


2210 


-13 


8.6 


53 


127.4 


119 


246.2 


185 


365. 


251 


483.8 


560 


1040 


1220 


2228 


-12 


10.4 


54 


129.2 


120 


248. 


186 


366.8 


252 


485.6 


570 


1058 


1230 


2246 


-11 


12.2 


55 


131. 


121 


249.8 


187 


368.6 


253 


487.4 


580 


1076 


1240 


2264 


-10 


14. 


56 


132.8 


122 


251.6 


188 


370 4 


254 


489.2 


590 


1094 


1250 


2282 


- 9 


15.8 


57 


134.6 


123 


253.4 


189 


372.2 


255 


491. 


600 


1112 


1260 


2300 


- 8 


17.6 


58 


136.4 


124 


255.2 


190 


374. 


256 


492.8 


610 


1130 


1270 


2318 


- 7 


19.4 


59 


138.2 


125 


257. 


191 


375.8 


257 


494.6 


620 


1148 


1280 


2336 


- 6 


21.2 


60 


140. 


126 


258.8 


192 


377.6 


258 


496.4 


630 


1166 


1290 


2354 


- 5 


23. 


61 


141.8 


127 


260.6 


193 


379.4 


259 


498.2 


640 


1184 


1300 


2372 


- 4 


24.8 


62 


143.6 


128 


262.4 


194 


381.2 


260 


500. 


650 


1202 


1310 


2390 


- 3 


26.6 


63 


145.4 


129 


264.2 


195 


383. 


261 


501.8 


660 


1220 


1320 


2408 


- 2 


28.4 


64 


147.2 


130 


266. 


196 


384.8 


262 


503.6 


670 


1238 


1330 


2426 


- 1 


30.2 


65 


149. 


131 


267.8 


197 


386.6 


263 


505.4 


680 


1256 


1340 


2444 





32. 


66 


150.8 


132 


269.6 


198 


388.4 


264 


507.2 


690 


1274 


1350 


2462 


+ 1 


33.8 


67 


152.6 


133 


271.4 


199 


390.2 


265 


509. 


700 


1292 


1360 


2480 


2 


35.6 


68 


154.4 


134 


273.2 


200 


392. 


266 


510.8 


710 


1310 


1370 


2498 


3 


37.4 


69 


156.2 


135 


275. 


201 


393.8 


267 


512.6 


720 


1328 


1380 


2516 


4 


39.2 


70 


158. 


136 


276.8 


202 


395.6 


268 


514.4 


730 


1346 


1390 


2534 


5 


41. 


71 


159.8 


137 


278.6 


203 


397.4 


269 


516.2 


740 


1364 


1400 


2552 


6 


42.8 


72 


161.6 


138 


280.4 


204 


399.2 


270 


518. 


750 


1382 


1410 


2570 


7 


44.6 


73 


163.4 


139 


282.2 


205 


401. 


271 


519.8 


760 


1400 


1420 


2588 


8 


46.4 


74 


165.2 


140 


284. 


206 


402.8 


272 


521.6 


770 


1418 


1430 


2606 


9 


48.2 


75 


167. 


141 


285.8 


207 


404.6 


273 


523.4 


780 


1436 


1440 


2624 


10 


50. 


76 


168.8 


142 


287.6 


208 


406.4 


274 


525.2 


790 


1454 


1450 


2642 


11 


51.8 


77 


170.6 


143 


289.4 


209 


408.2 


275 


527. 


800 


1472 


1460 


2660 


12 


53.6 


78 


172.4 


144 


291.2 


210 


410. 


276 


528.8 


810 


1490 


1470 


2678 


13 


55.4 


79 


174.2 


145 


293. 


211 


411.8 


277 


530.6 


820 


1508 


1480 


2696 


14 


57.2 


80 


176. 


146 


294.8 


212 


413.6 


278 


532.4 


830 


1526 


1490 


2714 


15 


59. 


81 


177.8 


147 


296.6 


213 


415.4 


279 


534.2 


840 


1544 


1500 


2732 


16 


60.8 


82 


179.6 


148 


298.4 


214 


417.2 


280 


536. 


850 


1562 


1510 


2750 


17 


62.6 


83 


181.4 


149 


300.2 


215 


419. 


281 


537.8 


860 


1580 


1520 


2768 


18 


64.4 


84 


183.2 


150 


302. 


216 


420.8 


282 


539.6 


870 


1598 


1530 


2786 


19 


66.2 


85 


185. 


151 


303.8 


217 


422.6 


283 


541.4 


880 


1616 


1540 


2804 


20 


68. 


86 


186.8 


152 


305.6 


218 


424.4 


284 


543.2 


890 


1634 


1550 


2822 


21 


69.8 


87 


188.6 


153 


307.4 


219 


426.2 


285 


545. 


900 


1652 


1600 


2912 


22 


71.6 


88 


190.4 


154 


309.2 


220 


428. 


286 


546.8 


910 


1670 


1650 


3002 


23 


73.4 


89 


192.2 


155 


311. 


221 


429.8 


287 


548.6 


920 


1688 


1700 


3092 


24 


75.2 


90 


194. 


156 


312.8 


222 


431.6 


288 


550.4 


930 


1706 


1750 


3182 


25 


' 77. 


91 


195.8 


157 


314.6 


223 


433.4 


289 


552.2 


940 


1724 


1800 


3272 

1 



572 



ENGINEERING THERMODYNAMICS 

Tablf, XXIX — Continued 

TEMPERATURES, FAHRENHEIT AND CENTIGRADE 



F. 


C. 


F. 


C. 


F. 


C. 


F. 


C. 


F. 


C. 


F. 


C. 


F. 


C. 


-40 


-40. 


26 


- 3.3 


92 


33.3 


158 


70. 


224 


106.7 


290 


143.3 


360 


182.2 


-39 


-39.4 


27 


- 2.8 


93 


33.9 


159 


70.6 


225 


107.2 


291 


143.9 


370 


187.8 


-38 


-38.9 


28 


- 2.2 


94 


34.4 


160 


71.1 


226 


107.8 


292 


144.4 


380 


193.3 


-37 


-38.3 


29 


- 1.7 


95 


35. 


161 


71.7 


227 


108.3 


293 


145. 


390 


198.9 


-36 


-37.8 


30 


- 1.1 


96 


35.6 


162 


72.2 


228 


108.9 


294 


145.6 


400 


204.4 


-35 


-37.2 


31 


- 0.6 


97 


36.1 


163 


72.8 


229 


109.4 


295 


146.1 


410 


210. 


-34 


-36.7 


32 


0. 


98 


36.7 


164 


73.3 


230 


110. 


296 


146.7 


420 


215.6 


-33 


-36.1 


33 


+ 0.6 


99 


37.2 


165 


73.9 


231 


110.6 


297 


147.2 


430 


221.1 


-32 


-35.6 


34 


1.1 


100 


37.8 


166 


74.4 


232 


111.1 


298 


147.8 


440 


226.7 


-31 


-35. 


35 


1.7 


101 


38.3 


167 


75. 


233 


111.7 


299 


148.3 


450 


232.2 


-30 


-34.4 


36 


2.2 


102 


38.9 


168 


75.6 


234 


112.2 


300 


148.9 


460 


237.8 


-29 


-33.9 


37 


2.8 


103 


39.4 


169 


76.1 


235 


112.8 


301 


149.4 


470 


243.3 


-28 


-33.3 


38 


3.3 


104 


40. 


170 


76.7 


236 


113.3 


302 


150. 


480 


248.9 


-27 


-32.8 


39 


3.9 


105 


40.6 


171 


77.2 


237 


113.9 


303 


150.6 


490 


254.4 


-26 


-32.2 


40 


4.4 


106 


41.1 


172 


77.8 


238 


114.4 


304 


151.1 


500 


260. 


-25 


-31.7 


41 


5. 


107 


41.7 


173 


78.3 


239 


115. 


305 


151.7 


510 


265.6 


-24 


-31.1 


42 


5.6 


108 


42.2 


174 


78.9 


240 


115.6 


306 


152.2 


520 


271.1 


-23 


-30.6 


43 


6.1 


109 


42.8 


175 


79.4 


241 


116.1 


307 


152.8 


530 


276.7 


-22 


-30. 


44 


6.7 


110 


43.3 


176 


80. 


242 


116.7 


308 


153.3 


540 


282.2 


-21 


-29.4 


45 


7.2 


111 


43.9 


177 


80.6 


243 


117.2 


309 


153.9 


550 


287.8 


-20 


-28.9 


46 


7.8 


112 


44.4 


178 


81.1 


244 


117.8 


310 


154.4 


560 


293.3 


-19 


-28.3 


47 


8.3 


113 


45. 


179 


81.7 


245 


118.3 


311 


155. 


570 


298.9 


-18 


-27.8 


48 


8.9 


114 


45.6 


180 


82.2 


246 


118.9 


312 


155.6 


580 


304.4 


-17 


-27.2 


49 


9.4 


115 


46.1 


181 


82.8 


247 


119.4 


313 


156.1 


590 


310. 


-16 


-26.7 


50 


10. 


116 


46.7 


182 


83.3 


248 


120. 


314 


156.7 


600 


315.6 


-15 


-26.1 


51 


10.6 


117 


47.2 


183 


83.9 


249 


120.6 


315 


157.2 


610 


321.1 


-14 


-25.6 


52 


11.1 


118 


47.8 


184 


84.4 


250 


121.1 


316 


157.8 


620 


326.7 


-13 


-25. 


53 


11.7 


119 


48.3 


185 


85. 


251 


121.7 


317 


158.3 


630 


332.2 


-12 


-24.4 


54 


12.2 


120 


48.9 


186 


85.6 


252 


122.2 


318 


158.9 


640 


337.8 


-11 


-23.9 


55 


12.8 


121 


49.4 


187 


86.1 


253 


122.8 


319 


159.4 


650 


343.3 


-10 


-23.3 


56 


13.3 


122 


50. 


188 


86.7 


254 


123.3 


320 


160. 


660 


348.9 


- 9 


-22.8 


57 


13.9 


123 


50.6 


189 


87.2 


255 


123.9 


321 


160.6 


670 


354.4 


- 8 


-22.2 


58 


14.4 


124 


51.1 


190 


87.8 


256 


124.4 


322 


161.1 


680 


360. 


- 7 


-21.7 


59 


15. 


125 


51.7 


191 


88.3 


257 


125. 


323 


161.7 


690 


365.6 


- 6 


-21.1 


60 


15.6 


126 


52.2 


192 


88.9 


258 


125.6 


324 


162.2 


700 


371.1 


- 5 


-20.6 


61 


16.1 


127 


52.8 


193 


89.4 


259 


126.1 


325 


162.8 


710 


376.7 


- 4 


-20. 


62 


16.7 


128 


53.3 


194 


90. 


260 


126.7 


326 


163.3 


720 


382.2 


- 3 


-19.4 


63 


17.2 


129 


53.9 


195 


90.6 


261 


127.2 


327 


163.9 


730 


387.8 


- 2 


-18.9 


64 


17.8 


130 


54.4 


196 


91.1 


262 


127.8 


328 


164.4 


740 


393.3 


- 1 


-18.3 


65 


18.3 


131 


55. 


197 


91.7 


263. 


128.3 


329 


165. 


750 


398.9 





-17.8 


66 


18.9 


132 


55.6 


198 


92.2 


264 


128.9 


330 


165.6 


760 


404.4 


+ 1 


-17.2 


67 


19.4 


133 


56.1 


199 


92.8 


265 


129.4 


331 


166.1 


770 


410. 


2 


-16.7 


68 


20. 


134 


56.7 


200 


93.3 


266 


130. 


332 


166.7 


780 


415.6 


3 


-16.1 


69 


20.6 


135 


57.2 


201 


93.9 


267 


130.6 


333 


167.2 


790 


421.1 


4 


-15.6 


70 


21.1 


136 


57.8 


202 


94.4 


268 


131.1 


334 


167.8 


800 


426.7 


5 


-15. 


71 


21.7 


137 


58.3 


203 


95. 


269 


131.7 


335 


168.3 


810 


432.2 


6 


-14.4 


72 


22.2 


138 


58.9 


204 


95.6 


270 


132.2 


336 


168.9 


820 


437.8 


7 


-13.9 


73 


22.8 


139 


59.4 


205 


96.1 


271 


132.8 


337 


169.4 


830 


443.3 


8 


-13.3 


74 


23.3 


140 


60. 


206 


96.7 


272 


133.3 


338 


170. 


840 


448.9 


9 


-12.8 


75 


23.9 


141 


60.6 


207 


97.2 


273 


133.9 


339 


170.6 


850 


454.4 


10 


-12.2 


76 


24.4 


142 


61.1 


208 


97.8 


274 


134.4 


340 


171.1 


860 


460. 


11 


-11.7 


77 


25. 


143 


61.7 


209 


98.3 


275 


135. 


341 


171.7 


870 


465.6 


12 


-11.1 


78 


25.6 


144 


62.2 


210 


98.9 


276 


135.6 


342 


172.2 


880 


471.1 


13 


-10.6 


79 


26.1 


145 


62.8 


211 


99.4 


277 


136.1 


343 


172.8 


890 


476.7 


14 


-10. 


80 


26.7 


146 


63.3 


212 


100. 


278 


136.7 


344 


173.3 


900 


482.-2 


15 


- 9.4 


81 


27.2 


147 


63.9 


213 


100.6 


279 


137.2 


345 


173.9 


910 


487.8 


16 


- 8.9 


82 


27.8 


148 


64.4 


214 


101 . 1 


280 


137.8 


346 


174.4 


920 


493.3 


17 


- 8.3 


83 


28.3 


149 


65. 


215 


101.7 


281 


138.3 


347 


175. 


930 


498.9 


18 


- 7.8 


84 


28.9 


150 


65.6 


216 


102.2 


282 


138.9 


348 


175.6 


940 


504.4 


19 


- 7.2 


85 


29.4 


151 


66.1 


217 


102.8 


283 


139.4 


349 


176.1 


950 


510. 


20 


- 6.7 


86 


30. 


152 


66.7 


218 


103.3 


284 


140. 


350 


176.7 


960 


515.6 


21 


- 6.1 


87 


30.6 


153 


67.2 


219 


103.9 


285 


140.6 


351 


,177.2 


970 


521. 


22 


- 5.6 


88 


31.1 


154 


67.8 


220 


104.4 


286 


141.1 


352 


177.8 


980 


526.7 


23 


- 5. 


89 


31.7 


155 


68.3 


221 


105. 


287 


141.7 


353 


178.3 


990 


532.2 


24 


- 4.4 


90 


32.2 


156 


68.9 


222 


105.6 


288 


142.2 


354 


178.9 


1000 


537.8 


25 


- 3.9 


91 


32.8 


157 


69.4 


223 


106.1 


289 


142.8 


355 


179.4 


1010 


543.3 



TABLES 

Table XXX 

HEAT AND POWER CONVERSION TABLE 



573 



Calorie 
Kilo °C. 


B.T.U. 
Lb. °F. 


Lb. ° C. 


Kilo °F. 


Calorie 
per Lb. 


B.T.U. 
per Lb. 


B.T.U. 
per Kilo. 


Calorie 
per Kilo. 


1. 

.252 
.4536 
.5556 


3.9683 
1. 
1.8 
2.2046 


2.2046 
.5556 
1. 
1.2261 


1.8 

.4536 
.8165 
1. 


1. 

.252 

.1143 

.4536 


3.9683 
1. 
.4536 

1.8 


8.7483 
2.2046 
1 
3.9683 


2.2046 

.5807 
.252 
1. 



Calorie 
per Cu. Ft. 


B.T.U. 
per Cu. Ft. 


Calorie 
per Liter. 


B.T.U. 
per Liter. 


1. 

.252 
28.317 

7.136 


3.9683 
1. 
112.37 
28.317 


.0353 
.0089 
1. 
.252 


.1402 

.0353 
3.9683 
1. 



Ft. -Lb. 



B.T.U. 



Calorie. 



Cent. Heat 
Unit, At. 



H.P. Sec. 



H.P. Min. 



H.P. Hour. 



1 

777.5 



1399.5 

550 
3.3X10* 
1.98X10 6 



1.268X10- 3 

1 

3.9683 

1.8 

.7074 

42.44 

2545 



324X10- 3 
.252 

1 
.4536 
.1783 
10.695 
641 



.18X10- 3 

.5556 

2.2046 

1 

.3931 

23.578 

1.413X10 3 



818X10- 

1.414 

5.61 

2.545 

1 

60 
3600 



.303X10" 4 
2.356X10" 2 
9.35 X10- 2 
4.24 X10- 2 
1.67 X10- 2 

1 

60 



5.05 XIO" 7 
3. 927X10-* 
1.558X10" 3 
.707X10" 3 
2.777X10" 4 
1.67 XIO" 2 
1 



574 



ENGINEERING THERMODYNAMICS 



Table 
SPECIFIC HEATS 







Atomi 


B cr -n 




Class. 


Substance. 


Weigh 
H=l 


t Specific 
Gravity. 


Authority. 




Aluminum 


26.9 


2.57 


Mallet 




Carbon (amorphous) 
Carbon graph. 


11.9 


9 






11.9 


9 2.10-2.32 


Smithsonian Tables 




Copper (cast) 


63.0 


7 8.8-8.95 


Smithsonian Tables 




Iron (pure) 


55.4 


1 7.85 


Smithsonian Tables 


Elements 












Iron (pure) 


55.4 


1 7.85 


Smithsonian Tables 




Lead (cast) 


205.4 


3 11.37 


Reich 




Mercury 


198,5 


14.18 


Mallet 




Nickel 


58.2 


I 8.65 


Smithsonian Tables 




Tin (cast) 


118.1 


7.29 


Mathiessen 




u Zinc (cast) 


64. Si 


S 7.05 


Smithsonian Tables 




Bronze 




8.75-9 


Smithsonian Tables 




Brass 




7.8-8.6 


Smithsonian Tables 




Brick work, Masonry 




1.84-2.3 


Smithsonian Tables 




Butter 




.865 


Smithsonian Tables 




Clay 




1.80-2.6 


Smithsonian Tables 




Coal 




1.2-1.5 


Smithsonian Tables 




Wood 




.4-1.2 


Smithsonian Tables 


Common substances • 


Glass 




2.4-2.8 


Smithsonian Tables 




Ice 




.9 


Smithsonian Tables 




Cast Iron 




6.8-7.5 


* 




Wrought Iron 




7.4-7.9 


* 




Marble 




2.5-2.8 


Smithsonian Tables 




Steel 




7.7-7.9 


* 




Sand 




1.45-1.6 


Smithsonian Tables 




Stone 




2.1-3.4 


* 



* Kent's Mechanical Engineers' Pocketbook. 



TABLES 



575 



XXXI 

OF SOLIDS 



Specific Heat. 


At Temperature. 


Specific Heat 

Calculated from 

Atomic 

Weights. 


Authority. 


C. 


F. 


.2089 
.2226 



20-100 
500 


-50 

+11 

977 

16-1000 

17 

300 



15 

300 

500 

720-1000 

1000-1200 

15 

200 

-78 to -40 

21-99 

500 

1000 

0-100 

16-197 

18 

200 

0-100 

15-98 


32 

68-212 

932 

32 

-58 

52 

1795 

61-1832 

62 

572 

32 

59 

572 

392 

1328-1832 

1832-2192 

59 

392 

-108 to -40 

69-210 

932 

1832 

32-212 

69-387 

64 

392 

32-212 

59-208 


.238 


Bontschew 
Bontschew 


.2739 




Bontschew 


.241 

.1138 

.1605 


.52 

.52 


Olsen 

Weber 

Weber 


.467 




Weber 


.310 

.0924 

.0985 


.102 


Dewar 

Naccari 

Naccari 


.1162 
.1091 


.117 


Olsen 
Naccari 


.1376 




Naccari 


.1765 

.218 

.1989 


.117 


Pionchon 
Pionchon 
Pionchon 


.0299 
.0324 


.031 


Naccari 
Naccari 


.0319 
.1084 
.1233 


.0323 
.11 


Regnault 

Voigt 

Tilden 


.1608 




Pionchon 


.0545 
.0538 


.052 


Bunsen 
Spring 
Naccari 
Naccari 


.0915 
.0996 


.099 


.0935 




Bunsen 


.0858 




Regnault 

Regnault 

* 


.0939 




About .2 








.55 






Siebel 


.197 








Regnault 

Regnault 
* 


.2-. 241 








.45-. 65 








.16-. 18 


- 






Regnault 
Regnault 
Regnault 
Regnault 
Regnault 

Regnault 

* 


.504 






.1298 








.1138 








.21 








.1165-. 1175 








.195 






.2-. 22 






* 











* Kent's Mechanical Engineers' Pocketbook. 



576 



ENGINEERING THERMODYNAMICS 





r/> 




P 




H-t 




P 




c 




1— 1 


1— 1 


hJ 


1— 1 






l=* 


y 


o 


X 


B 




I 


n 


w 


<1 




H 


u 




l=* 




t—1 




u 




w 




Ph 




au 



«g-g stj g^ 

02 ^ <J 



fl.g « 



O 

§1 



a ;S ^ ^ a 
s § I « 

aa AH 



p 

a 
.a & 



g .a .s 53 g g SbabSb h a S -8 "3d =i =i =3 =a 



fl a c 

a> 5 53 

QQ QQ 00 

I I S 

o o o 

5 £ F^ 



o o 



020 



<M 

CO 

a> oo <N 1 r* "* 



j H H ,1, 



cL 



CO 

CO 00 CO 00 00 

I> lO CO iO CO 



CO CO CO 
CO CO CO 



CO CO © 



§ o o 

CO O © _ C> iOrH 

CM 



^o 



00 o o o o 

*0 <N CO <N CO 

i-H o o o o 

<M i-l <N i-l <N 



W5 W5 lO 
l> l> t> 



COCO lO i> 00 CO OS <M CO 

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CO !>• »C 

CO © N >0 

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v 



TABLES 577 

Table XXXIII 

BAUME SPECIFIC GRAVITY SCALE 

Specific gravities are for 60° F. referred to water at same temperature as unity, at which 
temperature it weighs 62.34 lbs. per cubic foot. 
Tabular results are calculated from : 



Degrees Baume* = 



(145 \ 

145 — — ) for liquids lighter than water, 
specific gravity/ 

( — : 130) for liquids heavier than water. 

. \specific gravity / 



^specific gravity 
Relation between Specific Gravity and Baume 





.00 


.01 


.02 


.03 


.04 


.05 


.06 


.07 


.08 


.09 


Specific 






















G ravity 








Degrees Baume. 










.60 


103.33 


99.51 


95.81 


92.22 


88.75 


85.38 


82.12 


78.95 


75.88 


72. 90 1 


.70 


70.00 


67.18 


64.44 


61.78 


59.19 


56.67 


54.21 


51.82 


49.49 


47.221 


.80 


45.00 


42.84 


40.73 


36.68 


36.67 


34.71 


32.79 


30.92 


29.09 


27.30 1 


.90 


25.56 


23.85 


22.17 


20.54 


18.94 


17.37 


15.83 


14.33 


12.86 


11.411 


1.00 


10.00 




















1.00 


0.00 


1.44 


2. $4 


4.22 


5.58 


6.91 


8.21 


9.49 


10.74 


11. 97 2 


1.10 


13.18 


14.37 


15.54 


16.68 


17.81 


18.91 


20.00 


21.07 


22.12 


23.15 2 


1.20 


24.17 


25.16 


26.15 


27.11 


28.06 


29.00 


29.92 


30.83 


31.72 


32. 60 2 


1.30 


33.46 


34.41 


35.15 


35.98 


36.79 


37.50 


38.38 


39.16 


39.93 


40. 68* 


1.40 


41.43 


42.16 


42.89 


43.60 


44.31 


45.00 


45.68 


41.36 


47.03 


47.68 2 


1.50 


48.33 


48.97 


49.60 


50.23 


50.84 


51.45 


52.05 


52.62 


53.23 


53. 80 2 



Adapted from Smithsonian Tables No. 65. 

1 Specific gravity less than 1.00 particularly useful for liquids fuel, oils, and alcohols. 

2 Specific gravities greater than 1.00 particularly useful for non-freezing brines. 



578 



ENGINEERING THERMODYNAMICS 

Table 
SPECIFIC HEATS OF GASES: 



Substance. 


c P 


At Temperature. 


Authority. 


Cv 




°C. 


j^ 




Hydrogen, H2 


3.3996 

3.409 

3.410 


-28-+9 
12-198 
21-100 


-18.4-15.8 
53.6-388.4 
70-212 


Regnault 
Regnault 
Wiedeman 


2.4219 


Oxvcen. 0? 


.2175 
.2240 

.2300 


13-207 
20-440 
20-630 


55-405 
68-824 
68-166 


Regnault 

Holborn-Austin 

Holborn-Austin 


1603 






Nitrogen, N2 


.2438 
.2419 

.2464 
.2497 


0-200 
20-440 
20-630 
20-800 


22-392 

68-824 

68-1166 

68-1472 


Regnault 
Holborn-Austin 
Holborn-Austin 
Holborn-Austin 


1715 






Air 


.2377 
.2374 
.2375 
.2366 
.2429 
.2430 
.2389 


-30-+10 
0-100 
0-200 
20-440 
20-630 
20-800 
20-100 


32-50 

32-212 

32-392 

68-824 
68-1166 
68-1472 
68-212 


Regnault 

Regnault 

Regnault 

Holborn-Austin 

Holborn-Austin 

Holborn-Austin 

Wiedeman 






.1703 


Ammonia, NH3 


.5202 

.5356 
.5125 


23-100 
27-200 
24-216 


73-212 
80-392 
75-421 


Wiedeman 
Wiedeman 
Regnault 


.4011 






Carbon diox., C0 2 


.1843 
.2025 

.2169 


-28-+7 
15-100 
11-214 


—18-45 
59-212 
52-417 


Regnault 
Regnault 
Regnault 


.1558 


Carbon monoxide 


.2425 

.2426 


23-99 
26-198 


74-210 

79-388 


Wiedeman 
Wiedeman 


.1734 






Methane, CH4 


.5929 


18-208 


64-406 


Regnault 


.4505 






Benzole, CeHe 


.2990 

.3325 
.3754 


34-115 

35-180 

116-218 


93-239 

95-356 

241-424 


Wiedeman 
Wiedeman 
Regnault 


.2131 






Ethylene, C2H4 


.4040 


10-202 


50-396 


Regnault 


.3404 







TABLES 



579 



XXXIV 

RATIOS AND DIFFERENCES 



Determined from 


Cp-Cv 


777.52(CyC P ) 
1 (?Pj in ft.-lbs. 


Cp-r- Cv 

= T 


Wiedeman C p = 3.41 and 

-^ = 1.408 at 4° -16° C. by Lummer and Pringsheim 

Cv 


.9881 


768.267 


1.408 


Holborn and Austin C v = .2240 and 
- £ = 1.3977at5°tol4°C. 

Cv 


.0637 


49.528 


1.3977 


Holborn and Austin C v = .2419 and 

C P 

-ft = 1.41 by Cazin 

Of 


.0704 


54.737 


1.4105 


Wiedeman C v = .2389 and 

C p 

— = 1.4025 at 5° to 14° C. by Lummer and Pringsheim 

Go 


.0686 


53.338 


1.4028 


Wiedeman C v = .5202 and mean of 

(^ = 1.3172 at 0° C. and ^ = 1.2770 at 100° C. ) 

\Cv Cv ) 

= 1.2971 by Wiillner 


.1191 


92.603 


1.2969 


Regnault C p = .2025 and 

—- = 1.2995 by Lummer and Pringsheim 


.0467 


36.310 


1.2997 


Wiedeman C v = .2425 and mean of 

1-^ = 1.4032 at 0° C. and ^ = 1.3946 at 100° C. ) 
\Cp Cv ) 

= 1.3989 by Wiillner 


.0691 


53.726 


1.3985 


Regnault C v = .5929 

Cp 

- E = 1.316 at 30° C. by Miiller 

Cf 


.1424 


110.719 


1.316 


Q 

Wiedeman C v = .2990 and ~ = 1.403 at 60° C. by 

Cv 

Pagliani 


.0859 


66.789 


1.4031 


Regnault C p = .4040 and -^ = 1.1870 at 100° C. by 

Cv 

Wiillner 


.0636 


49.450 


1.1867 



580 



ENGINEERING THERMODYNAMICS 

Table XXXV 

COEFFICIENT OF LINEAR EXPANSION OF SOLIDS 



Substance. 



aXlO* 
per degree C. 



At Temp. 
C. 



xio* 



At Temp. 
F. 



Authority. 



Aluminum. . 

Antimony 
Carbon coke 
Carbon 

graphite. 
Copper. . . . 

Iron 

Steel 

Lead 

Nickel .... 
Platinum. . 
Tin....... 

Zinc 

Brasses and 

bronze ... 

Rubber. . . . 

Glass 

Solder 

Ice 

Paraffin .... 

Porcelain. . 
Wood 



Wax. 



Concrete. 
Masonry. . . 



.2313-. 3150 

,0882-. 1692 
.054 

.0786 

.1678 
1061-.1210 
1095-.1322 

.2924 

.1279 

.0899 

.2234 

.2918 

. 17-.21 

.770 

.058-. 0897 

.2508 

375 

1.0662-4.7707 

.0413 
.0325-. 0614 

2.300-15.227 

.1430 
.046-. 089 



40-600 

40 
40 

40 
40 
40 
40 
40 
40 
40 
40 
40 

0-900 

16.7-25.3 

0-100 

0-100 

-20 to -1 

0-16; 38-49 

20-790 
2-34 

10-26; 
43-57 



.1285-. 175 

.0311-. 094 
.03 

.0437 

.0932 
.059-. 0672 
.06085-. 0735 

.0975 

.071 

.05 

.1241 

.1621 

.0889-. 1167 

.4278 
.3222-. 0498 

.1338 

.2083 

5921; 2.6505 

.023 
.0181-. 0341 

1.278 
8.46 
.0795 
.0256-. 0494 



104-1112 

104 
104 

104 
104 
104 
104 
104 
104 
104 
104 
104 

32-1652 
62-77.5 
32-212 

32-215 

-4-30.2 

32-60.8 

100.4-120 

68-145.4 

35.6-93.2 

50-78.8 
109.4-134.6 



Fizeau and 

Le Chatelier 
Fizeau 



Limits of 

determination 
Kohlrausch 
Limits of 

determination 
Smeaton 
Brunner 
Rodwell 

Braun 
Limits of 

determination 
Kopp 

Clark 
Clark 



TABLES 



581 



Table XXXVI 
COEFFICIENT OF CUBICAL EXPANSION OF LIQUIDS 



Substance. 



aXlO* 
per ° C. 



At Temp 
C. 



aX10* 
per ° F. 



At Temp. 
F. 



Authority. 



Alcohol (methyl) 

Benzene 

Bromine \ 

Calcium chloride, CaCl 2 , 5.8 per cent 
Calcium chloride,CaCl 2 ,40.9 per cent 

Ether 

Hydrochloric acid, HC1+6.25 H 2 . 
Hydrochloric acid, HC1+50 H 2 0. . . 

Mercury 

Olive oil 

Phenol, C 6 H 6 

Petroleum, Sp.gr. .8467 

Sodium chloride, NaCl, 1.6 per cent.. 

Sulphuric acid, H 2 S0 4 

Sulphuric acid, H 2 S0 4 



1433 
,1385 
,1168 

0506 
,0510 
,2150 
.0489 
,0933 
.0179 
.0742 



.1039 
.1067 
.0489 
.0799 



-38-+70 
11-81 
-7-+60 
18-25 
17-24 
-15-+38 
0-30 
0-30 



3-157 
24-120 

0-30 
0-30 



.0796 
.0770 
.0649 
.0281 
.0283 
.1195 
.0272 
.0519 
.0099 
.0412 
.0500 
.0577 
.0593 
.0272 
.0444 



36-158 

32-178 

19-140 

64-77 

63-75 

5-100 
32-86 
32-86 



97-314 

75-248 

32-86 
32-86 



Pierre 

Kopp 

Pierre 

Decker 

Decker 

Pierre 

Marignac 

Marignac 

Spring 

Pinette 

Frankenheim 

Marignac 

Marignac 

Marignac 



582 



ENGINEERING THERMODYNAMICS 



Table XXXVII 

COEFFICIENT OF VOLUMETRIC EXPANSION OF GASES AND VAPORS AT 

CONSTANT PRESSURE 

(Heated without change of state.) 



Substance. 



'Pressure (Cm Hg) 



g^XIOO 

per Deg. C. 



0^X100 
per Deg. F. 



Authority. 



Air 

Air , 

Hydrogen 

Hydrogen 

Carbon dioxide 

Carbon dioxide 

Carbon dioxide 0°-64° 

Carbon dioxide 84°-100°. . . 

Carbon dioxide 0°-7.5° 

Carbon dioxide 64°-100°. . . 

Carbon dioxide 0°-64° 

Carbon dioxide 0°-7.5° 

Carbon dioxide 0°-64° 

Carbon dioxide 0°-100° 

Carbon monoxide 

Nitrous oxide 

Sulphur dioxide 

Sulphur dioxide 

Water vapor (steam) 0°-119 

Water vapor 0°-141° 

Water vapor 0°-162° 

Water vapor 0°-200° 

Water vapor 0°-247° 



76 
256 

76 
254 

76 
252 
17.1 
17.1 



atm. 
atm. 



24.81 atm. 

24.81 atm. 

24.81 atm. 

34.49 atm. 

34.49 atm. 

34.49 atm. 

76 

76 

76 

98 

1 atm. 

1 atm. 

1 atm. 

1 atm. 

1 atm. 



.3671 

.3693 

.36613 

.36616 

.3710 

.3845 

.5136 

.4747 

.7000 

.5435 

.6204 

.1097 

.8450 

.6514 

.3669 

.3719 

.3903 

.3980 

.4187 

.4189 

.4071 

.3938 

.3799 



.2040 

.2055 

.2034 

.20342 

.2060 

.2135 

.2855 

.2635 

.38885 

.3020 

.3446 

.6100 

.470 

.362 

.204 

.2065 

.217 

.2465 

.23261 

.23272 

.22617 

.21878 

.2111 



Regnault 

Regnault 

Regnault 

Regnault 

Regnault 

Regnault 

Andrews 

Andrews 

Andrews 

Andrews 

Andrews 

Andrews 

Andrews 

Andrews 

Regnault 

Regnault 

Reganult 

Regnault 

Hirn 

Hirn 

Hirn 

Hirn 

Hirn 



TABLES 
Table XXXVIII 



583 



COEFFICIENT OF PRESSURE RISE OF GASES AND VAPORS AT CONSTANT 

VOLUME 

(Heated without change of state.) 



Substance. 


Pressure (Cm Hg) 


« P X100 

per Deg. C. 


c^XlOO 

per Deg. F. 


:/ ! 
Authority. 


Air 


.6 
1.6 
10.0 
26.0 
37.6 
75.0 
76-83 
11-15 
17-24 
37-51 
76 
200 
2000 
10000 
76 

1 atm. 
1 atm. 
1 atm. 
76-104 
174 
793 

16.4 atm. 

16.5 atm. 
25.87 atm. 
25.87 atm. 
33.53 
33.53 

1 atm. 
1 atm. 
1 atm. 
1 atm. 
1 atm. 
1 atm. 
1 atm. 
1 atm. 


.3767 

.3703 

.3663 

.3660 

.3662 

.3665 

.3670 

.3648 

.3651 

.3658 

.3665 

.3690 

.3887 

.4100 

.3671 

.3670 

.3706 

.3726 

.3686 

.3752 

.4252 • 

.4754 

.4607 

.5728 

.5406 

.6973 

.6334 

.3667 

.3669 

.3656 

.3668 

.3676 

.3705 

.3674 

.3845 


.20915 

.2057 

.2035 

. 20335 

.20345 

.20360 

.20370 

.20265 

.20285 

.20320 

.20360 

.205 

.206 

.22775 

.20395 

.20290 

.2059 

:2070 

.20475 

.2085 

.2361 

.2641 

.256 

.3182 

.30035 

.38740 

.35190 

.2037 

.20353 

.2031 

.20375 

.20410 

.206 

.2041 

.21350 


Meleander 


Air 


Meleander 


Air 


MeleandeJ 


Air 


Meleande; 


Air 


Meleandet 


Air 


Meleander 


Air 


Magnus 
Regnault 


Air 


Air 


Regnault 
Regnault 
Regnault 
Regnault 
Regnault 
Regnault 


Air , 


Air 


Air 


Air 


Air 


Air 


Rowland 


Air 


Jolly 
Jolly 
Meleander 


Carbon dioxide 


Carbon dioxide 


Carbon dioxide 


Regnault 


Carbon dioxide 


Regnault 
Regnault 


Carbon dioxide 


Carbon dioxide 0°-64°. . . 
Carbon di xide 64°-100°. 
Carbon dioxide 0°-64°. . . 
Carbon dioxide 64°-100°. 
Carbon dioxide 0°-64°. . . 
Carbon dioxide 64°-100°, 

Hydrogen 


Andrews 
Andrews 
Andrews 
Andrews 
Andrews 
Andrews 
Regnault 
Regnault 




JoUy 


Nitrogen 


Regnault 




Regnault 


Nitrous oxide 


Jolly 




JoUy 


Sulphur dioxide, S0 2 


Jolly 



584 



ENGINEERING THERMODYNAMICS 



Table 

COMPRESSIBILITY OF GASES BY THEIR ISOTHERMALS. VALUES OF PV AT 

AND AT 1 ATMOSPHERE 



Pressure in Atmosphere. 


1 


100 


200 


300 


400 


500 


600 


f 32° F. 

Oxygen at { 211.1 

[391.1 


1.000 


.9265 


.9140 
1.4 
1.819 


.9624 
1.4529 
1.8849 


1.0516 

1.532 

1.96 


1.1560 

1.622 

2.05 


1.2690 
1.7202 
2.142 


[32°F. 

Air at {210.92 

[392.72 


1.000 


.9730 


1.010 
1.472 

1.886 


1.0974 
1.551 

1.9866 


1 .2144 

1.668 

2.096 


1.3400 
1.7825 
2.211 


1.4700 
1 908 
2.3298 


f 32° F. 
Nitrogen at { 211.1 
[391.28 


1.000 


.9910 


1.0390 
1.4890 
1.9064 


1.1358 
1.5903 
2.1045 


1.2568 
1.7060 
2.1324 


1.3900 
1.8275 
2.2575 


1.5258 
1.9548 
2.3838 


f 32° F. 
Hydrogen at { 210.74 
[393.5 


1.000 




1.1380 
1.5134 

1.884 


1.2090 
1.5858 
1.956 


1.2828 
1.6588 
2.030 


1.3565 
1.7310 
2.105 


1.4322 
1.8036 
2.1762 


f 32° F 
Carbon dioxide 1 2 i o ' 

at 1388! 9 


1.000 


.202 
1.03 

1.582 




.559 

.890 

1.493 




.891 
1.201 
1.678 




f 32° F. 

NH 3 aU 211.28 

[362.48 


1.000 




.9290 
.9750 


.8625 
.9555 


.832 
.9380 


.7450 

.8875 


.5850 
.8700 



Calculated from Smithsonian Tables Nos. 55 and 58, reporting Amagat's results 

Table XL 
VALUES OF THE GAS CONSTANT B 





Determined from 
Specific Heats by 

R = 777.52(C P -Cv) 

Section 5. 


Determined from 

Volume of One Lb. 

at 32° F. and 

29.92 ins. Hg. 

Section 8. 


Authority for 
Specific Volume. 


Hydrogen, H2 


768.267 
49.528 
54.737 
53.338 
92.603 
36.310 
53.726 

110.719 
66.789 
49.450 


765 . 893 
48.244 
55.981 
53.332 
90.467 
35.084 
55.135 
96.200 
Liquid at 32° 
54.153 


Rayleigh 

Rayleigh 

Rayleigh 

Rayleigh and Leduc 

Leduc 

Rayleigh 

Leduc 

Thomson 


Oxygen, 2 

Nitrogen, N2 

Air 

Ammonia, NH 3 

Carbon dioxide, C0 2 

Carbon monoxide, CO 

Methane, CH 4 

Benzole, CeHe 


Ethylene, C 2 H 4 









TABLES 



585 



XXXIX 

VARIOUS PRESSURES AND TEMPERATURES; THE VALUE OF PV AT 32° F 
TAKEN AS 1.00. 



700 


800 


900 


1000 




1.3853 

1.827 

2.2414 


1.5032 
1.9336 
2 . 3432 


1.6200 
2.0412 
2.4462 


1.7350 
2.151 


( Pressure 50 atm. ) 
Critical point \ \ Wroblewski 
[Temperature 180.4° F.J 


1.6016 
2.0328 
2.4514 


1.7344 
2.1592 
2.5752 


1.8630 
2.2896 

2.7 


1.992 
2.415 

2.828 


( Pressure 39 atm. "| 
Critical point \ \ Olszewski 
[ Temperature 220° F. J 


1.6618 

2.086 

2.5123 


1.7920 

2.22 

2.64 


1.9341 
2.3544 
2.7765 


2.0680 


f Pressure 35 atm. ] 
Critical point \ \ Olszewski 
[ Temperature 230 . 8° F. J 


1.5043 

1.876 

2.2484 


1.5776 
1.9552 
2.32 


1.6488 
2.1096 
2.3913 


1.7200 
2.093 


f Pressure 20 atm. 1 
Critical point \ \ Dewar 
[ Temperature 390. 1° F. J 








1.656 
1.999 


f Pressure 27 atm. 1 
Critical point \ [ Andrews 
[Temp. +87.66° F. J 


!87i5 


!9000' 


!93i5* 


".95" 


f Pressure 115 atm. 1 
Critical point \ \ Dewar 
[ Temp. +266° F. J 



and Table 62 Roth's results; also Table 218 reporting miscellaneous data. 



Table XLI 

DENSITIES OF GAS AT ONE ATMOSPHERE = 29.92" Hg AND 32° F., COMPARING 
EXPERIMENTAL. VALUES WITH COMPUTED VALUES FROM MOLECULAR 
WEIGHTS 















Lbs. 




Lbs. 












Molecular 


Cu.ft. 


Molecular 


Cu.ft. 


Gas. 


Sp.Gr. 


Lbs. per 
Cu.ft. 
Exptl. 


Cu.ft. 




Weight 


from 


Weight 


from 




Air=l. 


per Lb. 




Exact. 


Exact 


Approx. 


Approx. 










H=2. 


Molecular 


H=2. 


Molecular 














Weight. 




Weight. 


Hydrogen, H 4 . . 


.0696 


.005621 


177.9093 


Rayleigh 


2. 




2 




Oxygen, 2 


1.053 


.08922 


11.208 


Rayleigh 


31.76 


.08926 


32 


.08993 


Nitrogen, N 2 . . . 


.9673 


.07829 


12.773 


Rayleigh 


27.80 


.07813 


28 


.07869 


Air 


1.000 


.08071 


12.390 


Rayleigh 
and Leduc 














Ammonia, NH 3 . . 


.597 


.04758 


21.017 


Leduc 


16.9 


.04750 


17 


.04778 


Carbon dioxide 


















C0 2 


1 . 5291 


. 12269 


8 . 1506 


Rayleigh 


43 75 


12295 


44 


12366 


Carbon mon- 
















oxide, CO 


.9672 


.07807 


12.8090 


Leduc 


27.87 


.07833 


28 


.07869 


Methane, CH 4 . . 


.5576 


.04470 


22.349 


Thomson 


15.99 


.04494 


16 


.04497 


Benzole, C 6 H 6 . . . 




Liquid 














Ethylene, C 2 H 4 . . 


.9852 


.07951 


12.578 


Saussure 


27.98 


.07862 


28 


.07868 


Ethane, C 2 H 6 . . . 


1.075 


.08379 


11.9354 


Kolbe 


29.98 


.08426 


30 


.08431 


Butane, C 4 Hi . . . 


2.01 


. 16194 


6.1751 


Frankland 


57.96 


. 16289 


58 


. 16301 



Computed from data reported in Smithsonian Tables, Nos. 71 and 276. 



586 



ENGINEERING THERMODYNAMICS 
Table XLII 



INTERNATIONAL ATOMIC WEIGHTS 

Selected from Report of the International Committee on Atomic Weights, Journal Amer. 
Chem. Soc. f 1910. 



Substance. 



Symbol. 



Atomic Weight, 
0=16. 



Atomic Weight, 
H=l. 



Aluminum . 
Calcium. . . 
Caibon. . . 
Chlorine. . . 
Copper. . . , 
Hydrogen . 

Iron 

Lead 

Magnesium 
Manganese 
Mercury. . 
Nickel .... 
Nitrogen . . 
Oxygen. . . 
Platinum. . 
Potassium . 
Silicon .... 
Sodium. . . 
Sulphur . . . 

Tin 

Zinc 



Al 
Ca 

C 
CI 
Cu 

H 

Fe 

Pb 

Mg 

Mn 

Hg 

Ni 

N 

O 
Pt 

K 

Si 
Na 

S 

Sn 
Zn 



27.1 

40.09 

12.00 

35.46 

63.57 

1.008 
55.85 

207.10 
24.32 
54.93 

200.00 
58.68 
14.01 
16.00 

195.00 
39.10 
28.30 
23.00 
32.07 

119.00 
65.37 



26.9 
39.77 
11.99 
35.19 
63.07 
1.00 
55.41 

205.46 
24.13 
54.49 

198.50 
58.21 
13.90 
15.88 

193.40 
38.79 
28.20 
22.82 
31.82 

118.10 
64.88 



Table XLIII 
MELTING OR FREEZING-POINTS (at 29.92 Hg) 
Selected from Landolt, Bornstein, Meyerhoff, and Smithsonian Physical Tables. 



Class. 


Substance. 


Symbols. 


Freezing-point. 


Authority. 






C. 


F. 




Elements: 


Hydrogen 

Oxygen 


H 
O 

N 

CI 

Hg 

Br 

P 

K 

Na 


-258.9 
-230 
-210.5 
-102 

- 38.85 

- 7.3 
44.2 
62.5 
97 

113.5- 
119.5 


-432 

-382.5 

-347 

-151.5 

- 38 

45.2 

111.5 

144.5 

206.5 

236-247 


Travers, 1902 




Nitrogen 

Chlorine 

Mercury 

Bromine 

Phosphorus 

Potassium 

Sodium 


Fischer-Alt 

Olszewski 

Vincentini and Omodei, 1888 

Van der Plaats, 1886 

Helff, 1893 . 

Holt and Sims, 1894 

Kurnakow and Puschin, 1902 

Depending on form of S 




Sulphur 



TABLES 



587 



Table XLIII — Continued 

MELTING OR FREEZING-POINTS (at 29.92 Ho) 

Selected from Landolt, Bornstein, Meyerhoff, and Smithsonian Physical Tables 





Substance. 


Symbols. 


Freezing-point. 


» 


Class. 


C. 


F. 


Authority. 


Elements: 


Tin 


Sn 

Bi 

Ca 

Pb 

Zn 

Sb 

Mg 

Al 

Ag 

Au 

Cu 

Mn 

Si 

Ni 

Co 

Cr 

Fe 

Pt 

W 

NH 3 

CaCl 2 

CO 

C0 2 

NaCl 

S0 2 

ZnClo 


231.5 

269.2 

321 

326.9 

419 

624 

632.6 

657.3 

961 

1063 

1083 

1225 

1420 

1450 

1490 

1505 

1600 

1755 

2950 

- 75.5 

780 
-203 

- 57 
820 

- 76 
262 

-1922 

-171.4 

- 51 

- 31 

- 26 

- 12 

- 6 
+ 5 

10 
18 
22 
28 
32 

-169 
-130 


451 

517 

610 

621 

787 

1154 

1171 

1217 

1651 

1947 

1892 

2232 

2592 

2647 

2813 

2792 

2912 

3192 

5347 

-104 

1454 
-331.5 

70.8 
1510 

-105 
504 

-314 

-276.5 

- 59.8 

- 23.8 

- 14.8 
10.4 
21.2 
41 
50 
64.4 
71.6 
82.4 
89.6 

-272 
-202 


Kurnakow and Puschin,1902 




Bismuth 

Cadmium 

Lead 


Callendar, 1899 
Kurnakow and Puschin,1902 
Holborn and Day 




Zinc 


Holborn and Day 




Antimony 

Magnesium 

Aluminum 

Silver 


Fay and Ashley 
Heycock and Neville, 1895 
Holborn and Day 
Holborn and Day 




Gold 


Roberts and Austin 




Copper 


Roberts and Austin 




Manganese 

Silicon 


Day-Sosman 
General 




Nickel 


Carnelley, Pictet, 1879 




Cobalt 


General 




Chromium 

Iron 


General 

Roberts and Austin 


Inorganic com- 
pounds 


Platinum 

Tungsten 

Ammonia 

Calcium chloride 
Carbon monoxide 

Carbon dioxide . . 
Sodium chloride . 
Sulphur dioxide. . 
Zinc chloride .... 
Air 


Mean of three 
Waidner-Burgess, 

Waterburg 
Ladenburg and Krugel, 1 900 
Ruff and Plato, 1903 
Wroblewski, Olszewski 

(mean) 
General 

Ruff and Plato, 1903 
Faraday, 1845 
Braun, 1875 
Wroblewski, 1884 


Hydrocarbon 
constituents of 
liquid and 


Ethane 

Nonane 

Decane 


C 2 H6 
C9H2O 
C10H22 
C11H24 
C12II26 
C13H28 
C14H30 
C15H32 
C16H34 
C17H36 
C18H38 
C19H40 

C2H4 
C2H5OH 


Liquid Density 
.446 at 32° F. 
.733 at 32° F. 
.745 at 32° F. 


gaseous fuel 

Paraffine series, 
CnH2n+2 

Ethylene Series, / 
CnH2n \ 


Undecane 

Dodecane 

Tridecane 

Tetradecane .... 
Pentadecane .... 

Hexadecane 

Heptadecanc .... 

Octadecane 

Nonadecane .... 

Ethylene 

Ethyl'alcohol . . . 


.756 at 32° F. 
.765 at 32° F. 
.771 at 32° F. 
.775 at 40° F. 
.776 at 10° C. 
.775 at 18° C. 
.777 at 22° C. 
.777 at 28° C. 
.777 at 32° C. 

.610 

.806 at 32° F. 



588 ENGINEEEING THERMODYNAMICS 

Table XLIV 

BOILING-POINTS (at 29.92 Hg) 
Selected from Landolt, Bornstein, Meyerhoff, and Smithsonian Physical Tables. 



Class. 



Substance. 



Symbol. 



Boiling-point. 



Authority. 



Elements 



Inorganic com- 
pounds 



Hydrogen 

Oxygen 

Nitrogen 

Chlorine 

Mercury 

Bromine 

Phosphorus 

Potassium 

Sodium 

Sulphur 

Tin 

Bismuth 

Cadmium 

Lead 

Zinc 

Antimony 

Magnesium 

Aluminum 

Silver 

Copper 

Manganese 

Chromium 

Iron 

Ammonia 

Carbon monoxide 

Carbon dioxide . . 
Sulphur dioxide. . 
Zinc chloride .... 



II 

O 
N 
CI 
Hg 
Br 

P 
K 

Na 

S 

Sn 
Bi 
Cd 
Pb 
Zn 
Sb 
Mg 
Al 
Ag 
Cu 
Mn 
Cr 
Fe 
NH 3 
CO 

C0 2 

S0 2 
ZnCl 2 



Air. 



-252.5 
-182.7 
-194.4 

- 33.6 
357 

61.1 

287 
712 

750 

444.7 
2270 
1430 

782 
1525 

918 
1440 
1120 
1800 
1955 
2310 
1900 
2200 
2450 

- 38.5 
-191.5 

- 79.1 

- 10.8 
730 

-192.2 
-191.4 



-412 

-297 
-318 
- 28.5 

674 

142 

558 
1372 

1382 

837 
4118 
2607 
1440 
2777 
1686 
2622 
2047 
3272 
3552 
4192 
3452 
3992 
4442 
-37.4 
-313 

110.5 
12.6 
1347 

314 
312.5 



Dewar, 1901 

Holborn, 1901 

Olszewski 

Regnault 

Crafts-Regnault 

Mean of Thorpe, van der 

Plaats 
Schrotter, 1848 
Perman, Ruff, and Johann- 

sen 
Perman, Ruff, and Johann- 

sen 
Rothe, 1903 
Greenwood 
Barus, Greenwood 
Bams, 1894 
Greenwood 
Berthelot 
Greenwood 
Greenwood 
Greenwood 
Greenwood 
Greenwood 
Greenwood 
Greenwood 
Greenwood 
Regnault, 1863 
Mean of Wroblewski and 

Olszewski 
Villard and Jarry 
Regnault, 1863 
Freyer and Meyer 

Wroblewski 
Olszewski 



TABLES 
Table XLIV — Continued 

BOILING-POINTS (at 29.92 Hg) 
Selected from Landolt, Bornstein, Meyerhoff, and Smithsonian Physical Tables. 



589 



Class. 



Substance. 





Boiling-point. 


Symbol. 






C. 


F. 


CH 4 


-165 


-265 


C2HC 


- 93 


-135 


C 3 H 8 


- 45 


- 49 


C4H10 


+ 1 


33.8 


C 5 H 12 


36.3 


973 


C G Hi 4 


69 


156.2 


C7H16 


98.4 


209.1 


CsHis 


125.5 


257.9 


C9H20 


150 


302 


C10H22 


173 


343.4 


CnH24 


195 


384 


C12H26 


214 


417.2 


C13H28 


234 


453.2 


C14H30 


252 


485.6 


C15H32 


270 


518 


C16H34 


287 


548.6 


C17H36 


303 


577 


C18-H38 


317 


602 


C19H40 


330 


626 


C2H4 


-103 


-153.4 


C3H6 


- 50.2 


- 58.5 


C 4 Hs 


+ 1 


33.8 


C5H10 


36 


96.8 


C 6 Hi2 


69 


156.2 


C 7 H U 


96-99 


205-210 


CsHi6 


122-123 


251-255 


C9H18 


140-142 


284-288 


C10H20 


175 


347 


C2H2 


- 85 


-121 


CH3OH 


66 


150.8 


C 2 H 5 OH 


78 . 


172.4 


Mixture 




424 app. 


Mixture 




177 app. 



Authority. 



Hydrocarbon 
constituents of 
liquid and 
gaseous fuels 



Paraffine series, 

C»H2ra+2 



Ethylene series, 

C2H2re 



Methane 

Ethane 

Propane 

Butane 

Pentane 

Hexane 

Heptane 

Octane 

Nonane 

Decane 

Undecane 

Dodecane 

Tridecane 

Tetradecane 

Pentadecane. . . . 

Hexadecane 

Heptadecane. . . . 

Octadecane 

Nonadecane 

Ethylene 

Propylene 

Butylene 

Amylene 

Hexylene 

Heptylene 

Octylene 

Nonylene . 

Decylene 

Acetylene 

Methyl alcohol . . 
Ethyl alcohol. . . . 

Naphthas 

Benzines 



Young 

Ladenberg 

Young, Hamlen 

Butlerow, Young 

Thorpe, Young 

Schorlemmer 

Thorpe, Young 

Thorpe, Young 

Kraft 

Kraft 

Kraft 

Kraft 

Kraft 

Kraft 

Kraft 

Kraft 

Kraft 

Kraft 

Kraft 

Olszewski 

Ladenburg-Kriigel 

Sieben 

Wagner 

Wreden 

Morgan 

Moslinger 

Beilstein 

Beilstein 

Villard 



General 
General 



590 



ENGINEERING THERMODYNAMICS 



Table XLV 



LATENT HEAT OF VAPORIZATION AT ONE ATMOSPHERE PRESSURE 
Selected from Landolt, Bornstein, Meyerhoff, and Smithsonian Physical Tables. 



Substance. 



Ammonia 



Water . 
Benzol . 



Air .... 

Oxygen , 



Nitrogen 

Carbon dioxide 



Alcohol, methyl 

Alcohol, ethyl 

Alcohol + 5 % water . 

Decane 

Hexylene 



Octane. 



Symbol. 



NH 3 

H 2 6 

CeH6 



N 
C0 2 



CHOH 
C 2 H 5 OH 

C10H22 
CeHi2 

CsHis 



Cal. per Kg. B.T.U. per Lb. 



294.21 
291.32 
297.38 
296.5 
535.9 
532.0 
109. 
132.1 
154.5 
44.02 
45.4 
58.0 
60.9 
49.83 
72.23 
57.48 
56.25 
50.76 
31.80 
14.40 
11.60 
3.72 
267.48 
206.4 
214.25 
60.83 
87.3 

7i!i* 



530 

524.45 

535 

534 

964.6 

957.6 

196 

238 

278 
79.3 
81.7 

106.1 

109.8 
89.6 

130 

103.2 

10 .3 
91.5 
57.2 
25.9 
20.9 
6.7 

482 

372 

386 

109.5 

157.1 

128 



c. 



7.8 
11.04 
16.0 
17 
100 
100 

100 
210 



188 



-25 



6.5 
22.4 
29.85 
30 

30.82 
64.5 
78 
78.4 

159.45 
68 
70 

125 



F. 



4.6 
51.87 
60.8 
62.6 

2-12 

212 
32 

212 

410 



-306.4 



13 

32 

32 

43.7 

72.3 

85.7 



87. 
148. 
172. 
173. 
319 
154. 
158 
257 



Authority. 



Regnault 

Regnault 

Regnault 

Strombeck 

Andrews 

Scball 

Regnault 

Regnault 

Regnault 

Shearer 

Shearer 

Shearer 

Estreicher 

Shearer 

Cailletet 

Matthias 

Chappuis 

Matthias 

Matthias 

Matthias 

Cailletet 

Matthias 

Wirtz 

Schall 

Brix 

Louguinine 

Mabery 

Goldstein 

Goldstein 



TABLES 



591 



Table XLVI 



LATENT HEATS OF FUSION 
Selected from Landolt, Bornstein, Meyerhoff, and Smithsonian Physical Tables. 



Substance. 


Symbol. 


Cal. per Kg. 


B.T.U .per Lb. 


C. 


F. 


Authority. 


Aluminum. . . 


Al 


239.4 


432 


625 


1157 


Pionchon 


Lead 


Pb 


5.37 


9.66 


362.2 


619.2 


Person 


Iron 


Fe 
Cu 


6.0 
43.0 


10.8 

77.4 


1000-1050 


1832-1922 


Pionchon 


Copper 


Richards 


Nickel 


Ni 


4.64 


8.35 






Pionchon 


Zinc 


Zn 

Sn 

NH 3 


r 28.1 

14.25 
108.1 


50.5 
25.65 
195 


415 

233 

-75 


779 
451.4 
-102 


Person 


Tin 


Person 


Ammonia. . . . 


Massol 


Ice-water 


H 2 


79.25 


142.5 





32 


Person and Regnault 






79.06 


142.2 





32 


Regnault 






79.24 


142.5 





32 


Desains 






79.91 


143.9 





32 


Smith 






80.025 


144.3 





32 


Bunsen 


Benzol 


CeHe 


30.08 


55.5 


5.3 


41.6 


Fisher 



592 



ENGINEERING THERMODYNAMICS 
Table XLVII 



PROPERTIES OF SATURATED STEAM 

(Condensed from Marks and Davis's Steam Tables and Diagrams, 1909, by permission of 
the publishers, Longmans, Green & Co.) 









Total Heat Above 












Gauge 


Absolute 

Pressure 

Pounds 

per Sq.in. 


Tempera- 
ture, 
Fahren- 
heat. 


32° 


D. 


Latent 

Heat, 

L=H-h 

Heat-unita 


Volume, 

Cu. Ft. in 

1 Lb. of 

Steam. 


Weight of 

1 Cu. Ft. 

Steam, 

Pound. 


Entropy 
of the 
Water. 


Entropy 


Pressure 

Pounds 

per Sq.in. 


In the 

Water, 

h 

Heat-units 


In the 

Steam, 

H 

Heat-units 


of Evap- 
oration. 


















29.74 


0.0886 


32 


0.00 


1073.4 


1073.4 


3294 


0.000304 


0.0000 


2 . 1832 


29.67 


0.1217 


40 


8.05 


1076.9 


1068.9 


2438 


0.000410 


0.0162 


2.1394 


29.56 


0.1780 


50 


18.08 


1081.4 


1063.3 


1702 


0.000587 


0.0361 


2.0865 


29.40 


0.2562 


60 


28.08 


1085 . 9 


1057.8 


1208 


0.000828 


0.0555 


2.0358 


29.18 


0.3626 


70 


38.06 


1090.3 


1052.3 


871 


0.001148 


0.0745 


1.9868 


28.89 


0.505 


80 


48.03 


1094 . 8 


1046.7 


636.8 


0.001570 


0.0932 


1.9398 


28.50 


0.696 


90 


58.00 


1099.2 


1041.2 


469.3 


0.002131 


0.1114 


1.8944 


28.00 


0.946 


100 


67.97 


1103.6 


1035.6 


350.8 


0.002851 


0.1295 


1.8505 


27.88 


1 


101.83 


69.8 


1104.4 


1034.6 


333.0 


0.00300 


0.1327 


1.8427 


25.85 


2 


126.15 


94.0 


1115.0 


1021.0 


173.5 


0.00576 


0.1749 


1 . 7431 


23.81 


3 


141 . 52 


109.4 


1121.6 


1012.3 


118.5 


0.00845 


0.2008 


1.6840 


21.78 


4 


153.01 


120.9 


1126.5 


1005 . 7 


90.5 


0.01107 


0.2198 


1.6416 


19.74 


5 


162.28 


130.1 


1130.5 


1000.3 


73.33 


0.01364 


0.2348 


1.6084 


17.70 


6 


170.06 


137.9 


1133.7 


995.8 


61.89 


0.01616 


0.2471 


1.5814 


15.67 


7 


176.85 


144.7 


1136.5 


991.8 


53.56 


0.01867 


0.2579 


1.5582 


13.63 


8 


182.86 


150.8 


1139.0 


988.2 


47.27 


0.02115 


0.2673 


1 . 5380 


11.60 


9 


188.27 


156.2 


1141.1 


985.0 


42.36 


0.02361 


0.2756 


1 . 5202 


9.56 


10 


193.22 


161.1 


1143.1 


982.0 


38.38 


0.02606 


0.2832 


1 . 5042 


7.52 


11 


197.75 


165.7 


1144.9 


979.2 


35.10 


0.02849 


0.2902 


1.4895 


5.49 


12 


201 . 96 


169.9 


1146.5 


976.6 


32.36 


0.03090 


0.2967 


1.4760 


3.45 


13 


205.87 


173.8 


1148.0 


974.2 


30.03 


0.03330 


0.3025 


1.4639 


1.42 
lbs. 

gauge 


14 


209.55 


177.5 


1149.4 


971.9 


28.02 


0.03569 


0.3081 


1.4523 


14.70 


212 


180.0 


1150.4 


970.4 


26.79 


0.03732 


0.3118 


1.4447 


0.3 


15 


213.0 


181.0 


1150.7 


969.7 


26.27 


0.03806 


0.3133 


1.4416 


1.3 


16 


216.3 


184.4 


1152.0 


967.6 


24.79 


0.04042 


0.3183 


1.4311 


2.3 


17 


219.4 


187.5 


1153.1 


965.6 


23.38 


0.04277 


0.3229 


1.4215 


3.3 


18 


222.4 


190.5 


1154.2 


963.7 


22.16 


0.04512 


0.3273 


1.4127 


4.3 


19 


225.2 


193.4 


1155.2 


961.8 


21.07 


0.04746 


0.3315 


1.4045 


5.3 


20 


228.0 


196.1 


1156.2 


960.0 


20.08 


0.04980 


0.3355 


1.3965 


6.3 


21 


230.6 


198.8 


1157.1 


958.3 


19.18 


0.05213 


0.3393 


1.3887 


7.3 


22 


233.1 


201.3 


1158.0 


956.7 


18.37 


0.05445 


0.3430 


1.3811 


8.3 


23 


235.5 


203.8 


1158.8 


955.1 


17.62 


0.05676 


0.3465 


1.3739 


9.3 


24 


237.8 


206.1 


1159.6 


953.5 


16.93 


0.05907 


0.3499 


1.3670 


10.3 


25 


240.1 


208.4 


1160.4 


952.0 


16.30 


0.0614 


0.3532 


1.3604 


11.3 


26 


242.2 


210.6 


1161.2 


950.6 


15.72 


0.0636 


0.3564 


1.3542 


12.3 


27 


244.4 


212.7 


1161.9 


949.2 


15.18 


0.0659 


0.3594 


1.3483 


13.3 


28 


246.4 


214.8 


1162.6 


947.8 


14.67 


0.0682 


0.3623 


1 . 3425 


14.3 


29 


248.4 


216.8 


1163.2 


946.4 


14.19 


0.0705 


0.3652 


1.3367 


15.3 


30 


250.3 


218.8 


1163.9 


945.1 


13.74 


0.0728 


0.3680 


1.3311 


16.3 


31 


252.2 


220.7 


1164.5 


943.8 


13.32 


0.0751 


0.3707 


1.3257 


17.3 


32 


254.1 


222.6 


1165.1 


942.5 


12.93 


0.0773 


0.3733 


1.3205 


18.3 


33 


255.8 


224.4 


1165.7 


941.3 


12.57 


0.0795 


0.3759 


1.3155 


19.3 


34 


257.6 


226.2 


1166.3 


940.1 


12.22 


0.0818 


0.3784 


1.3107 


20.3 


35 


259.3 


227.9 


1166.8 


938.9 


11.89 


0.0841 


0.3808 


1.3060 



TABLES 
Table XLVII — Continued 



593 









Total Heat Above 
32° F. 










' 


Gauge 


Absolute 
Pressure 
Pounds 

perSq.in. 


Tempera- 
ture, 
Fahren- 
heat. 






Latent 

Heat, 

L=H-h 

Heat-units 


Volume, 

Cu. Ft. in 

1 Lb. of 

Steam. 


Weight of 
1 Cu. Ft. 
Steam, 
Pound. 


Entropy 
of the 
Water. 




Pressure 

Pounds 

per Sq.in. 


In the 

Water, 

h 

Heat-units 


In the 

Steam, 

H 

Heat-units 


Entropy 
of Evap- 
oration. 


















21.3 


36 


261.0 


229.6 


1167.3 


937.7 


11.58 


0.0863 


0.3832 


1.3014 


22.3 


37 


262.6 


231.3 


1167.8 


936.6 


11.29 


0.0886 


0.3855 


1.2969 


23.3 


38 


264.2 


232.9 


1168.4 


935.5 


11.01 


0.0908 


0.3877 


1.2925 


24.3 


39 


265.8 


234.5 


1168.9 


934.4 


10.74 


0.0931 


. 3899 


1.2882 


25.3 


40 


267.3 


236.1 


1169.4 


933.3 


10.49 


0.0953 


. 3920 


1.2841 


26.3 


41 


268.7 


237.6 


1169.8 


932.2 


10.25 


0.0976 


0.3941 


1.2800 


27.3 


42 


270.2 


239.1 


1170.3 


931.2 


10.02 


0.0998 


0.3962 


1.2759 


28.3 


43 


271.7 


240.5 


1170.7 


930.2 


9.80 


0.1020 


0.3982 


1.2720 


29.3 


44 


273.1 


242.0 


1171.2 


929.2 


9.59 


0.1043 


0.4002 


1.2681 


30.3 


45 


274.5 


243.4 


1171.6 


928.2 


9.39 


0.1065 


0.4021 


1.2644 


31.3 


46 


275.8 


244.8 


1172.0 


927.2 


9.20 


0.1087 


0.4040 


1.2607 


32.3 


47 


277.2 


246.1 


1172.4 


926.3 


9.02 


0.1109 


0.4059 


1.2571 


33.3 


48 


278.5 


247.5 


1172.8 


925.3 


8.84 • 


0.1131 


0.4077 


1.2536 


34.3 


49 


279.8 


248.8 


1173.2 


924.4 


8.67 


0.1153 


0.4095 


1.2502 


35.3 


50 


281.0 


250.1 


1173.6 


923.5 


8.51 


0.1175 


0.4113 


1.2468 


36.3 


51 


282.3 


251.4 


1174.0 


922.6 


8.35 


0.1197 


0.4130 


1.2432 


37.3 


52 


283.5 


252.6 


1174.3 


921.7 


8.20 


0.1219 


0.4147 


1.2405 


38.3 


53 


284.7 


253.9 


1174.7 


920.8 


8.05 


0.1241 


0.4164 


1.2370 


39.3 


54 


285.9 


255.1 


1175.0 


919.9 


7.91 


0.1263 


0.4180 


1.2339 


40.3 


55 


287.1 


256 . 3 


1175.4 


919.0 


7.78 


0.1285 


0.4196 


1.2309 


41.3 


56 


288.2 


257.5 


1175.7 


918.2 


7.65 


0.1307 


0.4212 


1.2278 


42.3 


57 


289.4 


258.7 


1176.0 


917.4 


7.52 


0.1329 


0.4227 


1.2248 


43.3 


58 


290.5 


259.8 


1176.4 


916.5 


7.40 


0.1350 


0.4242 


1.2218 


44.3 


59 


291.6 


261.0 


1176.7 


915.7 


7.28 


0.1372 


0.4257 


1.2189 


45.3 


60 


292.7 


262.1 


1177.0 


, 914.9 


7.17 


0.1394 


0.4272 


1.2160 


46.3 


61 


293.8 


263.2 


1177.3 


914.1 


7.06 


0.1416 


0.4287 


1.2132 


47.3 


62 


294.9 


264.3 


1177.6 


913.3 


6.95 


0.1438 


0.4302 


1 .2104 


48.3 


63 


295.9 


265.4 


1177.9 


912.5 


6.85 


0.1460 


0.4316 


1.2077 


49.3 


64 


297.0 


266.4 


1178.2 


911.8 


6.75 


0.1482 


0.4330 


1.2050 


50.3 


65 


298.0 


267.5 


1178.5 


911.0 


6.65 


0.1503 


0.4344 


1.2024 


51.3 


66 


299.0 


268.5 


1178.8 


910.2 


6.56 


0.1525 


0.4358 


1.1998 


52.3 


67 


300.0 


269.6 


1179.0 


909.5 


6.47 


0.1547 


0.4371 


1 . 1972 


53.3 


68 


301.0 


270.6 


1179.3 


908.7 


6.38 


0.1569 


0.4385 


1.1946 


54.3 


69 


302.0 


271.6 


1179.6 


908.0 


6.29 


0.1590 


0.4398 


1 . 1921 


55.3 


70 


302.9 


272.6 


1179.8 


907.2 


6.20 


0.1612 


0.4411 


1.1896 


56.3 


71 


303.9 


273.6 


1180.1 


906.5 


6.12 


0.1634 


0.4422 


1 . 1872 


57.3 


72 


304.8 


274.5 


1180.4 


905.8 


6.04 


0.1656 


0.4437 


1 . 1848 


58.3 


73 


305.8 


275.5 


1180.6 


905.1 


5.96 


0.1678 


0.4449 


1.1825 


59.3 


74 


306.7 


276.5 


1180.9 


904.4 


5.89 


0.1699 


0.4462 


1 1801 


60.3 


75 


307.6 


277.4 


1181.1 


903.7 


5.81 


0.1721 


0.4474 


1 . 1778 


61.3 


76 


308.5 


278.3 


1181.4 


903.0 


5.74 


0.1743 


0.4487 


1 . 1755 


62.3 


77 


309.4 


279.3 


1181.6 


902.3 


5.67 


0.1764 


0.4499 


1 . 1730 


63.3 


78 


310.3 


280.2 


1181.8 


901.7 


5.60 


0.1786 


0.4511 


1.1712 


64.3 


79 


311,2 


281.1 


1182.1 


901.0 


5.54 


0.1808 


0.4523 


1 . 1687 


65.3 


80 


312.0 


282.0 


1182.3 


900.3 


5.47 


0.1829 


0.4535 


1 . 1665 


66.3 


81 


312.9 


282.9 


1182.5 


899.7 


5.41 


0.1851 


0.4546 


1 . 1644 


67.3 


82 


313.8 


283.8 


1182.8 


899.0 


5.34 


0.1873 


0.4557 


1 . 1623 


68.3 


83 


314.6 


284.6 


1183.0 


898.4 


5.28 


0.1894 


0.4568 


1 . 1602 



594 



ENGINEERING THERMODYNAMICS 



Table XL VII — -Continued 









Total Heat Above 
32° F. 












Gauge 


Absolute 

Pressure 

Pounds 

per Sq.in. 


Tempera- 
ture, 
Fahren- 
heat. 






Latent 

Heat, 

L=H-h 

Heat-units 


Volume, 

Cu. Ft. in 

1 Lb. of 

Steam. 


Weight of 
1 Cu. Ft. 
Steam, 
Pound. 


Entropy 
of the 
Water. 


Entropy 
of Evap- 
oration. 


Pressure 

Pounds 

per Sq.in. 


In the 

Water, 

h 

Heat-units 


In the 

Steam, 

H 

Heat-units 


















69.3 


84 


315.4 


285.5 


1183.2 


897.7 


5.22 


0.1915 


0.4579 


1.1581 


70.3 


85 


316.3 


286.3 


1183.4 


897.1 


5.16 


0.1937 


0.4590 


1.1561 


71.3 


86 


317.1 


287.2 


1183.6 


896.4 


5.10 


0.1959 


0.4601 


1.1540 


72.3 


87 


317.9 


288.0 


1183.8 


895.8 


5.05 


0.1980 


0.4612 


1.1520 


73.3 


88 


318.7 


288.9 


1184.0 


895.2 


5.00 


0.2001 


0.4623 


1.1500 


74.3 


89 


319.5 


289.7 


1184.2 


894.6 


4.94 


0.2023 


0.4633 


1.1481 


75.3 


9P 


320.3 


290.5 


1184.4 


893.9 


4.89 


0.2044 


0.4644 


1.1461 


76.3 


91 


321.1 


291.3 


1184.6 


893.3 


4.84 


0.2065 


0.4654 


1 . 1442 


77.3 


92 


321.8 


292.1 


1184.8 


892.7 


4.79 


0.2087 


0.4664 


1 . 1423 


78.3 


93 


322.6 


292.9 


1185.0 


892.1 


4.74 


0.2109 


0.4674 


1.1404 


79.3 


94 


323.4 


293.7 


1185.2 


891.5 


4.69 


0.2130 


0.4684 


1.1385 


80.3 


95 


324.1 


294.5 


1185.4 


890.9 


4.65 


0.2151 


0.4694 


1.1367 


81.3 


96 


324.9 


295.3 


1185.6 


890.3 


4.60 


0.2172 


0.4704 


1.1348 


82.3 


97 


325.6 


296.1 


1185.8 


889.7 


4.56 


0.2193 


0.4714 


1.1330 


83.3 


98 


326.4 


296.8 


1186.0 


889.2 


4.51 


0.2215 


0.4724 


1.1312 


84.3 


99 


327.1 


297.6 


1186.2 


888.6 


4.47 


0.2237 


0.4733 


1 . 1295 


85.3 


100 


327.8 


298.3 


1186.3 


888.0 


4.429 


0.2258 


0.4743 


1.1277 


87.3 


102 


329.3 


299.8 


1186.7 


886.9 


4.347 


0.2300 


0.4762 


1 . 1242 


89.3 


104 


330.7 


301.3 


1187.0 


885.8 


4.268 


0.2343 


0.4780 


1 . 1208 


91.3 


106 


332.0 


302.7 


1187.4 


884.7 


4.192 


0.2336 


0.4798 


1.1174 


93.3 


108 


333.4 


304.1 


1187.7 


883.6 


4.118 


0.2429 


0.4816 


1.1141 


95.3 


110 


334.8 


305.5 


1188.0 


882.5 


4.047 


0.2472 


0.4834 


1.1108 


97.3 


112 


336.1 


306.9 


1188.4 


881.4 


3.978 


0.2514 


0.4852 


1 . 1076 


99.3 


114 


337.4 


308.3 


1188.7 


880.4 


3.912 


0.2556 


0.4869 


1 . 1045 


101.3 


116 


338.7 


309.6 


1189.0 


879.3 


3.848 


0.2599 


0.4886 


1.1014 


103.3 


118 


340.0 


311.0 


1189.3 


878.3 


3.786 


0.2641 


0.4903 


1.0984 


105.3 


120 


341.3 


312.3 


1189.6 


877.2 


3.726 


0.2683 


0.4919 


1.0954 


107.3 


122 


342.5 


313.6 


1189.8 


876.2 


3.668 


0.2726 


0.4935 


1.0924 


109.3 


124 


343.8 


314.9 


1190.1 


875.2 


3.611 


0.2769 


0.4951 


1.0895 


111.3 


126 


345.0 


316.2 


1190.4 


874.2 


3.556 


0.2812 


0.4967 


1.0865 


113.3 


128 


346.2 


317.4 


1190.7 


873.3 


3.504 


0.2854 


0.4982 


1.0837 


115.3 


130 


347.4 


318.6 


1191.0 


872.3 


3.452 


0.2897 


0.4998 


1.0809 


117.3 


132 


348.5 


319.9 


1191.2 


871.3 


3.402 


0.2939 


0.5013 


1.0782 


119.3 


134 


349.7 


321.1 


1191.5 


870.4 


3.354 


0.2981 


0.5028 


1.0755 


121.3 


136 


350.8 


322.3 


1191.7 


869.4 


3.308 


0.3023 


0.5043 


1.0728 


123.3 


138 


352.0 


323.4 


1192.0 


868.5 


3.263 


0.3065 


0.5057 


1.0702 


125.3 


140 


353.1 


324.6 


1192.2 


867.6 


3.219 


0.3107 


0.5072 


1.0675 


127.3 


142 


354.2 


325.8 


1192.5 


866.7 


3.175 


0.3150 


0.5086 


1.0649 


129.3 


144 


355.3 


326.9 


1192.7 


865.8 


3.133 


0.3192 


0.5100 


1.0624 


131.3 


146 


356.3 


328.0 


1192.9 


864.9 


3.092 


0.3234 


0.5114 


1.0599 


133.3 


148 


357.4 


329.1 


1193.2 


864.0 


3.052 


0.3276 


0.5128 


1.0574 


135.3 


150 


358.5 


330.2 


1193.4 


863.2 


3.012 


0.3320 


0.5142 


1.0550 


137.3 


152 


359.5 


331.4 


1193.6 


862.3 


2.974 


0.3362 


0.5155 


1.0525 


139.3 


154 


360.5 


332.4 


1193.8 


861.4 


2.938 


0.3404 


0.5169 


1.0501 


141.3 


156 


361.6 


333.5 


1194.1 


860.6 


2.902 


0.3446 


0.5182 


1.0477 


143.3 


158 


362.6 


334.6 


1194.3 


859.7 


2.868 


0.3488 


0.5195 


1.0454 


145.3 


160 


363.6 


335.6 


1194.5 


858.8 


2.834 


0.3529 


0.5208 


1.0431 


147.3 


162 


364.6 


336.7 


1194.7 


858.0 


2.801 


0.3570 


0.5220 


1.0409 






TABLES 
Table XLVII — Continued 



595 









Total Heat Above 












Gauge 


Absolute 

Pressuie 

Pounds 

per Sq.in. 


Tempera- 
ture, 

Fahren- 
heat. 


, >J~ 


X . 


Latent 

Heat, 

L=H-h 

Heat-units 


Volume, 

Cu. Ft. in 

1 Lb. of 

Steam. 


Weight of 
1 Cu. Ft. 
Steam, 
Pound. 


Entropy 
of the 
Water. 




Pressure 

Pounds 

per Sq.in. 


In the 

Water, 

h 

Heat-units 


In the 

Steam, 

H 

Heat-units 


Entropy 
of Evap- 
oration. 


















149.3 


164 


365.6 


337.7 


1194.9 


857.2 


2.769 


0.3612 


0.5233 


1.0387 


151.3 


166 


366.5 


338.7 


1195.1 


856.4 


2.737 


0.3654 


0.5245 


1.0365 


153.3 


168 


367.5 


339.7 


1195.3 


855.5 


2.706 


0.3696 


0.5257 


1.0343 


155.3 


170 


368.5 


340.7 


1195.4 


854.7 


2.675 


0.3738 


0.5269 


1.0321 


157.3 


172 


369.4 


341.7 


1195.6 


853.9 


2.645 


0.3780 


0.5281 


1.0300 


159.3 


174 


370.4 


342.7 


1195.8 


853.1 


2.616 


0.3822 


0.5293 


1.0278 


161.3 


176 


371.3 


343.7 


1196.0 


852.3 


2.588 


0.3864 


0.5305 


1.0257 


163.3 


178 


372.2 


344.7 


1196.2 


851.5 


2.560 


0.3906 


0.5317 


1.0235 


165.3 


180 


373.1 


345.6 


1196.4 


850.8 


2.533 


. 3948 


0.5328 


1.0215 


167.3 


182 


374.0 


346.6 


1196.6 


850.0 


2.507 


0.3989 


0.5339 


1.0195 


169.3 


184 


374.9 


347.6 


1196.8 


849.2 


2.481 


0.4031 


0.5351 


1.0174 


171.3 


186 


375.8 


348.5 


1196.9 


848.4 


2.455 


0.4073 


0.5362 


1.0154 


173.3 


188 


376.7 


349.4 


1197.1 


847.7 


2.430 


0.4115 


0.5373 


1.0134 


175.3 


190 


377.6 


350.4 


1197.3 


846.9 


2.406 


0.4157 


0.5384 


1.0114 


177.3 


192 


378.5 


351.3 


1197.4 


846.1 


2.381 


0.4199 


0.5395 


1.0095 


179.3 


194 


379.3 


352.2 


1197.6 


845.4 


2.358 


0.4241 


0.5405 


1.0076 


181.3 


196 


380.2 


353.1 


1197.8 


844.7 


2.335 


0.4283 


0.5416 


1.0056 


183.3 


198 


381.0 


354.0 


1197.9 


843.9 


2.312 


0.4325 


0.5426 


1.0038 


185.3 


200 


381.9 


354.9 


1198.1 


843.2 


2.290 


0.437 


0.5437 


1.0019 


190.3 


205 


384.0 


357.1 


1198.5 


841.4 


2.237 


0.447 


0.5463 


0.9973 


195.3 


210 


386.0 


359.2 


1198.8 


839.6 


2.187 


0.457 


0.5488 


0.9928 


200.3 


215 


388.0 


361.4 


1199.2 


837.9 


2.138 


0.468 


0.5513 


0.9885 


205.3 


220 


389.9 


363.4 


1199.6 


836.2 


2.091 


0.478 


. 5538 


0.9841 


210.3 


225 


391.9 


365.5 


1199.9 


834.4 


2.046 


0.489 


0.5562 


0.9799 


215.3 


230 


393.8 


367.5 


1200 . 2 


832.8 


2.004 


0.499 


0.5586 


0.9758 


220.3 


235 


395.6 


369.4 


1200 . 6 


831.1 


1.964 


0.509 


0.5610 


0.9717 


225.3 


240 


397.4 


371.4 


1200.9 


829.5 


1.924 


0.520 


0.5633 


0.9676 


230.3 


245 


399.3 


373.3 


1201.2 


827.9 


1.887 


0.530 


0.5655 


0.9638 


235.3 


250 


401.1 


375.2 


1201.5 


826.3 


1.850 


0.541 


0.5676 


0.9600 


245.3 


260 


404.5 


378.9 


1202 . 1 


823.1 


1.782 


9.561 


0.5719 


0.9525 


255.3 


270 


407.9 


382.5 


1202.6 


820.1 


1.718 


0.582 


0.5760 


0.9454 


265.3 


280 


411.2 


386.0 


1203 . 1 


817.1 


1.658 


0.603 


0.5800 


0.9385 


275.3 


290 


414.4 


389.4 


1203.6 


814.2 


1.602 


0.624 


0.5840 


0.9316 


285.3 


300 


417.5 


392.7 


1204.1 


811.3 


1.551 


0.645 


0.5878 


0.9251 


295.3 


310 


420.5 


395.9 


1204.5 


808.5 


1.502 


0.666 


0.5915 


0.9187 


305.3 


320 


423.4 


399.1 


1204.9 


805.8 


1.456 


0.687 


0.5951 


0.9125 


315.3 


330 


426.3 


402.2 


1205 . 3 


803.1 


1.413 


0.708 


0.5986 


0.9065 


325.3 


340 


429.1 


405.3 


1205 . 7 


800.4 


1.372 


0.729 


0.6020 


0.9006 


335.3 


350 


431.9 


408.2 


1206.1 


797.8 


1.334 


0.750 


0.6053 


0.8949 


345.3 


360 


434.6 


411.2 


1206.4 


795.3 


1.298 


0.770 


0.6085 


0.8894 


355.3 


370 


437.2 


414.0 


1206.8 


792.8 


1.264 


0.791 


0.6116 


0.8840 


365.3 


380 


439.8 


416.8 


1207.1 


790.3 


1.231 


0.812 


0.6147 


0.8788 


375.3 


390 


442.3 


419.5 


1207.4 


787.9 


1.200 


0.833 


0.6178 


0.8737 


385.3 


400 


444.8 


422 


1208 


786 


1.17 


0.86 


0.621 


0.868 


435.3 


450 


456.5 


435 


1209 


774 


1.04 


0.96 


0.635 


0.844 


485.3 


500 


467.3 


448 


1210 


762 


0.93 


1.08 


0.648 


0.822- 


535.3 


550 


477.3 


459 


1210 


751 


0.83 


1.20 


0.659 


0.801 


585.3 


600 


486.6 


469 


1210 


741 


0.76 


1.32 


0.670 


0.783 



596 



ENGINEERING THERMODYNAMICS 



Table XL VIII 
PROPERTIES OF SUPERHEATED STEAM 

(Condensed from Marks and Davis's Steam Tables and Diagrams) 
v =specific volume in cubic feet per pound, h =total heat, from water at 32° F. in B.T.U. per pound, 









n 


= entropy 


, from water at 32°. 










Pressure 
Absolute, 


Temp. 

Sat. 


• Degrees of Superheat. 


Pounds 






















per Sq.in. 


Steam. 





20 


50 


100 


150 


200 


250 


300 


400 


500 


20 


228.0 


V 20.08 


20.73 


21.69 


23.25 


24.80 


26.33 


27.85 


29.37 


32.39 


35.40 






h 1156.2 


1165.7 


1179.9 


1203 . 5 


1227.1 


1250.6 


1274.1 


1297.6 


1344.8 


1392.2 






n 1.7320 


1 . 7456 


1.7652 


1.7961 


1.8251 


1.8524 


1.8781 


1.9026 


1.9479 


1.9893 


40 


267.3 


v 10.49 


10.83 


11.33 


12.13 


12.93 


13.70 


14.48 


15.25 


16.78 


18.30 






h 1169.4 


1179.3 


1194.0 


1218.4 


1242.4 


1266.4 


1290.3 


1314.1 


1361.6 


1409.3 






n 1.6761 


1.6895 


1 . 7089 


1.7392 


1.7674 


1.7940 


1.8189 


1.8427 


1.8867 


1.9271 


60 


292.7 


v 7.17 


7.40 


7.75 


8.30 


8.84 


9.36 


9.89 


10.41 


11.43 


12.45 






h 1177.0 


1187.3 


1202 . 6 


1227.6 


1252.1 


1276.4 


1300.4 


1324 . 3 


1372.2 


1420.0 






n 1.6432 


1.6568 


1.6761 


1.7062 


1.7342 


1.7603 


1.7849 


1.8081 


1.8511 


1.8908 


80 


312.0 


v 5.47 


5.65 


5.92 


6.34 


6.75 


7.17 


7.56 


7.95 


8.72 


9.49 






h 1182.3 


1193.0 


1208.8 


1234.3 


1259.0 


1283.6 


1307.8 


1331.9 


1379.8 


1427.9 






n 1.6200 


1 . 6338 


1.6532 


1.6833 


1.7110 


1.7368 


1.7612 


1.7840 


1.8265 


1.8658 


100 


327.8. 


v 4.43 


4.58 


4.79 


5.14 


5.47 


5.80 


6.12 


6.44 


7.07 


7.69 






h 1186.3 


1197.5 


1213.8 


1239.7 


1264.7 


1289.4 


1313.6 


1337.8 


1385.9 


1434.1 






n 1.6020 


1.6160 


1.6358 


1.6658 


1.6933 


1.7188 


1 . 7428 


1.7656 


1.8079 


1.8468 


120 


341.3 


v 3.73 


3.85 


4.04 


4.33 


4.62 


4.89 


5.17 


5.44 


5.96 


6.48 






h 1189.6 


1201 . 1 


1217.9 


1244.1 


1269.3 


1294.1 


1318.4 


1342.7 


1391.0 


1439.4 






n 1.5873 


1.6016 


1.6216 


1.6517 


1.6789 


1.7041 


1.7280 


1.7505 


1.7924 


1.8311 


140 


353.1 


v 3.22 


3.32 


3.49 


3.75 


4.00 


4.24 


4.48 


4.71 


5.16 


5.61 






h 1192.2 


1204.3 


1221.4 


1248.0 


1273.3 


1298.2 


1322.6 


1346.9 


1395.4 


1443.8 






n 1.5747 


1.5894 


1.6096 


1.6395 


1 . 6666 


1.6916 


1.7152 


1.7376 


1.7792 


1.8177 


160 


363.6 


v 2.83 


2.93 


3.07 


3.30 


3.53 


3.74 


3.95 


4.15 


4.56 


4.95 






h 1194.5 


1207.0 


1224.5 


1251.3 


1276.8 


1301.7 


1326.2 


1350.6 


1399.3 


1447.9 






n 1.5639 


1 . 5789 


1.5993 


1.6292 


1.6561 


1.6810 


1 . 7043 


1.7266 


1.7680 


1.8063 


180 


373.1 


v 2.53 


2.62 


2.75 


2.96 


3.16 


3.35 


3.54 


3.72 


4.09 


4.44 






h 1196.4 


1209.4 


1227.2 


1254.3 


1279.9 


1304.8 


1329.5 


1253.9 


1402.7 


1451.4 






n 1.5543 


1.5697 


1.5904 


1 . 6201 


1 . 6468 


1.6716 


1.6948 


1.7169 


1.7581 


1 . 7962 


200 


381.9 


v 2.29 


2.37 


2.49 


2.68 


2.86 


3.04 


3.21 


3.38 


3.71 


4.03 






h 1198.1 


1211.6 


1229.8 


1257 . 1 


1282.6 


1307.7 


1332.4 


1357.0 


1405.9 


1454.7 






n 1.5456 


1.5614 


1 . 5823 


1.6120 


1.6385 


1 . 6632 


1.6862 


1.7082 


1.7493 


1.7872 


220 


389.9 


v 2.09 


2.16 


2.28 


2.45 


2.62 


2.78 


2.94 


3.10 


3.40 


3.69 






h 1199.6 


1213.6 


1232.2 


1259.6 


1285 . 2 


1310.3 


1335.1 


1359.8 


1408.8 


1457.7 






n 1.5379 


1.5541 


1.5753 


1 . 6049 


1.6312 


1.6558 


1.6787 


1 . 7005 


1.7415 


1 . 7792 


240 


397.4 


v 1.92 


1.99 


2.09 


2.26 


2.42 


2.57 


2.71 


2.85 


3.13 


3.40 






h 1200.9 


1215.4 


1234.3 


1261.9 


1287.6 


1312.8 


1337.6 


1362.3 


1411.5 


1460.5 






n 1.5309 


1.5476 


1.5690 


1.5985 


1.6246 


1 . 6492 


1.6720 


1 . 6937 


1 . 7344 


1.7721 


260 


404.5 


v 1.78 


1.84 


1.94 


2.10 


2.24 


2.39 


2.52 


2.65 


2.91 


3.16 






h 1202.1 


1217.1 


1236.4 


1264 . 1 


1289.9 


1315.1 


1340.0 


1364.7 


1414.0 


1463.2 






n 1.5244 


1.5416 


1 . 5631 


1 . 5926 


1.6186 


1 . 6430 


1.6658 


1.6874 


1.7280 


1 . 7655 


280 


411.2 


v 1.66 


1.72 


1.81 


1.95 


2.09 


2.22 


2.35 


2.48 


2.72 


2.95 






h 1203.1 


1218.7 


1238.4 


1266 . 2 


1291.9 


1317.2 


1342.2 


1367.0 


1416.4 


1465.7 






n 1.5185 


1.5362 


1.5580 


1.5873 


1.6133 


1.6375 


1 . 6603 


1.6818 


1.7223 


1.7597 


300 


417.5 


v 1.55 


1.60 


1.69 


1.83 


1.96 


2.09 


2.21 


2.33 


2.55 


2.77 






h 1204.1 


1220.2 


1240.3 


1268.2 


1294.0 


1319.3 


1344.3 


1369.2 


1418.6 


1468.0 






n 1.5129 


1.5310 


1.5530 


1 . 5824 


1 . 6082 


1 . 6323 


1 . 6550 


1 . 6765 


1.7168 


1.7541 


350 


431.9 


v 1.33 


1.38 


1.46 


1.58 


1.70 


1.81 


1.92 


2.02 


2.22 


2.41 






h 1206.1 


1223 . 9 


1244.6 


1272.7 


1298.7 


1324.1 


1349 . 3 


1374.3 


1424.0 


1473'. 7 






n 1.5002 


1.5199 


1 . 5423 


1.5715 


1.5971 


1.6210 


1.6436 


1 . 6650 


1.7052 


1.7422 


400 


444.8 


v 1.17 


1.21 


1.28 


1.40 


1.50 


1.60 


1.70 


1.79 


1.97 


2.14 






h 1207.7 


1227.2 


1248.6 


1276.9 


1303.0 


1328 . 6 


1353.9 


1379.1 


1429.0 


1478.9 






n 1.4894 


1.5107 


1.5336 


1 . 5625 


1 . 5880 


1.6117 


1.6342 


1.6554 


1.6955 


1.7323 


450 


456.5 


v 1.04 


1.08 


1.14 


1.25 


1.35 


1.44 


1.53 


1.61 


1.77 


1.93 






h 1209 


1231 


1252 


1281 


1307 


1333 


1358 


1383 


1434 


1484 






n 1.479 


1.502 


1.526 


1.554 


1.580 


1.603 


1.626 


1.647 


1.687 


1.723 


500 


467.3 


v 0.93 


0.97 


1.03 


1.13 


1.22 


1.31 


1.39 


1.47 


1.62 


1.76 






h 1210 


1233 


1256 


1285 


1311 


1337 


1362 


1388 


1438 


1489 






n 1.470 


1.496 


1.519 


1 . 548 


1 . 573 


1 . 597 


1.619 


1.640 


1.679 


1.715 



TABLES 



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215 250 255 



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130 135 110 115 150 155 



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550 555 500 505 570 575 
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535 540 545 550 



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255 200 205 270 275 280 325 330 335 340 345 350 500 505 510 515 

Temperature in Degrees Fahr. 
Chart A.— Steam, Pressure-temperature (Table XLVII). 



598 



ENGINEERING THERMODYNAMICS 



14 - 16 


18 20 








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Upper Horizontal Scale ^Pressures inXbs.Ter Sq. In. Abs. 
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200 225 250 



205 215 225 
8 '9 10 11 12 



190 200 350 







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Chart B— Steam, Heat of the Liquid (Table XLVII), 



TAtf-LUSS 



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580 590 

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60 55 280 290 300 305 455 465 ' ' 475 ' 480 630 610 650 655 

Lower Scale ^Temperature in Degrees F. Upper Scale =Pressure in lbs. per sq.in. abs. 

Chart C— Steam, Latent Heat (Table XLVII). 



600 



ENGINEERING THERMODYNAMICS 





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Temperature, Fahrenheit 



370 390 410 

103 110 120 130 140 150 160 170 





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290 310 330 
30 34 , 38 42 46 50 54 



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■ / K r- 




- EL 
















U\ T 


























- 7 












V-v 
















h" E 




























1r G" 












-{\b Gu 


ide-Gurve-- 






t £- 


at 


L £ 


. .. _J .1.. 



30 120 210 300 

Temperature. Fahrenheit 

lS5v 1400 1450 1574 



250 270 290 

21 22 23 24 25 26 27 28 29 29. 



50 60 70 
0.1 0.125 0.15 0.175 



*■ a 


x i- 






=.-1=0-- 




























^ 








jlL 





30 40 



230 240 250 

1195 



.230 560 



580 590 600 
1152 1200 1250 1300 1350 



11 


IS 




10 


17 


i ¥i 


19 s 


M 


! 


] 


































,' 

































































































































KJ j 1 1 1 M 1 1 1 1 i 1 II 






-^ 




v s 






:fct \ 





Chart D— Steam, Total Heat (Table XL VII), 



TABLES 



601 



CO 3 

GO 1 

COD 

CO. 6 
1 

CO G 

CO. 7 

COS 

CO. 9 

CI 

CI 

01.1 

61.2 

CI 3 

61. 4 

] 

61.4 

CI 5 

ei c 

01 7 
120 

61.7 







A- 


/ 


/ 


/ 


- - -/- A 


/ 


/ 



























m 



{7: 



7 



ICO 170 


i; 


-14* 




y 
























./. x_ 




„3 _±_ 




/ 




y ' T 1 




/ ° 


E 



1 1 | i | | . j 1 ■ pj-r 


y 




/ 














mm y 




X 








1 : 1 'J-' 












X 


~\- 








61. S 
G1.9 
63.0. 
62.0 
62.1 

02.2 

I 

C2.2 

C2 3 

02 4 

I 

62.3 
G2.4 



58.0 
58.2 
58.4 
58 



+++ 



^= 



^H 



:5 



120 ' 2C0 

5S.C 
53. S 
.39 



300 50.8 
51.2 



51. C 





>y 


^^ 






~Z ' 


>^ 


< I \/ 


y^-4- 1 v 


>-T 1 71- 


^ .— r .:. 



100 



+= Mil Mil 111 


J ±y^ 


±_± 

-, - r 


l-h-— C3^ 


— .-= 




■-j-3^-r — 


j 1 1 ii 


T 






Tfc^n t 






■I ■ 


-dlLU L. - .... 





210 



^ 



59.2 
59.4 



±B;x 

i 



59.0 



X _T 


i u- 


T 3 


'' 


^ x^ 




z 




^ 












y? "! t 




n T i 








■*" 





220 



! 

62 3 

















50 








GO 




























- 


- 


- 


















^ 


























62.1 










































































_J_ 





59.6 | | | | | | || i 


1 1 i | | 1 1 | | j j 






















^n 




ca n ,2 






iJ 








:.±:.. 



a 

MB3 51 - 8 

52.2 
52.6^ 
52.6 
53.0 
53.4 



240 54.4 

54.2 
54. G 
55.0 



2z 



Uf- 



440 



^i ii ' I m r M +H 



09.0 

cu.o 
J 

co.c 

60.5 

G1.0 
61.5 
G2.0 
62.5 

42.6 
43.2 



* 



250 300 350 100 



xbhHxt^ 
Guide-Cur-ve 



i 



50 100 150 200 

Temperature Degrees Fahr. 













^** 


-^ R~ - 






^: : i ... 



580 590 600 

*■ 

8 ^xtti H-H-t- 



210 220 360 370 380 560 

Vertical Scale ^Density, Pounds per Cu.Ft. 
Horizontal Scale = Temperature -Fahrenheit 



Chart E. — Steam Specific Volume and Density of the Liquid (Table XLVII) 



602 



ENGINEERING THERMODYNAMICS 



2.5 3 


*. 












;? 01 



















































o 
o 

^ 210 

o 

2 
a 
o 



ss eo t5 



55 



460 5*00" -'^50 - 




z 








/• j-1.2 








,/ 












7 


.90 


/ 




/ -1 1 


-.11 






























































8 310 

% 105 

O 

CM 400 



35 40 45 




i " x x x : 




--r 
















^Tfe" 4- 




,$ X 




S I 




3. 1:. 





0035 255 



■Hoo3 15 

0024 



M 



•S '00 



"§ 1200 



O 1450 



CO 

jg 1800 

.•2300 



ffl 



3,5 45 



W 



14 1,6 18 20 1.6 
05 




270 290 320 


4- 3t 47' 


XX 2i 


? r X * 







































.035 405 430 

200 220 250 



1350 1460 



tf 



.030 

2.3 
,025 

2 
3 



M 



F.. 605 



1150 


1250 




i - _ i; ; - o 




_^ 3~ * 




-- - -<« 






„■=" 








- "4" 








' .____ _ 





155 1J0 



1050 



m 



1 1 


"-±~Z1 




<< 






4X ;a 




:xt :? 




-XT ,^ y 




M ' 








iZL.. ...... 





I 



w 



& 



019 
3.5 



110 125 140 50 


1 750 820 530„ 


1 l'l 1 ' 1' 




-= * 










T 


,^ 








~Z- M " 


X - 






X -Jit 


- 




t i?-I 




^ -.26 : 


X ,*- 












x* 




? .24 •° : 


"Z 




s j 


,x 






-j- 





305 80 


90 1Q0 565 620 


685 




" - ,^ I I 




it 


/ „„ X . 


:z 










* ?:::::: 


z 




,z: 


/ 




? 


s 










/ •' 






' . /; 










z 






:?_ 












:?± 












^ X 








-vl7 g Z 











hl.5 
1.4. 



JO 55 155 180 305 330 480 

Lower Scale = Temp.in Degrees "Fahr., Upper Scale =Press. in Lbs.per Sq.In.Abs- 
Chart F.— Steam Specific Volume and Density of the Vapor (Table XLVII). 



TABLES 



603 






N(NN(NNiMN(NNINNNOOMOOMOOr)(05^0»OOiOOcOH 
MMIMINHHOOOlOJOOGONNtOcDiOiO^^^MCOINNHH 
^^^^^^^^MWMCOMroMWCOCOCOWCOMmWWCOCCl 






^OOCKNOOOINOrJfCOOO^OOINNOOcO^OOOcOINOOOOlN 
MOOCOmOOOcO^HOJO^NONMNOOOCOCOHOJ©^^ 
©cOiOiOiOiOt}(^t)<^MCOC<3WW(NN(N(NhhhhOOOO 






(NH>C(N05CO -tf O (MCONCOHCDNCO^ 00 "<* <M 05 l> rjn <M 
M^»ON050MCO'*CON(X)000(N(N(N'*'OCDC005H(MTt*cDOO 

OONCOiO^^CONHOOJOONNCOiOTtinNHOOlCiOONOiO 



t! ^ °4 

aj -p a) 



OO^OOOCONOC^OOO^OO^OJ^OOIMCOONCOOOHMCDOO 
HWiOONOaWiONOOOHW^cONffiOlMW^iONOOOiO 

»O>C»OiOiO«DCO(£>COcOcONNNNNNN00 00 00 00 00 0000 00ai 

^i x^H T^ *^^ T^ ^^ ^^ ^^ ^P ^^ ^T' *^ ^^ ^^ "^^ ^^ "^^ ^T* ^^ ^^ ^^ ^^ "^^ *^ ^H "^H "^H 



•53 cgO 

0381, 

o;t— Ip^ o> 

Q ft 



NTt<HOOiO(NaKOMOCOWON^HOO^HOOiOHNTt(HN 
M(N(N(NHHH000005aiOJ(X)OOOONNNO©<OiOiOiOT}< 



(NMININ(N(N(NININIMM 



^H ^1 ^1 ^^ ^1 ^Jl TJ1 H/H H^ T^ H/i ^H TJH ^H ^JH ^H 



O 3 M. O 



»0 00 HlNM^iONCOHCO'^COOOOHMCOOOOHCO'ONOH 

COCONNNNNNNNOOOOOOOOOOGG05005000000H 
COCOCOCCCCCOCOCOWWWWCClWMCOCOCOWCO'^'^TjH'^^^r^ 
(N(NIMMIMN(N(N(N(NININN(N(N(N(N(N(M(N(M(N(NMN(NN 
OOOOOOOOOOOOOOOOOOOOOOOOOOO 






00WN05H(NC0OH^NO^©e0NM00NTt*©(0NONTtiN 

OOOHiN^iOcOOOaiOHW^iOMXlOHCOiONOOHCCl^tDOO 

OOOOOOOOOOOOOOOOOOOOOOOOOOO 






co 



(NOOOOOOOOOcO^cOCOONcOMMNOOOO^iOO^iMO 
NOOOMtOH^oocOOOWOOCOOl^OiOHNlMOOiOONCOON 

iO^-*MN(NHOOai©00(»Nt>NC0C0»O»C^-*^c0C0f0N 

CNICNIC^tNtNC^C^CNtNT-^THTHrHrHrHrHTHrHrHTHl^rHrHrHrHrHrH 



rj(CONH^©00CO©O5(N©© 






OlOOJOOOHHHlNINNINWMmcO^^^^iOiOiOtOtDCD 
(NININWCOMCOPOMCOCOCOMMMMCOWMMCOnMCOWCOCO 
lOiOiOiOiOiOiOiOOiOiOiOiOiOiOiOiOiOiOiOiOiOiOiOiOiOiO 



0"C 0> • 



to to to to to 10 »o 

lONO^NO^N COOOO^NOiOMcOOO rj< N CO O O 
COIN(NHOOO>00(X)NCOiOiO^COMMH005aoONNCOiO^ 

000000050500505005010505050)0000000000)000000 

OOOOOCO»OiOiOiO»0»0»OtOtoiOiOtOtO»OiO»OiO»OiOtoiO 


MM(N(N(NNINMM^Tt!^rJ<WMN(N(N(N(NHHHHOO 



■*MNHO0)00NC0iO^MINHO05MNC0iOi(C0NHO0500 

I I I I I I I I I I I I I I i I I I I I I I I I I I I 



v m d . 
$ 3GG <3 



toot^oooo too too 

OOcOMHOOiOHNTtfOONNMHiOOiOOWO^OiiOiOH 



I I I I I I I I I I I 



OOOOHHNNiNCOMTtirtiiOffl 
I I + 






CO i> rH. tO 

0)OOOOHH(N(NINMM 



itTj(»oiOcOfflNNOOOOOOOO 

rH rH H rH, rH rHrHrHrHrHrHrHCNCN| 



"3 6^ 

coh 



OOQONCOiOT(iW(NrtOOOONcO»OT)<M(NHOOOONtOiO^ 
<*WC0COWCO«COWMCO(NIN(NN(N(NNINIMINhhhhhh 

I I I I I I I I I I I I I I I I I I I I I I I I I 



I I 






OHIMMHioONOOOOHiNM^iCcONMOOHMM^iOCO 









604 



ENGINEEEING THERMODYNAMICS 



3 ^ 



8 



X 
i— i 

M 

pq 
H 





NMOOCOOi^OiOHtOiNNWCTi^OcOHNlNOO^OlOHNM 
OOffiOJOOOOOONNCDCOiOiO^'H^cOCONfNHHHOOQOi 
COCO(N(M(NlM(N(NNIN(N<N(N(N N (M (N(N(M(M(N(N(M(N(Nhh 


ft -g 
a ••$ 


OOOcOOOOO^O(NcOiMOOO(X)^OcOiMiO(NOcOt)IcOOON 
ON^MHOOC-^COOJNWiMOOOCOCOHOlNiOMOOOO^H 
OQOIOOIOOOOOOGONNNNNCOCOCOCOIOIOIOIOIO^-^^-^ 
rHOOOOOOOOOOOOOOOOOOOOOOOOOO 




*h V 03 

■+> oShh 


OOSOOHcOO^^IMfNIMNNNIMMM^iOcDOOaiOlM^cDOO 
OHMOCDOJONW^COcOOOOM^iOcONOOOONM'^iOiO 


lO^mMHOOOOONOiO^^eONHOOiOONNOiO^MCl 


pi d^ 


OHlN^^iOCcOOOOOOOOOOOOOOOOONOiO^iMHOOOOTtlN 
(NCO^'OCDNCOOiOHNM'^iOCOMXlOOHiMeO'^^iOcON 

ffiffiOOffiffiffiffiOOOOOOOOOOHr- It— It-It— 1 t-H t— It— It— 1 

■^■*T}iTtirtl^^^»0>OiO>0»OtOiOiOiOiO»0»0>OiOiOkO»0»0»0 


Density of 
Liquid, 
Pounds 

per Cu.ft. 


■*HNTjiONC0Oc0(N05iOH00^ONM05^HN"*OCD(Na) 

T)iTjimMMiMN(MHH0000505050000NNNcO(OCO>OiOTji 

tHt-Ht— It— It-It— ItHtHt— It— It-It— ItHOOOOOOOOOOOOOO 


Sp. Vol. 
of Liquid, 
Cu.ft. per 

Pound. 


COiONOSHMCNOlMeOCOOONiOCDMO^NOOMcDOOOi 
HHHH(M(MNNmccmmcO-*'*THx)i^iOiO»Oi.OCDCOCDcDcD 

TJH ^H "^H "^H ^i *^"i tji "^H ^H TJH ^i ^"' ^i Tt^ **?P ^"1 "^^ ^"1 "^^ "^^ ^jH *^^ ^^ ^H ^^ "^^ TJ^ 

N(N(NM(MiM(NN(N^fqiN(NW(N(N(NNiM(N(NlN(N(N(N(N(N 
OOOOOOOOOOOOOOOOOOOOOOOOOOO 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


OC^rJHcOGOO<N^I>>OiT;cOcOOOT-iTtHc005CClTtil>0(NCOOt)(N»0 

OOOOOOOOOOOi050>050iC OOOHHHHiNNINMmcOMT)i^ 

OOOOOOOOOOT-.rH^-I^HTHTHTHTHTHrHTHrHrHT-lrHTHTH 


Total Sp. Vol. 
Heat. j of Vapor, 
Above , Cu.-ft. per 
32° F. ) Pound. 


iOH00iOC0O00iOC0HO55O^(NO00©^lNO00MOm(NOO) 


(MMHHHHOOOOOOJOOiOOOOOOOOOaiNNNNNNcO 


(MiQOiHCOCOOi(NiOa)H(NiO00(M^NO5(N^ffl00HCOCD00O5 


NNN(X)OOQOOOOiOSOOOOOHHHH(NlMlM(NMMmMM 

COCOCO(X)WCO(^COCOCOTHTtl^TtlTfl^T^TtlTtl^TfTtlTtH^'^Ttl^ 


-♦a . 

3« 


N>OOOOMCOOi(N^C0050M© <M -<tf CO 00 (NtPCDOO (MM 
Tji«M(NHOffiaiQONffl©»Oii^M(MHOOai(X)NCO©»OT)i 


Heat of 
Liquid 
Above 
32° F. 


OOffi0300000iOOOOOOOOOOOOOON©CD©©CDiO>OT)iTHTfi 


NOTtiMro(NHOOON©iOTtiMNH00500N©iOT)i«(NHO 
^r)l^Tji^^^^MMMMWMMCOW(NNNiM(N(NN(NN(N 

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


NC005iQH00TtlONC0ONTjlO00>ONO00{0^M(NHO 05 


ONNOOOiaiOHHWMM'^vOiOCONOOOOOOHNMTtliOiO 

HHHHHHHHHHHHHHIMN(NNIN(N(N 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


TtH CO <N 00 iO t-h r-~ rt< t^rtHT-Ht^»0(M05l>iOCOTH Ci 00 I> l> CO 

H(N(NCOCO"*»0»OCONN(»ffi050HH(NM^iOcOcDNOO©0 
NNNNNNMN(N(N(N(NlNNWCOCOCOfOCOMMCOCOMMTji 


o ft . 

02H 


CONHOOiOONCOld^MNHOHlNM^lOCDNOOOiOHINCO 

i i iT i i i i i i i i \ 




SOOOOHNM^iOtriNOOOOHNMTiiWCDNOOOSOHNM 
■^TtiTjiiCiOiOiOkOi^iOiOiOiOcDCDtOCOCCOcOcDtOcDNSNN 

Tp t^i "^i "^H *^jH ^H ^f* ^H "r^* n*^ ^H ^H t^H t+4 ^fH ^H T+i t^H t^H ^H ^H *^H t^ *^i tji ^i t^i 



TABLES 



605 



>> 

Q, U 


00-^OcO(NNM05iOHNCO(»T)*OtO(NOO^OcO(NOOTtiOcO(N 
Hr 1 HH, 1 HH rH 1 r 1 HHHHH 1 Hr 1 T 1 H . H 1??°9° 




®(NOOOtO^(NOiMOCO(NOOcO^(NOO ^cD(NO(N^(DcO 
OJNiQNOOOffl^HOJNiOlNOOOcD^IN N^N05HCC>ON 
COWCOWCOININNIMHHHHHOOOO OOOOi-ti-Hi-Hi-1 

ooooooooooooooooooooooooooo 




>- o> ej 
as-g a> 


OM^OOONiOOOH'^NO'^NO^OOINCOO^OONOO'*'* 
NNOOOOOOH(N^iOiONOOOOOOOHHiN(NNCO(NINHH 




External 
Latent 
Heat. 


OOOONOOCOlNOlCOMOOMOiOlMOC^OCDtNOO^OcOH 


HHH(NW(N(NNaNN(NN(NN(N(NNCOMMMCOCOMCOM 


Density of 
Liquid, 
Pounds 

per Cu.ft. 


^0©(MOO-*OCOHNW05»OOcO(NCOM05tOOcO^MrOOOTl< 
'-f'^COCOfNfMlNrHr-iOOOXJJOJOOOO^NCOCOCOiOkO^^cOcO 


OOOOOOOOOOOOC50iaiOi050505ffiOiOi050)0500) 


Sp. Vol. 
of Liquid. 
Cu.ft. per 

Pound. 


(MiOOOOMiONOiMi-OOOOWCOQHWCOOOlMiONOCOCOaKN 
NNNODOOOOOOOOOQOiOOOOhhhhiminWMMCCiMt)* 

(NINW!NlN(NMNN(M(MIN(N(NNNlN(NIMlNN(N(M(NO(N(N 

OOOOOOOOOOOOOOOOOOOOOOOOOOO 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


OOlNiOOONOOMNO^OOIN'OaiCONHioaiCONINCOINtOH 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


iCOiOINCO'^OOOcO'^COMMNINMM'^cONCOOhMtJIcoiX) 

NCO^MHOOiNtOiO^MiMHOOlOONcOiO^^eONHOOS 


<CcDcOOcO©»OiO»OiOiOiO»C»OkO'*^^ , *'^'*'*'*Th'^TtiM 


Total 

Heat 

Above 

32° F. 


<N^OI>. (N^cOOOHNiOOO (M-^iOOOO (N M © © N 00 OS 
^^^TH»C>OiOiOiOOcOCDcONNNNI>NOOOOOOOOOOOOOOOO 


Latent 
Heat. 


iCtDOOOHM^tOOO HM1OCO00O5 HNM^iOtOOcOCOffl 
M'MHOOOCONfflcOiOTHMlNHOOOOONCOiO^WINHO 


• Heat of 
Liquid 
Above 
32° F. 


CO(N(NNHOOOOCDO)OONCOCOIO»OIO^NIX)QOOOH(NM 


0300N©iO^WNH©OONcOiO^M(NHOOH(N^iOCONGO 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ++ + 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


©O©05O05OOOhNM^C000h^«005N>0OMNH«0h 


CNOOOlOHW^iOtONOOOiOHM^iOOOOffiONWiOCDOO 
(MC^(N<MCOCOCOCOCOCOCOMCO^^^^^TJH^Tt*lOiO»0«5lC>0 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


C0C0C0©O©NNN00ffi HM»000HC0®05NC0O^00W00 


HNcC^iOtflNOOOlOHCO^iOCONOOHMTttiONOOffiHiM 


Scale, 

Temp. 

°F. 


Ti<»otoNooo)OHeqfOT)<ictoNooaoH(NeoT)i>o{ON[)OffiQ 

HHHHHH(M(M(N(N(N(N(N(M(N(MCO:CWWIWCOCOCOCOW^ 




^lOcONQOffiOHfMM^iOcONOOOOHNM^iOtONOOOiO 
^f^t^b-t^t^OCOOOOOOOOOOOOGOOOOOCS500i0505050i0500iO 

*^H "^ "^ Tp "^i -^H "^H ^J4 ^H TJH -^ ^) ^H ^i ^j ^H t^ ^i ^H ^H ^^ TT* *^ "^ ^^ ^^ JO 



606 



ENGINEERING THERMODYNAMICS 



I 



pq 



3 Q 



>> 

ft fc 

°«« s 

»" O ft 

£ > 


00^0©MOO^O>0(NOOCOO)iOHNCCi05iOHOO^OCO(NOO^ 
NNMOcDiOiOiO^^MMiNtNNHHOOOOSoaiOOoONN 
OOOOOOOOOOOOOOOOOOOO05050i05O0i05 


& -3 


OON^-*iOOO(X)(»ONM(N(NO(NOOOOH^»OcOOOcDOOO 
OM^COOOON^cOOlOM^NCSHirOCOOOOlN^coooONU: 

OOOOOOOOOOOOOOOOOOOOOOOOOOO 




c fl +j 
E o> o3 

3m w 


■^OiOJOl'^OiiOOiSNOlOlMffiCOMONiONkOOOCOOMNH 


»0^cO(MH00300NcOiO^M(MHOOi(X)N©i0^ei:(MHffiOO 


03 +» . 

h © « 


©HOHCOHiOO^OOHTjfoOH^sowiOoOOlN^iONOOrt 
»OCO!0NN0000Ci)©O)OOOHHH (M(M(MCOCOCOCOCOOOCO 




Density of 
Liquid, 
Pounds 

per Cu.ft. 


05iOO©HNINOOC005'*0>OOiOOiOO'00'00>00(DH(D 
(MlMO<HH000505(X)a)tX)NNcOi©iO»OTf^MfC(M(MHHO 


©oiOiOiOiQOioooooooooooooooooooooooooooooC'OOoooooooo 

MMCOCOD5WMCOMMWMWCOi.-jMMMMMCOMCOCOMMM 


Sp. Vol. 
of Liquid, 
Cu.ft. per 

Pound. 


■^OOh^NOIINtHNO^NOMiOO'^NO^NO'^QOH^OO 
^■^iOiOiOiOcOCOcONNNOOOOOO©OlOiOOOHHH(N(NN 

OOOOOOOOOOOOOOOOOOOOOOOOOOO 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


iCcOCDNNOOOCOJOOOHHlNMM^iOiOtDcONOCOOOJOH 
NNNN(NN(NlNNnWCOCOMMMMWMCOMWCOMM'*T}i 




Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


H^NO^NH^oONOHiOOOWNHCOONiMNMOOCC©'* 
0500r-NOiOiO'^COCO(M(N'-i0005CROOOOI>l>©OiOiO^'<* 


MCOmcOMCOMMMMMMnMMlN(N(N(N(N(N(NNiN(N(NIN 


Total 
Heat 
Above 
32° F. 


MOOOOOOiaiffiOlffiOOOOOOOOOHHHHHHHH 


If ® 

3* 


!0>0>OCOCOCDCOCO<0«00<OCOffl©©CiiOfflCD»OiOiO^'^MeO 


QOONcOiO^MNHOOiOONfflkO^COMHOOJOONCOiiJ^M 
COMCOMCOfOO-iCOCOMNM(N(N(N(N(NIMINNHHHHHHW 


Heat of 
Liquid 
Above 
32° F. 




OJOHClCO^iONOOOlOHNniOCONOOGOHNM^iONW 

HHHHHHHHH(N(M(M(M(NlM(N(NMCOCOCOWCOMCOW 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


lOOtOOJiCMOliOM OOCO^MNIMHHHHrooOOOOOWOJH 


03H(MMiONOOO(N'^iON05HfOiCN03HCO^C>OOON'*N 
lOcOcDcOcDCiiCNNNNNNOOOOOOOOOOOCiOJOJffiOOOO 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


MN(NOMO>©(NON>OWH 03©0000000050»OiC»OiOCOOO 


■*iON00OHMlON00O(N^c0NO5HMiONaiHM>ONO5H 
NNNNOOOOOOOOOOOO©0)0)0505©OOOOOhhhhhM 


<o ft . 
o <u 


H(MMT)(iOcDNQOffiOHNCO'*iOcDNOOaiOHMCOTtfiOtON 


m ft • 


H(NM^>OcONOCl030H(MM^iOfflN00050HMMTjiiOCON 



TABLES 



607 



'B 
3 


ONC0O3iOHQ0THO©N00»OHNCCO©(Na)iCHNMOffi(N 
NCOcOiOiOiO^Tj*^coMMN(NHHHOOOJ05050000(X)NN 
C5050©05050505050305(J50i050505C)OQOO(X)OOOOOOOOa)CO 




ft Tj 

-tog- 


ONONNNNMWTPfflOCDMOOOONN^^CNOOOOOOOO 
N050MiON03HCOiON05HCOiOOOO(N^fflOOON^(COOO 
NN00000000000503Oi0505OOOOhhhhhN(N(N(N(NC0 

O O O O O O O O O O © © i-H iH i-H tH .H l-H 1-4 1-4 ,H rH »H H r-J ,-1 ,-H 


a CI +3 
m <U sj 


O^OOMNNOCO^OCONNOOOOOHN^iONOOON^cOOO 
<X)^^O^^^COCMi^OC}OOCOTtico<NOOit>.QO"<tiTriC^OOO<0 


NCOO^WfNHOOSOONiO^MNHOOJN^kO^COINHaiQO 




OH(NMMM^'*'*^COM(Nhh00500COiOCO(NOOO©tJ((N 


Tt^ *^ ^* ^^ "^ ^ *^ "^H" '^ ^* ^ ^ ^^ "^ ^^ "^ ^^ "^ ^ ^^ "^ '^ ^^ T^ "^ ^4 ^4 


Density of 
Liquid, 
Pounds 

per Cu.ft. 


HtOHOHCOlNNiNNWNINNMcOHiOOiOO^OJMOOlNN 
005®OOOONNOOiO»0^^mM(N(NHH00050000NN© 


oor^l>t>r^t-t^i>-t^i>l>r^r^r^r^r^r^r^r^r^r^cocDcocOcocO 

MCOWCOCOnMWCOCOMMMMCOMMCOMMnCOMCOCOWCO 


Sp. Vol. 

of Liquid, 

Cu.ft. per 

Pound. 


HiOQ0M»O05N^00HiO(»lNiO0iM00O>Oa>CCCDOiO05'*N 
MCOCO^Tti^iOiOiOcOOCONNNOOOOOJOaiOOHHHiNIN 
cOcOcOcDcOcDcDcOcOcOcOcDcOcD©CDcOcOcOcONNNNt>NN 
(N(N(N(NM(N(N(N^INM(N(NW(N(N(N(NIN(NlNN(NiN(N(NCl 
OOOOOOOOOOOOOOOOOOOOOOOOOOO 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


HMM^^iOcDNOOOOOJOHMCO^iOiOcOMOOaiOHlNM-* 




Sp. Vol. 

of Vapor, 

Cu.ft. per 

Pound. 


Oi0HNC0OO(N00i0H(X)^HiX)i0(N0)OMO00i0NON^ 

^MCONWNHHOOOOOJOOOOOOONNNNcOOCCOiOiO 

(NINlNINNlNlNlNINlNINHHHHHHHHnHHHHHHH 


Total 
Heat 
Above 
32° F. 




a> cs 


(MrHrHO OSOiOONCOiO^cOHaiNO^MH OOMOCOH© 

(MH00500COiO^C«5(NHOO)OOCO»O^M(NHO[)ONOiO^IN 
HHHOOOOOOOOO050i0i05Oi05O3®O)00iX100M0000 


Heat of 
Liquid 
Above 
32° F. 


(N^iOCOOOOOOOHiNM^iOcOOOOOlNMiONOiMM^iO 


OiOH(N»^iCN00030H(N«l^iOCOOOOiOHlNWiOcON(X) 
W'*'^'^^^Tt(^TtH^kO>OiO»0'OiOiOiO»OcOcO©COOi©<C« 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


N^NHiCOMOOMOO^OkOOOMHHiOiO^MWHHWiO 


0>H«COOOOMiOOOOMOOOOM005lM^NOMCOaiiM»000 
OHHHHiMlNlNNMCOCOM^^TH^iOiOiCCOcOCOCONNN 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


OS t-H -^ 00 <M © *C iOHN(NCOMOOOOO(N(NH OOOO <N 
M©COOMiOOOOMiCOOOMiOGOHMtDOlM»OOOHM<OOM 


oTft . 


COOJOHNcO^iQcONoOaOHiMCOTtiiOCDNlXlOiOHNCOTti 
COfflNNNNNNNNNNOOOOOOOOOOOOOOOOOOOOOJOJOiOSOl 


• ft . 

SB* 


OOffiOH(MM^iO(ONOOOlOH(NMT)<iO?ONOOaiOH(MWTfl 



608 



ENGINEERING THERMODYNAMICS 



§ 
O 



X 
i— ( 

X 

m 
H 



5? 
O 

A 

s 

H 

GQ 

O 

m 

W 
i— i 

H 

Ph 

O 



>> 

a fn 
o«_, o 
£ o §< 

1 > 


©C0cO'O»O^^'*MC0MM(NlMHHOOO©O3©000000NN 
00 00 00 00 00 00 00 00.00 OOOOOOOOOOOOOOOOOOOONNNNNNNN 




Entropy 

of 
Liquid. 


OOOOOiOOINNINNININININlMINlMCOCOW^^MMCOCOlNIN 
N^cDOJHOOiONOlHCOiONCTiHMiONaiHMiONOJHCOiO 
roccrocO^^^^^*OiOOiOiOOC©COCC>CC>r^l-~l>t^l>000000 


E cd 03 

-P 03HH 


OMiOCOHCOCD03NiOOOtO'*NO'*OOHioaiW(NHiOO'*0 
iOCOHON^N003CO^«®N>0(MaiNi<HanDcoaiCO(N05 


NtOiO^IMHOOJNiOiO^iMHOOJNtOiO^INHOaiNcOTtl 

NMMM(N(M(Nr- It- It-Ht— It-It- It— li—lOOOOOOOOOiC^ClC} 


External 
Latent 
Heat. 


ONIO(MON'*HOO»OW030COOCOIMOJ10HNM03»OOOH 
C<| tH tH-tH OOOOOJ05000000000NNCDcOcOiOiO'T)iTtiTliMfCi 


^^^TH^il^TtlfOCOMMMCOMCOCOCOfOMMCOMWMCOCO 
!00»0»0»OiOiO»OiO>OiO»OiO»0»OiO»0»0>OiO»OiO'C'OiOiOiO 


Density of 
Liquid, 
Pounds 

per Cu.ft. 


H<0OiiffinNMOO^aMN(NC0O'*OMNH>003e0NH 
cOiOiO^WM(NiMHHOOia>OOOOI>NCDiOtO'*TtiM(M(MHH 




Sp. Vol. 
of Liquid, 
Cu.ft. per 

Pound. 


(NiOOlCONMCOHlOOiCacONHCDHCOOiOOl^OOMOlMOO 
MCOCO^^^iOcOtONNNOOOOffiOOOHHHlNINCOCO^ii 
NNNNNNNNNNNNNNNNOOOOCOOOOOOOOOOOOOOOOO 
M(NMM(N(N(NCqiN^MMMlMMM(N(N(N(NMMIN(N(N(N(N 
OOOOOOOOOOOOOOOOOOOOOOOOOOO 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


lOcOOOOiOHNCOiiiOtDOOaiOlNCOTHONCCOHCqiiiONCO 
OCOCOCONNNNNNNNNOOOOOOOOOOOOOC010505 05 050 05 




Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


N05N^NO00c0^NO00©rjl(MO00OilC0H®00cDTt<C0H 

O^T^T^T^T^COCOCOCOCOCaCQtNC^CqT-HTHT-lTHT-HOOOOOO 


Total 
Heat 
Above 
32° F. 


■^cODSHOOONNNTH^MiNlNHOlOOcDiOCOlNaiOOtO^OOJ 


HHHHHOooooooooooioirooioiffiooooooooooN 

I0i0i0»0»0'0i0i0»0»0i0i0»0i0i0i0i0>0'0i0i0"0i0i0i0>0'0 


a> o3 


N»OMHOOiOCOHOJOt)IhcOCOM t^Tt<TH00»OT-l00Tti COCO 

H00500COkOiiroHOOiOOcDiOTJicOHOO)NcO»OCOlMHC6X 
OOOONNNNNNNNcOcOcDcO©CDCDCDiO"OiC"OiOiO»Oii^ 


Heat of 
Liquid 
Above 
32° F. 


N00OO(MM^c0(X1Q0O(Nt1I(0C005H(Nt}Ii0N00OHM'*© 


OJONCO^iOONOOOlHlNCOiiiOiOOOOOHlNCOiOcONOOOl 
©NNNNNNNNNOOOOOOOOOOOOOOOOOJQOlOlOiOJOJOiC 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


MMNOJW00MO0M00C0 00MMCOMO5MMCOMM00MWCOCO 


HTtlNO^NOCONO^NH^OOMfflOilOOlMCOOiOCBMN 
OOQOOOOiOiffiOOOHHHiMMiMCOCOiiTji^ioiOOcDcONN 
HHHHHH(N(NIMIN(N(NINIM(MNM(N(N(M(N(N(N(MIN(MIN 


Pressure, 

Pound 
per Sq.in. 
Absolute. 


t^ CO »o »o *0 lO lO 

©©(MiCONiOOOlNWOlMOOiCONHiOOSMNHiOO^OON 
030)OOOhhh(MM(NMCOCOt)i^iO»OiCcDcONN00000001 
HH(M(MN(N(NNNN(MN(M(NN(MIM(M(N(N(N(NNWN(N(N 


Scale, 
Temp. 


iOtONOOO)OH(NCOr)iiOcDNopO)OH^i>3iiiOONOOOiOH 
OiOSOSOiOiOOOOOOOOOOrHT-iT-i-T-HT-iT-HT-iT-iT-i^HCNC^ 


^ a . 

SB* 

1 ^° 


lOcDNOOOiOHMM^iOcONOOOlOHiMCO^iOiONOOGOH 
iOiOiOiOiO©CDcDCDcOcOcOCOCO©NNNNNNNNNNQ000 
»OiOiOiOtO»OtO>OiOiO»OiOtOlOtOiO>OiOiOiOiOtOiOiO>OiO»0 



TABLES 



609 



.« 


S 


fi 


<1 




P 




w 


1 


H 


i— i 


2 


h-l 


P 


X 


H 




<1 


s 


C/J 




fe 


< 


O 


H 


CO 




W 




HH 




H 




tf 




w 




P-l 




O 




tf 




P-. 



1 > 


O3iOH00^Oc0M©C0N00iOH00rt<HNMONMO©C0O)i0 
00©iOiO»O^^COCOMiM(MNHHHOOO©05C»(»OONN 
NI>NNNNNNNNNNt>l>NI>NNNNCOCDOCD<£)0?0 




a t3 


(NW«MMMMMMM(N(N(NMmMmMNCOMN(NlN(N(N(N 

N0)HMiON05HWiONOiHC0iON©HC0iON05HM»ON© 
Q000©0)050)03OOOOOhhhhhnNNNNMMMMM 
HHHHHHH(NIM(N!NIMM(N(N(NMIN(NN(N(N(NININ(NIN 


u V 03 


■^05COOOT)<aJiOOiOCDN^ONCCOOO»0(M>ON03SOCOO>0 
NMHOOcO(MOcD030N^NOOiO(M03CDM05iOHI>^OCDiN 


nwHoiooso^MiNoaioocoio^WHOooNco^roMoa 

COWMMCOmcOCOCOWCOMMCOCOrocOCOCOMCOMMMWCOM 


"3 43 . 
C fl-P 

(- <u d 


COHNNcOHiOOi005McOOCONON»OOOOM©t)00(M-*iO 
(NiMHHOOOiaoONNOOiO^^MIMHHOOiOOtlONOiO 


MMMMMMINNNNNINNNNNNNNINNHhhhhh 


Density of 
Liquid, 
Pounds 

per Cu.ft. 


lOffiMSHiOOiMSO^OONiOffiMOO^OOMiOOlMCDOM 
00050000NCOCOiOiO'*COM(NHH00050000NCOCO'OiO^ 


^^Tt<T+iTt<^TtiTt<TttTtiTf*Tti^T+irt<^^T)HfOMCOCOMirOCOirOCO 


Sp. Vol. 
of Liquid, 
Cu.ft. per 

Pound. 


CONCONWNHNMKXNOiTtfCJiOHOOHiOMNlNOO^OiOIN 
iO<OOcONN0000030500HHiNCOM'*^iOiOcOCONOOOOOJ 

oooooooooooooooooooooioiai05CT)Oi050J050305©05CBOioiai 
ooooooooooooooooooooooooooo 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


OOOOOOOOhhhhhMIM(N(M(NCOCOMCOCOM'*^^ 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


o ioo»ooooo io ic m 

O05O)0305O0i030000000000000000NNNNNNNNNI>CD 


1— 1 


Total 
Heat 
Above 
32° F. 






43 . 

d-P 
a> c3 

-P & 


(OMONMOiON OO-iO HOO^OOMOSiOOcOHCONNMOO 


NiO^MHOffiNO^COMOOlQOCD^OMiMHOlOOCOiOMlNO 
•*^^^^^COMMCOMMCONIM(N(NM(NIMHHHHHHH 

"^ "^ "^f "^ "^T *^ ^T* '^ "^ *^ "^H *^ "^ ""nT* TT" "^ "^ Tt^ TT" ^ TT" ^H '^ Tf "^ "^ *^ 


Heat of 
Liquid 
Above 
32° F. 


OHM^iOONffiOHlMcO-iHCONOOOSOHM^iOCONC&OH 
OOOOOOOOt-h^Ht-i^h^^h^h^h^ho1(MC^(N(N(M(MC^COCO 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


MMM(X)MMMWMeOMMmMWWOOmoOMCOMMMMD3M 


H^OONNHioO'OO'OOiCO'CO'OO'CHtOHMMNm© 
OOOOOOOlOiOOr-iHiMMCOM^'^iOiOOcDNNOOOOffiOOO 
(NNNNNMMMMMMCOCOWMCOMMMMMMCOWMilT+i 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


^®MNN(OO>OO>OOiOO>OOiOOiOOcDHOlNNN00^ 
OJffiOOHH(N^COMM*TtHOiOcDcONN0000050500HH(N 


<a a . 

« (Co 


SK5 l °' :0N000iOH(NM '* li 5 i: C | N0005OH(NM'*iOCON00 




^<£^£$gt^coa>OT^c<i<rc^iocor^aoc^a^c^roTt<io<or^oo 

ooooooooooooMGoaioiaaoicsaaiooiOOOoooooo 

iOiCiOOiOiOiOiOiOiOiOiOiOiO»OOiQiOcO©cOcOCO©CO©© 



610 



ENGINEERING THERMODYNAMICS 



X 



>> 
©«« ° 


(NOO»OHOOT)(HOO^OSCOO©C005CO(NOO»CHNrHOcOC005 
N©Cc0»O1010^tJ(tJ<MC0M^NHHHOOO0J01050000N 
0©0©COCOCOCOCOCOCO<DCO©fflCOCOcDCOOiOiOkOiO»CiO 






(NlNOOOHOOJOOOOOOOOOOOOONCOCO^iUMOOOOOOOO 
HMiONOiHCOTlHcOoOO(N^CDOOOW^COOOON^cOOOaiH 
^■^tH'^'* iOtOiO»0»OcOCO©COONNNNN 00 00 00 00 _O0 00 O) 
N(N(N(N(N(N(NIN(N(N(NIN(N(NNN(N(N(N(N(NNIN(N(N(N(N 




K <U 03 


nfNHOOOOOOOOOHM^ifJNOOON^(OOOONiOOO 
OO^ONCOOJ^OtOHiOHNlMcOHiOOiOO^OOlNNHiOOJ 


COMMMCOMWMCOMCOMMCOC005COMWWCOCOCOMMCOCO 


el fl+5 


CO00O5OOOOOOOOO0500c0i0C0<NO00OTt<(NO00iO(N 
ThCONiNHOOlOONCOiO^MHO^OONO^OONHOOONcO 

HHHHHHOOOOOOOOOOiOiOffiOiOOiOaoOOOOO 


Density of 
Liquid, 
Pounds 

per Cu.ft. 


NOMNOCONOM©OMiOCOOMCfiOOHT)(NON>ONOM 

MMNHHOOiOSOONOOiO^^MNHHOfflflSOONCO©^ 

co co co co co co cm in eq <n <n cm' cm cm cm cm <m cm cm CM i-i tH t-1 ,-J r-i i-! i-H 


Sp. Vol. 

of Liquid, 

Cu.ft. per 

Pound. 


OOCOOcOHOO^aiOW(»iO(NOcOT)<OOiOlMa)iOM05NU5H 
OOHHiN(NMC0Tt<iOiOON000005OOH(N(NWTj(^iCcDN 

05000000000000000i-lTHrHr-(r-^i-Hi-(T-li-li-lT-l 
(NCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO 

ooooooooooooooooooooooooooo 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


NffiH^©00ON^C000NiOC000(Ni000H^N05(N»ONOM 
•^■^ioioiOiOCO©COOCONNNN00000005050iaiOOOHH 

,_i ^ ,_; ,-i ^ ,-J ,-J ,-J ^ ,-J ,-J ,-i th" ^ ,-J ,_)' ^ tH rH r-i r-i rH C* C<J (N* <N* N 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


HN^tOOOO^OOOOOOOO(N(NTH©00(N^OO(NCOO 
OONcOiO^MMHOOOOONOOiO^MiMHOOaiOOOONN 
CDcD©cOfflcOcOtOcOcOiO>OiOioiO»OiO»OiO>OiO>0'^^^ , T(<^ 




Total 
Heat 
Above 
32° F. 


HOOiONONMONMON^OOlMOO^OcDNNHiOOiOH 


NHHHHOOOCiaooaoOOONNcdcOcdiOvO^ilcOMNN 
^^rJH^T^TjHTjHTJHCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO 


-P . 
d-f5 

0J 03 


C000M©TttO5C000C0NOiOO-^l>H^00HiO00HTl<NOC0c0 


OlN©^c0HO00N»O'*(NHaiNC0'*lNHffiNc0'*(NHaiN 
OOOOOOOOOOOiOOOOOOOOOOOOOONNNNNNffl© 
^^^^^H^^COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO 


Heat of 
Liquid 
Above 
32° F. 




COCOCOCOCOCO^tirtHrH^^TH^TtlTtiuo»OiO>OiOiO>0>OcOcOcCiCD 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


COCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO 


^O>OMNiM00rHO«OHNma)>OM00TJ<HN'^H00iOOCnCi 

HNNMMtH^iOcOiONNOOOOOJOOhNNW'^^iOOcON 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


ffi»0000(NNCO C5JlO HCDNOO^OOOMOlCOlNOJCOCOOiO^O 
(N CO rt< rjl to lO CO COV- 000005©OHH(MNmr(l^iOcONNOOO 


■aah 

o <i) 
03 H 


CSiOHNfO^iOcONoOOJOHiNM^iOCDNOClOOHNCO^iO 
^>CiO'OiOiO>OiO«:iO»OcOcOcDcDcOCDcOffl<OcDI>NNNNN 


,sa^ 


©OH(Nc0rtliOC0N00c3>OH(NC0TttiOcDN00a)OH(Nf0rtl»O 
OHHHHHHHHHH(N(NN(N(N(NCN(N(NcNICOrtMMMM 
COiOcOcOC05DCOcOCOcO(OtOOcDcD©©cOcO©tOOcOcflCOcOcB 



TABLES 



611 



n_ o 


iOMOO^ONCCOHOINOOCOCDNOO^OCOINOOtHOCOIN 

NNtococo»0'o , *'* , *ncom(N(NHHHOOo>©a>cooo 




°<4-.'3 

1 a 


©COO^OMOOCiOOOOOOOOOOCO^Ttc^^rJi^^rHTjfcO 

0)05ffiG)OOOOOHHHHH(N(N(N(NiNWroC0C0Mrtt 
iN(N(N(NCOlXiCOCOCOCOWCOCOCOWirOWCOCOMCOCOCOWCO 




a dp 

M « e3 
-|J c3hh 

d>_;« 


O-*NO^00(MC0iO'*00(NCDO»OO'*O5Ttf00C000C0CTHO 
lONHiOOOOWCOCONOOOHfO'^iOiOCONiOOOOOOJOOO 


NCOMCOCOCOCOCOCOCO(NlMN<N(N(N(NW(M(M(N(N(NiM(N 


03 4J . 

d d^ 


OC0WOc0M»^OO(N00^Oi0OOHt0(NN(NNHiO 
IOC*5(NH0500CDIO'^IMH0300NIO'*(MH0300C010M(NO 


OOOOOOOONNNNNNNcOcOCOCOCO(OCOiOtO>OiOiOiCiO 

"^H ^H ^H ^H ^^ "^^ ^^ ^P ^"* ^^ "^* ^^ ^"* "^^ ^^ "*^ ^^ "^T "^P ^H ^3^ * , 5^ ^^ ^T^ ^J^ 


Density of 
Liquid 
Pounds 

per Cu.ft. 


IQOOOINIONOINIONNIN^OOOOHCO^CONOOOOOIO 
^MM(NHOOOiOON(OtO»O^MM(NH00500N(0»OiO 


,-i,-i,-irH,-i,-i,-i©©0©©©©0©©©©C50>0>a>CiO> 


Sp. Vol. 
of Liquid, 
Cu.ft. per 

Pound. 


©NiOMOOOtO^HOOlCO^CONOOOlOaaiOOOO 
NOOOJOHHCTro^iOiOcONOOOlOHHtNM^CONOOO 
HHH(N(M(M(M(N(MNNN(N(N(NMeOMMWMMWirOW 

MCOMCOMMMfOMCOCOMMMMMCOWCOCOCOCOCOCOCO 

ooooooooooooooooooooooooo 




Density of 
Vapor, 
Pounds 

per Cu.ft. 


OOOON^NOMOOOiOOMiONMNOiOOONCOO^OO 
HHNMNNMMCO^^^mioiOCO^NNNOOOOOiOJO 


N(N(NN(NIN(N(N(N(N(N(NM(NNMN(NININN(NINNIN 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


^OO^OOO^OOOCOOONCOIMCOO^O^O^O^O^ 
cOiOiOiO^^cOINIMHOOaiOJOOOONNOCDkOiOrtt^CO 




Total 
Heat 
Above 
32° F. 


NOiOO-*NO(N>OOOOMiOOOO(NfOONOOO)H(NMiO 


"HHOOOJOOOONOiOiO^MlNMHOaiOONCOCOiO^W 
MCOWCO(M(N(NlN(MN(NN(N(NNM(NHHHHHHHH 
iO'OiOiOiO>OiOiO»0»0>0»0»OiOi^>OkOiOiO>OiO»C'0»CiO 


+3 . 

ej-g 

<D 03 


OHrPOOOOiOOOOOOOOOOiOOOONffliO^MHO 


CO^NOOOcOiONHONiOMHOiCD^INOOOO^CqOOO 
MfOMCOMMMMMMMMMMCOMMMCOMMMMMCO 


Heat of 
Liquid 
Above 
32° F. 


NOSHTj<toooOM»OOOOM>OOOOMiOOOOlNr((N©(NiO 


<OO00O5OHC0^iOC00005OHMrJ(iO<O0Cl0)OH(NTl*iO 
COCOOcONNNNNNNNOOOOOOOOOOOOOOOOOJOiOiOiO 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


MMMCOMMMMMMMMMMCOMMMMMmroCOCOM 


COON»OHanOMHffi>O^H05NiOrtlNH©NNiOiOM 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


00»C(NOcO^OOOCO^Offlffl^(MO©OOffl^(NNOCCC 
®Oh(N(NMt)<^iOCONNOOOOHHNM^iOCONOOCO 


Scale, 
Temp. 


ON000JOH(NM^»OCDN000iOH(MMTiHiOfflN0005O 

NNNNoooooooooooooooooooooiOiaiOffiOiaiOJaioso 

t-HtHt— I»-Ht— li— li— It-HtHtHi— ItHi-HtHt-Hi— It— lr-(THTHr-lrHi-H^-lC<l 


• a 

2B* 


©NOOOOHiMM^iOcONOOOiOHlMMTtfiOtDNOOOSO 
COCOCOCO^rtiT^TjHTti^T)H'#-rHTtHiOiOiOiO>OiO»0>O^O l O^O 



612 



ENGINEERING THERMODYNAMICS 



■+3 

3 
•o i 

4 
I- 



t 



950 



mmmi 


— - -k 

Inch Absc 


SllplHI 


^ s 


:::::::£:::|::::| 


EEEEE£EEEEEEE]EEEE<E 


■HIli 


2::::::::::::::::::: a. 



v 20 



f 



100 900 

850 

3 800 



J 200 
•S770 



15 



-A 



60 7 


80 


T" 




J 


I 








- t 








' / 




./ 




t 






































































































'' / 













' 












-Ini- 
































































x 




/ 












A 




^L 





44- -l- 4-4> 


itji 


m i 
























7" D 













250 



H 



EE 



E Hct jeeBjII 



$100 



™flf 



00 170 18 


zr , 


+** 


i ^ 


J .gi 
















s"'\ 



Guide Curves 



uo 












150 








I 





















































































































































'• 





















































































































50 60 ""100 110 

'Temperature in Degrees Fahr, 




Chakt G. — Ammonia, Pressure-temperature (Table XLIX). 



TABLES 



613 



00 , 05 , 70 


i 75 ( 80 , 85 , 








/ 




^- 




- - -£ 




2 




2 


























































/ 




Y 


t 







340 300 | 380 ( 400 t 420 ( 








2 




/ 




/ 




-/ - 








- -,Z - 




2 




/> 








_/ 




7 






/: 


















7 




2 








/ 




y 














70 170 

Upper Scale lor Pressures and Lower Scale for Temperatures 
Chart H — Ammonia, Heat of the Liquid (Table XLIX). 



614 



ENGINEERING THERMODYNAMICS 



'no r rao ? 


130 140 150 




tl 




z 
























J 




J 




Z 




7 


010 _ ji 




2 




2 












010 - * 




_ f 








/ 






~X- 




r„ 



75 80 85 


90 95 TOO 105 












/. 
















/ 




















































"t" 




i:: 



400 r .4120 r 


440 1 460 . IJSOj 








Z 




z 








t 




7 




/ 




2 








.: 




z 












z 










Z 












Z 




/ 




f 












/ 




Z 
















/. 









40 S 
50 55 


60 
00 65 '70 






B 


4 




4 


5 


/ 




4 




4 




4 














H - ^ 




M 7 




















•R -^ / 




<3 ,? 


v)- 




1 __- 



P ,20 



lr^560 



140 J 
300 ( .320 


50 100 

310 360 380 

i i i 








Z 




z 












z 




z 












z 
















Z 










Z 
















Z 




z 












Z 




/ 




/ 


1 


450 £_ 


,.,1,,. 







300- 










1 




7 




£ 


350- 






zf: 








X 




fv 


400- 


Z* 




__ 




/J 




~? 






450- + 






-7 i 




?H 




7 


25 


500- % - 




















550- > |j 




"2 ■ 




I^S 




rfS 


juide-Ourye-- 


^T 




!JJ 1 UJ 



¥i i , ,^°, , i , 4 , 5 , 





20 _ 


10 20 


"575 - 






^ 




? 






































^ 




-S 


D" 




»._i__. 


-20 


10 



10 


-, * , r 




.7 




y 




* 




s 




? 


595 


7i 




s 






"2^ 




? 






















- A 




-t 



400 
465 
.'470 
475, 

475 
,480 
485 

.41)0 

) 
495 

500 



| 220 ( 240 r 260 ( 280 




z 


7 


/ 


,/ 


z 


/_ 




t 


z 


I 




If / 


z 


z 


/_ 




J 


z 


z 




y 


-^ U- 


Z M 


/ 


^ 



640 660 080 7 


)0 720 740 760 780 




UJ _£, 












7 




E 








Z 








^ 








Z 








2 








2 








> 




t 








t 




4 




t 








t 














L 




> 












z • 








t 
















t : 
















t 


"IT 


i 









-2Q. 80 



160 170 ,180 ,190 200 210 






























z 


z 












Z yj\ 


2 a. 


^ X 


-L — 



100 160 



500,520 540, 


jOO^SO 600 _ 620 








Z 




_/ 




7 




_/ 




z 








z! 












7 








Z 








z 




7 








Z 










z 




















z 












2 








z 










l£- 


z 


X 




— ± 



Chart I. 



JTeniperature Degrees Fahr. 

-Ammonia, Latent Heat (Table XLIX). 



TABLES 



615 



75 , 80 , 85 p 


90 95 100 105 








































_ — *~ 










- w 










t 






40 


60 
60 65 70 














ftf>0 














— — — 




"^~'\ 




^^^\ 










^^ 












9 "L 


J 


§ 




fL, 20 l 


40 
i0 , . «. . 


p 


















if*" 




" _,* 


.3 




OS ^ 




Em «,' - 








3 ■^'" - 








> " " 












3 


^L 


^* 1 

q 

W 20 


l) 20 

25 
i I » 1 i 










o 






^ 


























s* 




S 




S 














_x 


-20 -1 
10 



















^ 




^ 




„" 




^ 
































A 









400 


180 


440 ,. 


460 


480 






















































*>"! 






















































































\ 








Rift 










s s 
















< 


















s 
















s 






















5 






























140 


150 

300 320 


310 360 


160 
380 


















_,» 
























































































































































^^ 








jj,. 










s v, 


































^^ 






1 






























120 


130 
20 240 


260 























































































































































__- 
















ooo 




























































^ 






H " 






























100 
1 


60 1 


i 


110 

180 

i 


190 


200 


12 
211 




> 


















































































































































































































« 


























V 
















| 






80 
110 


90 
120 , 130 


L40 


1C 
150 





































































































































— 


























































































c. 
































■L-. 







/ 




E F 



t' = 



-40 40 80 120 160 200 



640 660 680 70 


720 740 760 780 




i 


















\ 




V 




V 




s 








V 
















s 




\ 




v 




s; 










3 




£ 




Y 




5 




. V 




\ 




^ 








$ 




l 




5 




£ 




\ 




$ 






M" 


5 


-t- 





540 


500 




V, 







540 




5( 







580 




30 





6 


20 


























































































. 














































s 














































s 












































































































































s 














































\ 




























535 






















































































































































































































































































































































































































































530 




















































































































































L 




































































1 


50 


















r 





















1{ 






-30 -20 60 70 80 

Xower Scale Temperature Degrees Fahr. Upper Scale-Absolute Pressure Ebs. Per Sq. In. 



Chart J.— Ammonia, Total Heat (Table XLIX). 



616 



ENGINEERING THERMODYNAMICS 

15. , 17 



;s 



42 % 



40 30 

18 , 20 , 



41.0 



20 

m 


, , ,35 


L0 

40 45 
























or 
















I -.0245 






















































- N '- 








± . 








V 








^ 










s s 


40.2 .. 






:_i 







10 


20 





ICO 




17C 






180 


( 1?0 




200 


210 


































































| 




































(V 


































































































































































































































-.027 



























































































































































































































































































































































































































































































































































































; 025 35.1 



S 



I 



m 



40 20 20 40 00 80 100 140 180 200 

.028 G-uide Curve 



2. 




75 




80 








30 
,9,0 








LOO 


40 
105 












































































































tf 






































r, 






















































































































































































































s 






































































































































































































































































































































































A 




-.020 




































^ 





100 1 
t 300 


L0 120 
? 350 






^ 




s 




35 ^ 




M \ 




\ 




V 




V 




S 




^ 




\ 




\ 




^ 














\ 




\ 




V 




S 




s 




\ 




\ - 




V 




S - 




S, 






120 1 


30 140 



50 


\ 1 1 l 55 ^, ! 


600 














s 






v 














S 






S 






\ 












S 






\ 






V 






S 


















5 






\ 






















































































5 









4 



IK 






120 




50 
130 


140 




CO 
150 




























































33. 
























r 










































































































































































\ 
































V 






































































































































.0205 
























































































































































»i 4 































400 450 








■s -iy 


__5 -^ 
















5 J-.030 


Vy 




V 


S 


S. 


\ 


s _ 


:::::::::::::::^:::- 


S^---.030 


5 


\- 





1C0 


L70 180 
P° , , ,750 , 














31 S 




\ 








s 




\ 






-0325 


5 








!s 




S3 








s 














:$ 




Y 




: S 












\ 




\ -.035 




: s 








3 




\ 








: s 








: S 






29.5 ..._. 




180 ] 


90 200 



Left Hand Scale =lbs. "per cu. ft. Density Right Hand Seale=cu. ft. per Lb. Specific Volumes 
Lower " ^Degrees F. Temp. Upper " —lbs. per sq. in. Pressures 

Chart K. — Ammonia, Specific Volume and Density of the Liquid (Table XLIX). 



TABLES 



617 



\ 



I 



.04 



.05 



20 
A 

3.5 
3 
40 
2.8 




i06 



2.5 
2.3 



-30 -20 


X 


^ & It 


A » 


\ 407 


A _ . . 




5 


v 






» 




S, -.08 














5 i09 






^ JT 






S j 


t. 


3 11 





M 



-10 



100 



S 



1:12 .8 



-.11 



1.0 



120 



:15 

il6 



+10 



140 



u 









]-' J 








. it 


V 


rr 


*•* - x 


j- 


s ^ x 




\^ 


" ^.3 


5»* x 




si. 




\, 




X s 




x 


^d- 


4.X " 


;:^_ 



50 



t X Tr 


s * i • j- 


s ^ ' 


S v I -T 


V i- 


V 3 


v i 






















, ■ ?■ 







1 ! 


" T 1 i 


Ai 


_l'° 


s* M4- 




\ r T^ 




% X 


; 3 


s ^ l 




^ X X 


- 3 „ 


s s 1 I 


4.6 


S X 




i^X 




x x» 




1 1 


--S lJr» 


X X- 


- ^5-7 







90 



100 













i f 






j. i 


•A « 


£X "*"*" 


x 




4^ 








^ 8 




x ^" s ^». - 


i.O 


si 


X 


=»^ 4 




X 


""-^ ^-q 








1 - 


.j 


: :2_L 



110 



120 



1 



% 



\ 



40 80 120 160 200°F 
Guide Curve 



130 



140 



^ 



-1.2 





X 4 


< TF 417 




""•^ J 


^ T 




""«! 2 




::=^2^ 



180 



170 



3-1.5 



150 



160 



.4 

.3 
180 



180 

B 



:::::f.5 



190 



200 



Curve Showing- Effect of Temperature Veriation upon the 
Specific Volume and Density of Saturated Ammonia Vapor 
Left Hand Scale — Specific Volumes Right Hand Scale = Density 

Chart L— Ammonia, Specific Volume and Density of the Vapor (Table XLIX), 



618 



ENGINEEEING THERMODYNAMICS 



►J 

m 

e3 



p 

1 

CO 

a 

I 



o o 

1 ► 


«OHU50>CO«30iOfflCOHMNMaiiOO^OitOitHCOH 
00N>O'*(NHai00©TttC0(NO00NiO'*MHO00NiO'^(NH 
NNNNNNcO©0©CO©tOiO'OiO»0'0>0»0'*T)i'^TiH'^'* 
IM(N(N(NININNIN(N(N(M(N(N(N(NN(N(M(N(NWIN(NINIM(N 




>> -J 
p. 12 


©00(N(MaiNM©HNCOlN(N»O00>ON(»HO00H«)cOC0 

N®"500<0010000)H©NON^NTHHiicDNiO?OOOiON 
HO0300Nc0i0»0WMHHO0500N©'O^C0(NHOai00N 

oooooooooooooooooooooooooo 




"3-g . 


rH (N •«* CO !>• 00 (MOl©00^0©00 NOIOHNM^ CO 
OO^OONOO^HNIMiXHOO^IMOOiOOOlNOOCOaiiOH© 


NNN©O»Oi0W'*'*MMM(N(NHHHOOfflb000000N 
0000000000000000000005050050505 




00500©T|<MiMOOO©^MHO©00>OnHOOOON©iOTp 


^S^^^^^^^^^S^^^^^^^^^^^^^^ 


Density 
of Vapor, 

Pounds 
per Cu.ft, 


^NN00O0)©O©M^O)HN(N00rfiOiNO©N©NOiCi 
•*©T)<0©HNMOO^O>OH©(MNC003'*0»OHNmO© 
CDON00000505OOHINNWC0^t)< iO lO © N N 00 00 ffi O O 


N<N<NWN(NNCOMWirOCOCOCOCOCONMCOCOWCClMCO'*^ 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


OOOM©0>(N©O^OOiM©HN(NNCOOO^O©(NOOt|<0© 
NN©iO^^MM(MHHOOOOOOOONM>©©>OiOiOr}( 




Density 
of Liquid, 

Pounds 
per Cu.ft. 


iONNNNNO»OCO^COO(MHOOOOOOOO©iOiOiOiOrJlM 
•*M(NH00500N©>0'*CO(NrtOON©iO^CO(NHOaiOO 

©©©©©©©©©©©©©©©©©©©©©©©©©© 


Sp. Vol. 
of Liquid 
Cu.ft. per 
Pound. 


t)<0© 00©HOO©i<(N 00 CO -* <N ©oo^kknoo 

ICHO'0»010>0>0'0>0»0©©©©©©©©©©©©©©©© 
t— 1 7— It— 1- t— H t— It— It— IrHi— Ir It—It— li— 1 t-H t— It— It-It— It— It— Ii— It— It— If- It— 1 t— 1 

OOOOOOOOOOOOOOOOOOOOOOOOOO 




Total 
Heat 
Above 
32° F. 


OCOO©.COCDO©OOOOCOOCD©NNNNOOOOOOOOOOOO 


00005000500050000005000050000500 


4= . 
1=1 43 

I" 


©NOO^O©(NOO^©>OlM©(NOO^O©(MNCOOO^OiOH 


(M(NHHHOOOi®00(X)00NN©©©»O>O^^MC0C0MiM 

<NC<IC^C<lC<l<N<NTHi-li-l,-li-lrHi-l,-l,-lrH,-l,-lTHTH,-l,-l,-lTH,-( 


Heat of 
Liquid 
Above 
32° F. 


(OM00^OffllN00P30i>0OiO(M00MO^O»0OOHNM 


TPmM(N(N(NHHOOO)O)C»00(»NNC0tOtOiOiO^rHcClC0 

(N(NINlNlN(NlMINNINHHHHHHHHHHHT-IHHrtH 

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


0000(MOtHOCOtHCOOCOCOOtHCO»0 lO^-i OOOHCOMOO 

^OOCONHNMOl^OiiOOSCONHiOO^O^OOiiO^dni 
OOHH(N(NMCOii^iOiO©©NNOOOOOOai©OOHH(N 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


lO O CO 00 CO 00 CO 1>00 (NN(NO0N«3NO0 lO 

©WNlMNlNOOCClOi^O^NHcb^OJnoOmOOMOlTjloi 
HlNINMMrPTjHiOiOaNNNOOOOOlOOlOOHHiNIMCOCO 
(N(NNN(N(NIMINNIN(NINiMiN(N(NiNNMMCOCOMCCiC<3CO 


<u ft . 


OOOONCiO^MNHOOOONCiO^COIMHOHlNM^O 

77777777777 i i i i i i i i i 


^ ft • 


OH(NC0'"#»O©N00 05OHCJC0'^i0©N00aiOH(NM'^iO 
T^T+irtlT^r+iTtlT+i'^'<*TtHlO>OlOlOlOlOlO»OlO»OcOCOCOCOCOCO 
^^ *^H ^^ ^H ^H ^H ^H t^ T^ ^^ ^^ ^H ^i "*^ ^Ji ^H ^"1 ^i ^i ^i ^^ ,, q^ "^i ^i ^^ '^^ 



TABLES 



619 



P«4H 2 


(OOiOO©HNH(OHNHiOO«OONOOTttiflai(N©aiM 
OOOCO»OM(N005NcO'*MHOTON'OM(NOOO(OiOmHO 
MCCMMMMCO(N(N(NIN(NN(NHHHHHHOOOOOO 
MN(N(N(N(M(N(N(N(N(NNINIM(N(N(N(M(N(N(N(N(N(N(N(N 




>> 

ft -g 

a 3 


00 l> CO <M rt< OS »0 

CON«»OOONHO ©iOCOOOCOOOHtjHtHcOcO^MOcOhcO 
0005CONOOOM©NC)OOa)iOiOcOcONNlM(M(N(N(MNHHO 
(OiO^MINHOCTiOONCDiO^MIMHOOlOONcOiOTtHMWH 

(NtMtMfNC^lMfNrHrHT-HrH^Hl-HT-lT-lrHrHOOOOOOOOO 

OOOOOOOOOOOOOOOOOOOOOOOOOO 




B Q as 

3*" 


(NCO00(X)0000 00M0000M00aiHCOHC0iONTtlH»O©CONO5 
(NN(MNW00,'^CTiTjH0>iO0i'^Oi0O^00C0NHiO05'^0C(N 


NCDCD»OiO^'fCOCO(MlMHHHOOaiOOOONNcDiOiO'^^ 

oi05050)0500JCi05050505ai050503oO(X)oo(X)oooooooooO(X) 


03 +o 

0+; 


W(NNMNIN(N(N(M(M(N(NH05N^(NOOOCO^OONOO^ 


Tt^ ^J^ ^^ ,< ^ ^^ ^^ ^^ ^t^ ^^ ^^ ^^ ^^ ^^ "^^ ^^ ^^ ^^ ^T^ ^^ "^^ ^^ *^ ^^ ^T' ^Cp ^^ 


Density 
of Vapor, 

Pounds 
per Cu.ft. 


• 
lOOMNN TjtHCTiNiO^MOiNCOOiOHiONiO^-H lO'i-t 
(NOiiOHOO^HOO^HOOiOMHOJN^CCHOJNCOiO^fOM 
HHiNWM^iOiOcONN000500H(NCO-*Tt<iOON(X)050 


rt<T^Tt'T^T^T^^TtiTtt^T^'<!tirtii0^0>0»O>O>O»O»0>O»O^0i0cD 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


mNHtO(NON^lNH0005NiOCOiO(OOOOCO»OIX)HiO«) 
(NOOiCHOOiOHOOiONOiCONOiCDMON^NOlCOmHOOtO 
T)<CClMW(NlN(NHHHOOOOl050i05000000NNNNcOiO 

C<l(^C^(N<^(^C<l(N(NC<IC^(N(NT-HT-(rH^H^H^-(T--I^HrHT-HT-^T-HT-l 


Density 
of Liquid, 

Pounds 
per Cu.ft. 


HOOOONtOlNOOOiOCOOcO«00(NO©(NNM03^kOHiOO 
Ni0-*MIMHOOONCO^OOW00100COIOM^005NCO^M 


OOOOOOO05OJOJO1OS050500000000000000NNNNN 
©COcOcOcOCD©iOiOiO»OiO»0»0»0>OiO'OiO»OiO'C"OiO»OiO 


Sp. Vol. 
of Liquid, 
Cu.ft. per 
Pound. 


NO»c005(NOOMNO^00(Nc0Oii00(NNHOHiOOi0 
T)*iOiOiCiO©cONNNQOOOOO©GOOOHH(N(NMCO^^ 
CO CD CO CO© OCOCDCOCOCOCOCOCDCONNNNNNNNNNN 

OOOOOOOOOOOOOOOOOOOOOOOOOO 




Total 
Heat 
Above 
32° F. 


0000©O3O3O50000000000NNNNNcO'O»Ot)<M(NH ooo 


00cX)CX)(X)CX)C«cX)GOcXDcX)0000O0GOO0cX)O0O0O0CCO0O0O000l>t^ 

ffi050505ffi050)C7)050)050)050501005030)®OiO)Oi©0505 




io »o o o»o lOOO iq»o»o io »o >o»o lO lO 
coHNNooeco5cooi-*05'^03'*aiWNHco mnf- hooico 


HHOOO)ffiOOOONNcDcOiO»O^TtiCOCClNCNHOOO)(Xl(X) 
i-Hi-i^H^HOOOOOOOOOOOOOOOOOOO05050i 


Heat of 
Liquid 
Above 
32° F. 


OOMOOCOO)T)iO>OOcOHCDHN(McDHcOHCDOiOOiOOiO 


NNHHOOOOOJOOOONNCOO^IO^^CCWNCNHH 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


M05MWMHC0M(XlC000«OMC0W»MroMcX)C0MQ0M00 


O'OHNmOiiOHN^ON'cHON^HOOiOMffiN^HOOiO 

coco^Tt>oiocoi>Noooi©OHHNcoco^ioiocoNooooai 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


CO 00 >o.ioi>.io »o lOiO 

iOOtONOOMO©INC35>ONOOiONOicOMON^NOtOCod 
•*iO»OCOONQOOOOiffiOHHMMM^iOCOcON0000050H 


OJ ft . 

^a^ 

CQ£h 


ONOOOiOH(NCO^iOcONOOOOH(NM^>OcONOOaiOH 

HHHHHHHHHHCMN(NCN(N(N(M(N(N(NCOCO 


^ ft • 

.sa^ 


(ONOOOOHiMM^'OcONMOiOHiNM'cHiOcCiNOOOiOH 
COCOCOCONNNNNNNNNNOOOOOOOOOOGOOOOOOOOOOI© 



620 



ENGINEEEING THERMODYNAMICS 



1 
| 

•to 

i 

pq 



O 

59 

< 
u 

Q 
H 

P 

H 
«1 

to 

Pn 
O 

CO 

H 
i— i 

H 

? 

O 

Ph 



>> 

1 ► 


NHlM^iONOHlNINWINfMNHOONiCCOCOHaiNrJtOCO 
OONiOCOHOOOO^fNOOOO^NOSMOMHacO^INON 
OlOlOlO3CSi000000000000NNNNCOtD©CCi32iOiOiOiOiO"* 


CI "1 


OJH^HOiCO^CO© 

N»OCOOOHN^iOCO lOOJ^COHOOOiNOCOOCClOO^M 

OH(NH^10CDNOOOHIN^(00005HMCOOOO^IO^N 

OOOOOOOOOi-i'-iT-ir-i,-i,-irHr-i<M<N<N<N<N<N<N<N 

oooooooooooooooooooooooooo 




d P)-p 

C <U 03 

►S 1-1 ■ 


cOHiooococoioffi^^oocoeo^^iOHooooiooiNiocoH 

I>iM'OOi(M»003(N»OOOHCO©03HiM'HcONOHMiOCONN 


MCOINHHOOOJOONNcOiO^^MMHOOaiGONcOiO'* 


"o3+3 

Pi (3-P 

(-. 0) 53 


Oi^OiOO^NO^COCiNNNN ©»iO rt<(NOcO(NOOiO(Nai 

OJ050iOOOOI>CO©»O^CO(MH00500N©»0^(MHaiOONiO 


COCOMMMKMCOMMMMMM(MN(M(M(N(M(NNHHHH 


Density 
of Vapor, 

Pounds 
per Cu.ft. 


MNN^MN03OON^OiTHM(NiC00O00MN00HONC 
NINHH(NIN(N^ii!OOOC»INiOOilNCD'HiOO©HOO»OHO 
HIMM^iO©NOOOJOH(M^iOCi0005H(M^iONOOO(MM 


COCOOCDCOCOCOCOCONNNNNNNN00000000000003050 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


MNWOr)<OcONHN(NON^HOOkOeOHOOONcOiO»OiO 
C0O00»OC0H00c0^HO3N^NOM0MH05Ci^iMO00© 
COCDiOiOiO»O^^i<Tj(C0 CO CO COCO(N(N(NINhhhhhOO 


Density 
of Liquid, 

Pounds 
per Cu.ft. 


HOOONHMCO^COCOHCJOOONIO^NOOOWCOOOHH 
H03NiO^(MO00©^iM0iN(O(MO00iOroH00>ON0lN^ 


I0i0ici0i0i0»0i0»0'0»0»0i0i0i0'^ic»0i0i0i0»0»0i0»0i0 


Sp. Vol. 
of Liquid, 
Cu.ft. per 
Pound. 


0>0(NOC1^^^00^0N^iCtXcOCO©iOOOOOOCOCOiOOO 
O'CHCO(M00'*ON'*H00iOC0H05NCDtD>O'HC0MWW-* 
kO«:©©NNOO©OJOHHND5^^iOCONOOaiOHNM'!(< 
NNNNNNNNNOOOOOOOOOOOOOOOOOOOOOOa)©0105 0)05 

OOOOOOOOOOOOOOOOOOOOOOOOOO 




Total 
Heat 
Above 
32° F. 


>o >o >o »C O !0 *o io o *o *o io »o io 

NcD^COIN005N»Oa3HOO!O^HN^ON^ON^O©H 


O5O5O>O5O5O5OJ050505O30505O)O5©O5O50iOll3iO505ffi03OJ 




iQiOtOiOOO *o io io o »o o OiOO 

NH^NOMtOOOHMlOCOOOOHHlMNCO^^iOiOiC^M 


NNcOiOiO^MMINHOOiOOOONCDiO^CONHOOJOONCO 
010iOi050505050505050iOOOOOOOOOOOOOOOOOOOOOONt^NN 


Heat of 
Liquid 
Above 
32° F. 


OOOONNMOO^OOINOO^OOMOO^OCDNCBIONOO 


HHMNMCO^WJiCOCONOOOOOSOJOHHNNM^Tt* 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


»ocoococococooiococooo>ocooooocoococococooooococ» 


COH05N'*lMOa)»CCOHOOCO'*lMOaiOO<0'*(NOOO<0'0'* 
OHHlNCO^»OiOcONGOOOOJOHCQIMM-*iOcONNOOffiO 
iCOiOiOU0iO»OiOiOiO>OiOiOcD©COcOCOCD5C©COCD<OCi3N 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


<N t- t> C^J iO(NiO>01> lOiOiO 

00©fOHffiN»0(NOiX)©MH05NiO^C<lH©NiOMHOO 
HiMCO^^iOfflNOOOOOlOHHlMeO^iOOONOOOiOHH 


<o ft . 
£ So 


(NMTHiOONOOOOHMCO^iOfflNOOOOHNCO^iOON 


• ft . 


MM^iOONOOOJOHiNCO^iOONOOaiOHNM^iOON 

0505050i05050505OOOOOOOOOOT-Hr- trHrHT-Hr-lTHi— 1 



TABLES 



621 



g S 

n P 
•< Eh 

CO 

O 

CO 

W 

t— t 

g 

Ph 
O 
« 
P-. 



1 ► 


H Ift H 00 05 H ■* 

lOMON^HGCOMON^HOO^HOO^HN^oN 

Tt*T(in*C0MMNMNNHHHOOO©0i0300(»00N 

^H^HrH^Hr-l^i-li-H^-lrHi-lrHT-Hr-tT-li-IOOOOOOO 


£ o o* 


IONHNCO«©NOOMNMN OCO^OOhoOOOcO^ 
OMNO^OONcOH(OHOOWOtD^»OHO©OiOH 
H(NWiQCOI>050(NC«0>OCO(!OOHCO^N030lN»ON 

ooooooooooooooooooooooo 




E « <a 


O)NCD<OHCOO3C0N(NlN(NCiCOM00COOlO5»ON-*00 
©OCOiO^INOCftNiOMOfflMOJkOHOHtOH^oO 




"3 ^ 
a d*i 
m © ca 
812 « 


©M0^05^©NOOOOOOOOCO(MN(NNHCOOMM(N 
TttCOHOOONiOMHONiOMHOOcOMHOOOnON 


HHHH00000050)00)OiOOOOOOOONNNN(0 


Density 
of Vapor, 

Pounds 
per Cu.ft. 


ONO(NiNHO)OONNONN(NiOiON^COH^MTH 

N^Tt*Tj*iON©W00Tt<(NOO(N>OO»OiN050CHNM 
»ON©hM>CNOIN>C00h^NO^NHtHQ0MNM 


050)050000hhhh(^(N(NCOCOCO^^^ioiO© 


Sp. Vol. 
of Vapor, 
Cu.ft. per 

Pound. 


iOCO©COCOCO©CDCOCDCDCOCOcOCOCDNOOO'Mm'^CO 
TJHINOXtD^lNOOOtO^iNOOOtO^MOaiN'OMH 
0000505050505COOOOOOOOONNNNNCDOCOCDO 
i-irH,-iOOOOOOOOOOOOOOOOOOOO 




Density 
of Liquid, 

Pounds 
per Cu.ft. 


NM(NONSOOCONOO>OOOOOOONiOOOOOOOi(N 


HOOOOaOiOOOOCONNNOC^iO^rJt^MNIM 


Sp. Vol. 
of Liquid, 
Cu.ft. per 
Pound. 


O<N<MOtH0000O<Mt^c0OOOOOO(M<MOOOO 

cONOKN^fflO^OJ^OHO^OOOOCONINOOOO 
iOCDN050HNTfiONOOON^COOOO(N'*NO(NO 
050)0i0JOOOOOOOHHHHH(NiM(NINC0C0m 

OOOOOOOOOOOOOOOOOOOOOOO 




Total 
Heat 
Above 
32° F. 


I0i0»0»0 10 lO lO to >o lO 
©HOHCOHiOO^OOMiOOOOMfflOKN^COOOW 


OOOlOiOOOONNCiOiO^mcONHOOOOONNffl 
OffiOOOOOOOOCOOOOOOOOOOOOOOOOOOOOOOONNNNN 




rH O 00 O CO COC00510H©0^00(NIOOOO(NM^CO 


lOTHMHOffiNCO^MINOfflNiO^lNOOiNiOMH 
b.NNNNffl©fflOiO©iOiO»OiOiOiOiO'^'^'^Tjf^ 


Heat of 
Liquid 
Above 
32° F. 


iOHXiQMH05NiOCOr-i05N©>O^Tt(^T}<^iOCON 


lOCOCONOOOiOJOHNMM^iOONOOOOHlMM'* 
HHHHHHHiNN(N(N(NN(NIN(N(NMWCCiMCOM 


Pressure, 

Pounds 

per Sq.in. 

Gage. 


CO00MMCOCOM00 00 00O0M00O0O5COCOCOO0COCOO0CO 


^«COMMMCOCO-*iOiOCOCDiO©NNNNOOC6aiO 
HfNM^iOONOOfflOHiNM^iOcONOOaiOHN^ 
NNNNNNNb.N0C000000000000000000O5OiO5O5 


Pressure, 

Pounds 

per Sq.in. 

Absolute. 


OOOOOOOOOOOOOOOOOOHHOHMlNINNCO^^iO 
(NM^iOCDN00O5 O <M CO^iO©NQOO)OH(NCO^iO 

t>i>i>t>i>i>i>i> cxf oo 00000000000000000505005 




OOOiOHlNM^iOONOOOSOHlNM^iOCNOOOlO 
U5iO(OtDCO©cOcOcOcOcOfflNNNNNNNNNNOO 


~ 2, . 


OOOiOHNM^iOcONOOaOHMM^iOCONOOOO 
HH(N(N(N(N(N(N(N(NM(MCOMfOCOCOMMCOMCOTH 



622 



ENGINEERING THERMODYNAMICS 



215 
235 



205 
fl95 
185 

"o 
^170 



::C: 



W 



m 



t 



495 

485 

415 

165 

155 

145 

435 

1 425 
"o 
ja 
"Ul5 



2 4.05 


at 

^395 

CO 

1385 



•S375 
I 365 



105 
95 
85 
75 
65 
65 
55 
15 



/ 






600 



I 



700 



£660 

.a 

1650 



= =fE 



2 610 

! 

£600 



M* 



0-, 

t 
I 

£100 



S 



-10 40 
Guide Curve 



870 



I 



t70 r-10 



50 70 



+10 30 40 

.Temperature in Degrees Fahr. 

Chart to Show Pressure Temperature Relation of Carbon Dioxide Vapor 
Chart M. — Carbon Dioxide, Pressure-temperature (Table L). 



TABLES 



623 



320 345" 3 


rO 385" 430 


- 6 'i 


L *« 


B 




4_ 


;?* 


-in X 




10 T j 


5": 


X .<= 




T **- 




X-' 




i 




15.5 :i 




i 

220 24? 2" 


D 20 
rO 295 






JaL 






i 




i*' 




2 




_r 




± 


'+*_ 


± 







600 650 


TOO 








S 




S 




,«= 




„j 




;;^ 






10 ?£ 




_<«: 




S 




ZZ 




r. ,' 


35 







Ordinate, Heat of Liquid 
Co. in B.T.U.above 32 F. 



Abscissa, Temperature 
Degrees Farenheit 
On.Lower£cale,Pounds"per 24 
sq.inch. Abs. on Upper J3cale. 
750 800 8£0 900 95,0 16 



40 5 
445 4Y0 490 


60 

30 
520 5'45 57D 


+5 ||' 


- - ,3 


H- 






?* n 25 




~?f 




r ' 










«.* 




=-5- 




B =-? 




-6 S 





uideCuriV.e 



ffffffF 



I 



^-1Q 20 30 ». .40 60 s 70 80 20 

Chart N. — Carbon Dioxide, Heat of the Liquid (Table L). 



624 



ENGINEERING THERMODYNAMICS 



450 475 500 525 550 575 



10 

225 250 275 



122 



/ 



/ 

















i 




1 




t 




r 








i 




1 




r 








1 




t 








4 i- 




A-E- 












7 




_.: 




1 




t 








t 








to 


JV 












L 






100 -/( 




Z v 
















Zd 












/ 




4 












- ' i ,L 


1 1 i 1 1 1 1 



20 3 
*& 325 350 I 


40 
75 400 425 












Z 


Ph 






t 




4 




t 








t 


P 








E-< 








ffl 




























rt -V 




w 1XU f- 








c3 Z 




|J > 




(1 




7 




r 




7 




_/ 




7 








2 








t 




4 




7 










X 




J_ 



20 20 40 CO 80 
10 C25 650 075 700 725 750 



75 800 825 850 875 900 925950 



Lower Scale -Temp Deg\ Fahr. Upper Scale -Absoute Pressure 
Chart 0,— Carbon Dioxide, Latent Heat (Table L). 



TABLES 



625 



1 

§80 
PL, 

© 70 
Dh-20 

^.99 



Guide 



i^urve 



20 40 GO 



o 

3 



600 C25 G50 


075 7,00 725 750 






















% 








96 V 








s 








95 V 




yo ^ 












s 














$ 








% 








% 








5 








i 








$ 








$, 








£ 








L 




5 








$ 








"D 













225 . 25C 


.10 

275' 


300 


20 


































S9 


































































































1 








r 








11 







""±0 5 

±50 475 50 


GO 

) 525 550 575 






























"s.. 










v^ 




















s 




\ 




























T- 







775 800 825 850 875 900925950 



Lower Scate-Temperature Fahrenheit Upper Scale-Absolute Fxessure 
Chart P.— Carbon Dioxide, Total Heat (Table L). 



626 



ENGINEERING THERMODYNAMICS 



(.02170 



.01770 



.01690 



450 4?5" 500 525 550 575 



7 



7 



7 



/ 



/ 



.01770 



58 AIM 




20 20 40 60 







225 




2 


50 


10 

275 




300 


2C 












































.01610 
















/ 


































































A 




.01570 


y 





















.01810 



600 625 650 675 700 7p 7£j0 

1 


X- t. 


I 


7 




t * 


r 


i 


r 


i T 


/ 


_. Z 


-/- 


f 


I 


f 


T D •« 


z 





775 


800 


825 850 875 


900 


92E 


950 
























.02340 




















l 




























< 














i 
























.02300 




















\ 








































I 




















\ I 


.02260 


















/ 
























































/ 






.02220 
















/ 




















/ 






















/ 




















/ 






















/ 






















/ 






















/ 




















/ 










.02140 












/ 




















/ 




















/ 










- 












/ 












.02100 










/ 


















I 


/ 




















/ 






















/ 














.02060 








r 


















/ 






















/ 








* 








s 




1 


/ 
















.02020 




/ 




















/ 




















/ 






















/ 
















E 




.01980 


/ 





















20 10 40 50 60 60 70 

Left Hand Scale = Specific Volt? me. Right Sand Scale = Density: 

Lower Scale = Temperature Degrees Tahr. Upper -Scale c=? Absolute Pressure. 

Chart Q. — Carbon Dioxide, Specific Volume and Density of the Liquid (Table L). 



TABLES 



627 



..37 



225 250 275 300 



450 475 5Q0 525 550 575 



/ 



0.6 

MJ.5 M 
6.4 
6.3 .18 



-6.2 



6.1 

-6.0 .26 
5.9 



.22 



5.S 



.:;o 



.31 



,3.7 
[-5.6 

5-5 .06 
-5.4 
-5.3 
-5.2 .07 









4 


_ _ __^|_ 




^_ 


/ ®~ 


7 1 


Z 


X^ 




::::::^:::::::::::: 




::: :z^: ::: :::: : 


:-::2::::::::::::::: 


:::^:::::::::::::::: 


7 


f 


_j 


:?:::::::::::::::::: 


y -A Gu-^e CuifVe 


:::::::::::i:::::: 



20 20 40 



20 30 40 

325 350 375 400 425 



.09 



775 800 825 8 


SO 875 990 9^5 950 












2-16.0 




















y -14.8 








^ -14.4 




7 




/ -14.0 








j -13.6 




t 




J 13.2 
















7 




r -12.4 


/ 




r 




7 




t 




7 








2 




_' 




7 


-11.2 






7 




r 


-10.8 


7 








7 








2 








7 













-4.3 .11 



-4.1 



.14 



.146 



625 650 675 700 725 75 



8.S 

8.4 
8.2 
8.0 
1-7.8 
7.6 
7.4 
7.2 
■7.0 



Chart R. — Carbon Dioxide, Specific Volume and Density of the Vapor (Table L). 



628 



ENGINEERING THERMODYNAMICS 



Table 

SOLUTIONS OF 

RELATION BETWEEN PRESSURE, TEMPERATURE, 



•£ 

s >>■* 

X&A 


in . 

MS 

Qffl 


"3 > 

ted 


Pounds pee Square Inch Gage 







5 


10 


15 


20 


25 


30 


35 


40 


45 


50 


55 


1 






206.3 


223.6 


234.9 


247.4 


256.2 


263.8 


270.4 


277.1 


282.8 


288.1 


292.9 


297.5 






204 


219 


232 


242 


251 


260 


267 


274 


280 


286 


291.5 


297 


1.84 


11 


.993 


201.4 


219.3 


231.5 


243.3 


251.7 


259.4 


266.4 


272.7 


278.4 


283.7 


288.5 


293.1 


198.5 


214 


226 


236.5 


245.5 


254 


261.5 


269.5 


274.5 


281 


286.5 


292 


2 






201.1 


218.5 


230.8 


242.1 


250.9 


258.6 


265.5 


271.9 


277.6 


282.8 


287.7 


292.2 






194 


212.5 


225 


235.5 


244.5 


253 


260.5 


267.5 


273.5 


280 


285.5 


291 


3 






195.8 


213.2 


225.5 


236.6 


245.6 


253.3 


260.2 


266.8 


272.3 


277.5 


282.4 


286.9 






191 


206 


219 


229 


238 


246.5 


254 


261.5 


267 


274.5 


280 


285 


3.80 


12 


.986 


191.5 


208.8 


221 


232.3 


241 


248.7 


255.7 


262 


267.7 


272.9 


277.8 


282.4 


186.5 


200.5 


214 


224.5 


233 


241.5 


249.5 


256 


262.5 


269.5 


274.5 


280.5 


4 






190.5 


207.7 


220 


231.2 


240 


247.6 


254.7 


260.9 


266.7 


271.8 


276.1 


281.4 






185 


200 


213 


223 


232 


240.5 


248 


255 


261 


268 


273.5 


279.5 


5 






185.2 


202.4 


214.6 


225.8 


234.6 


242.2 


249.3 


255.6 


261.4 


266.5 


271.4 


276.1 






180 


195 


207.5 


217.5 


226.5 


235 


242 


249 


255 


262.5 


268 


273.5 


5.30 


13 


.979 


183.5 


200.7 


212.8 


224.1 


232.8 


240.5 


247.5 


253.8 


259.6 


264.8 


270.2 


274.1 


178 


192.5 


206 


216 


225 


234 


240.5 


252.5 


254 


261 


266 


272 


6 






180 


197.1 


209.2 


220.5 


229.2 


237 


243.9 


250.2 


256.1 


261.2 


266.7 


271.2 






175 


189.5 


202 


212.5 


221 


229.5 


237 


248.5 


249.5 


257 


262.5 


268 


6.80 


14 


.972 


175.8 


193 


205 


216.2 


224.9 


232.6 


239.6 


246.0 


251.8 


257 


262.1 


266.7 


171 


185.5 


198.5 


208.5 


217 


225 


232.5 


239.5 


245.5 


252.5 


258 


263.5 


7 






170 


192.1 


204 


215.3 


223.9 


231.7 


238.6 


245.1 


250^8 


256.1 


261.1 


265.8 






170 


184.5 


197.5 


207.5 


216 


224 


231.5 


238.5 


244.5 


251.5 


257 


262.5 


8 






168.8 


187.2 


199.1 


210.3 


218.9 


226.9 


233.7 


240.1 


245.9 


251.2 


266.2 


260.8 






165.5 


180 


193 


203 


211.5 


219.5 


227 


233.5 


239.5 


246 


252 


257.5 


8.22 


15 


.966 


165.4 


185.8 


197.8 


209 


217.7 


225.4 


232.4 


238.6 


244.2 


249.3 


254.1 


258.7 


164.5 


179 


191.5 


202 


210.5 


218.5 


226 


232.5 


239 


245 


250.5 


256.5 


9 






160.8 


182.5 


194.5 


205 


214.3 


222 


229 


235.2 


240.8 


245.9 


250.7 


255.3 






161 


175.5 


188.5 


198.5 


207 


215 


222.5 


229 


235 


241.5 


247 


252.5 


10 


16 


.960 


156 


177.7 


189.6 


200.6 


209.2 


216.9 


223.9 


230.1 


235.5 


240.6 


245.4 


250 


156.5 


171.5 


184.5 


194.0 


203 


211 


218 


225 


230.5 


237 


242.5 


247.5 


11 






156.4 


173.2 


185.1 


196.1 


204.7 


212.4 


219.4 


225.6 


231 


236.1 


240.9 


244.5 






152.5 


167.5 


179.5 


190 


198.5 


206.5 


213.5 


220 


226 


232.5 


.237.5 


242.5 


12 






151.9 


168.9 


180.6 


191.9 


199.6 


208.3 


214.8 


221 


226.4 


231.5 


236.4 


240.0 






149 


163 


175.5 


185.5 


194.5 


202.5 


209.5 


216 


222 


228 


233 


238 


12.17 


17 


.953 


151 


168 


179.9 


191.0 


199.6 


207.3 


213.6 


219.6 


225.0 


230.3 


234.4 


239.0 


147.5 


162 


174.5 


184.5 


192.5 


201.5 


208.5 


215 


221 


227 


232.5 


237 


13 






147.5 


164.4 


176.4 


187.4 


196.1 


203.7 


210.1 


216.1 


221.4 


226.8 


230.8 


235.5 






144.5 


159 


171 


181.5 


190 


198 


205 


211.5 


217.5 


223.5 


228.5 


233.5 


13.88 


18 


.946 


143.7 


160.5 


172.3 


183.4 


192 


199.7 


206 


212.1 


217.6 


222.7 


227.2 


231.8 


141 


155 


167.5 


178 


186.5 


194.5 


201.5 


207.5 


214 


219.5 


224.5 


230.0 


14 






143.2 


160 


171.8 


182.9 


191.5 


199.2 


205.5 


211.6 


217.1 


222.2 


226.7 


231.3 






140.5 


154.5 


167 


177.5 


186 


193.5 


201 


207 


213.5 


219 


224 


228.5 


15 






139 


155.8 


167.6 


178.7 


187.3 


195.0 


201.3 


207.4 


212.9 


218.0 


222.5 


227.1 






137 


151 


163 


173.5 


182 


190 


197 


203 


209.5 


215 


220.0 


225 


16 






134.8 


151.6 


163.4 


174.5 


183.1 


190.8 


197.1 


203.2 


208.7 


213.8 


218.3 


222.9 






132.5 


147 


159 


169.5 


178 


186 


192.5 


199 


205 


211 


215.5 


220.5 


16.22 


19 


.94 


133.8 


150.6 


162.3 


173.3 


181.4 


189.5 


196 


201.8 


207.1 


212.3 


217.1 


221.7 


131.5 


146 


157.5 


168.5 


177 


185 


192 


198 


204.5 


210 


215.0 


220.0 


17 






130.6 


147.4 


159.1 


170.1 


178.2 


186.3 


192.8 


198.6 


203.9 


209.1 


213.9 


218.5 






129 


143 


155 


165.5 


174 


182 


188 


195 


201 


207 


211.5 


216.5 


18.03 


20 


.935 


126.2 


142.9 


154.6 


165.6 


174.2 


181.9 


188.9 


195.1 


200.7 


205.7 


209.5 


214.1 


125 


139 


151 


161.5 


170 


177.5 


184.5 


191 


197 


202.5 


207.5 


212.5 


19 






122.3 


138.9 


150.7 


161.6 


170.3 


177.9 


185.0 


191.1 


196.8 


201.7 


205.6 


210.1 






121.5 


135.5 


147.5 


157.5 


166.5 


173.5 


180.5 


187 


193 


198.5 


203.0 


208.5 



TABLES 



629 



LI 

AMMONIA IN WATER 

AND PER CENT NH 3 IN SOLUTION 



Above One Standard Atmosphere 




Qffl 


!&3 


60 


65 


70 


75 


80 


85 


90 


1 

1 95 


100 


105 


110 


115 




301.9 


306.3 


310.4 


314.4 


318.2 


321.8 


325.2 


328.5 


331.7 


334.8 


337.8 


340.7 






1 


301.5 


306 


310 


315 


318.5 


322 


325.5 


329 


307.5 


335.5 


339 


341.5 






297.5 


301.8 


306 


310 


313.8 


317.4 


320.8 


324.1 


327.3 


330.4 


333.4 


336.3 


.993 


11 


1.84 


296.5 


301 


305.5 


310 


313.5 


317.5 


321 


324.5 


330.5 


331 


334 


337 


296.7 


300.9 


305.2 


309.2 


312.9 


316.6 


320 


323.2 


326.5 


329.6 


332.6 


335.4 






2 


295.5 


300 


304.5 


309 


312.5 


316 


320 


323.5 


327 


330 


333 


336 






291.4 


295.6 


300 


303.9 


307.6 


311.3 


314.7 


317.9 


321.2 


324.3 


327.3 


330.1 






3 


289.5 


294.5 


299 


303 


307 


311 


314.5 


317.5 


320.5 


324 


327.5 


330 






286.8 


291.1 


295.3 


299.3 


303.1 


306.7 


310.1 


313.4 


316.6 


319.7 


322.7 


325.6 


.986 


12 


3.80 


284.5 


290 


294 


298.5 


302 


306.5 


310 


313 


316 


320 


323 


325.5 


285.7 


290.1 


294.2 


298.3 


302.1 


305.6 


309.1 


312.4 


315.5 


318.7 


321.6 


324.5 






4 


284 


289 


293 


297.5 


301 


305.5 


309 


312 


315 


318.5 


326.5 


324.5 






280.4 


284.8 


288.9 


293 


296.3 


300.3 


303.8 


307.1 


.310.2 


313.4 


316.3 


319.2 






5 


278.5 


283 


287.5 


292 


295.5 


299.5 


303 


306 


310 


313 


316.5 


319 






279.2 


283.5 


287.1 


291.7 


295.5 


299.1 


302.5 


305.8 


309 


312.1 


315.1 


318 


.979 


13 


5.30 


276.5 


281.5 


285.5 


290 


294 


298 


301 


304.5 


307.5 


311 


315 


317.5 


275.6 


280 


284.1 


288.2 


291.9 


295.5 


299 


302.2 


305.5 


308.5 


311.6 


314.4 






6 


273 


277.5 


281.5 


286 


290 


294 


302 


300.5 


304 


307 


310.5 


313.5 






271.1 


275.4 


279.6 


283.6 


287.4 


291 


294.4 


297.1 


300.9 


304 


307 


309,9 


.972 


14 


6.80 


269 


278.5 


277.5 


281.5 


285.5 


289.5 


303 


296 


300.5 


303 


306.5 


309 


270.1 


274.5 


278.6 


282.7 


286.4 


290.1 


293.5 


296.7 


300 


303 


306.1 


308.9 






7 


267.5 


277.5 


276.5 


281 


284.5 


288.5 


302 


295 


299.5 


302 


305 


308 






265.2 


269.6 


273.7 


281.7 


281.5 


285.2 


288.6 


291.7 


295.1 


298.1 


301.2 


303.9 






8 


262 


267 


271.5 


275.5 


279.5 


283.5 


287 


290 


293 


296.5 


300 


303 






263.1 


267.4 


271.6 


275.6 


279.4 


283 


286.4 


289.7 


292.4 


296 


299 


301.9 


.966 


15 


8.22 


261 


266 


270 


274.5 


278 


282.5 


286 


289 


292 


295.5 


296.5 


301.5 


259.7 


264 


268.2 


272.2 


276 


279.6 


283 


286.3 


289.6 


292.6 


295.6 


308.5 






9 


257 


262 


266.5 


270.5 


274.5 


278 


282 


285 


282 


291.5 


294.5 


297.5 






254.4 


258.7 


262.9 


266.9 


270.7 


274.3 


277.7 


281 


284.2 


287.3 


230.3 


293.2 


.960 


16 


10 


252.5 


257.5 


261.5 


265.5 


269 . 5 


274 


277 


280 


277 


287 


290 


293 


249.9 


254.2 


258.4 


262.4 


266.2 


268.8 


273.2 


276.5 


279.7 


282.8 


285.8 


288.7 






11 


247.5 


252.5 


256.5 


260.5 


264.5 


268.5 


272.5 


275 


272 


282 


285 


288 






245.4 


249.8 


253.9 


257.9 


261.7 


264.3 


268.7 


272 


275.2 


278.3 


281.3 


289.2 






12 


242.5 


247.5 


251.5 


256 


259.5 


264 


267.5 


270 


267 


277 


280 


283 






243.4 


247.7 


251.9 


255.4 


259.7 


263.3 


266.7 


270 


273.2 


276.3 


279.3 


282.2 


.953 


17 


12.17 


242 


246.5 


251 


255 


253.5 


263 


266.5 


269 


266.5 


276 


279 


282 


239.9 


244.2 


248.4 


251.8 


256.2 


259.8 


263.1 


266.5 


269.6 


272.8 


275.7 


278.6 






13 


238 


243 


247 


251 


255 


258 


263 


266 


262.5 


272.5 


275.5 


278.5 






236.2 


240.5 


244 


248.7 


252.5 


256.1 


259.8 


262.8 


266 


269.1 


272.1 


275 


.946 


18 


13.88 


234.5 


239 


243.5 


247 


250.5 


255 


259.0 


261.5 


258 


268.5 


271.5 


274.5 


235.7 


240 


243.5 


248.2 


252 


255.6 


259 


262.3 


265.5 


268.6 


271.6 


274.5 






14 


234 


238.5 


242.5 


246.5 


250 


254.5 


258.5 


261 


257.5 


268 


271 


274 






231.5 


235.8 


239.4 


244 


247.8 


251.4 


254.8 


258.1 


261.3 


264.4 


267.4 


270.3 






15 


229.5 


234 


238.5 


242.5 


246 


250 


254 


256.5 


260 


263.5 


266.5 


270 






227.3 


231.6 


235.1 


239.8 


243.6 


247.2 


250.6 


253.7 


257.1 


260.2 


263.2 


266.1 






16 


225 


230 


234 


237.5 


241.5 


246 


249.5 


252 


255.5 


259 


262 


265 






226.1 


230.4 


234.6 


238.6 


242.4 


246 


249.4 


252.7 


255.9 


259 


262 


264.9 


.94 


19 


16.22 


224.5 


229 


233.5 


237 


241 


245 


248.5 


251.5 


254.5 


258 


261 


264 


222.9 


227.2 


231.4 


235.4 


239.2 


242.8 


246.2 


249.5 


252.7 


255.8 


258.8 


261.7 






17 


221 


225.5 


230 


233 


237.5 


241.5 


245 


248 


251 


254.5 


257.5 


260.5 






218.5 


222.8 


227 


231 


234.8 


238.4 


241.8 


245.1 


248.3 


251.4 


254.4 


257.3 


.935 


20 


18.03 


217 


221.5 


225.5 


229.5 


233 


237.5 


241 


243.5 


247 


250 


253 


256.5 


214.6 


218.8 


223.1 


227 


230.9 


234.4 


237.9 


241.1 


244.4 


247.4 


250.5 


253.4 






19 


213 


217.5 


221.5 


225 


229 


233 


237 


239.5 


243 


246 


249 


252 







630 



ENGINEERING THERMODYNAMICS 



Table 

SOLUTIONS OF 

RELATION BETWEEN PRESSURE, TEMPERATURE, 



4» 

r %XA 


in . 

Qffl 


<a.t3 
















Pounds per Square Incb 


Gage 







5 


10 


15 


20 


25 


30 


35 


40 


45 


50 


55 


19.87 


21 


.928 


119.4 


135.9 


147.6 


158.6 


167.2 


174.4 


181.5 


187.2 


192.5 


197.5 


202.3 


206.9 


118 


132 


144 


154 


163 


170.5 


177 


184 


189.5 


195.5 


200.5 


205 


20 






118.9 


135.5 


147.1 


158.2 


166.7 


174.4 


181.1 


186.7 


192.1 


197 


201.9 


206.4 






117.5 


131.5 


143.5 


153.5 


162.5 


170 


176.5 


183.5 


189 


195 


200 


204.5 


21 






115.2 


131.8 


143.4 


154.5 


163.0 


170.7 


177.4 


183.0 


188.4 


193.3 


198.2 


202.7 






114 


128 


140 


150 


158.5 


166 


173 


179.5 


185 


191 


195.5 


200 


21.75 


22 


.921 


112.9 


129.4 


141 


151.9 


160.5 


168.2 


174.6 


180.1 


185.3 


190.3 


195.1 


199.7 


111.5 


125.5 


137.5 


147 


155.5 


163.5 


170 


176.5 


182.5 


188 


193.0 


197.5 


22 






112 


128.5 


140.1 


151.0 


159.6 


167.3 


173.7 


179.2 


184.4 


189.4 


194.2 


198.8 






110.5 


124 


136.5 


146 


154.5 


162.5 


169 


175.5 


181.5 


187 


191.5 


196 


23.03 


23 


.915 


108 


124.5 


136.1 


147 


155.6 


163.3 


170.0 


175.4 


180.2 


185.2 


190.0 


194.6 


107 


120.5 


132.5 


142.5 


150.5 


158.5 


165 


171.5 


177.5 


183 


187.5 


192.5 


24 






114.8 


121.3 


132.9 


143.8 


152.4 


160.1 


166.8 


172.2 


177.0 


182 


186.8 


191.4 






103.5 


117 


129 


138 


147 


154.5 


161.5 


168 


174 


179 


184 


188.5 


24.99 


24 


.909 


101.5 


117.8 


129.3 


140.1 


148.6 


156.3 


163 


168.4 


173.6 


178.6 


183.2 


187.8 


99 


113.5 


125.5 


135 


143.5 


151 


158 


164.5 


170 


175.5 


180 


185 


26 






98.3 


114.6 


126.2 


136.9 


145.5 


153.1 


159.8 


165.3 


170.4 


175.5 


179.9 


184.7 






95.5 


110.0 


122.0 


131.5 


140 


147 


154 


160.5 


166.5 


171.5 


176.5 


181 


27 






95.1 


111.4 


123.1 


133.7 


142.3 


150.0 


156.6 


162.1 


167.2 


172.4 


176.7 


181.5 






92.5 


106.5 


118.5 


128 


136.5 


143.5 


150.5 


157 


162.5 


168 


172 


177.5 


27.66 


25 


.904 


93.0 


109.4 


121.0 


131.7 


140.1 


147.9 


154.5 


159.9 


165.1 


170.3 


174.4 


178.9 


90.0 


104.0 


116.5 


126 


134 


141.5 


148.5 


154.5 


160.5 


165.5 


171 


175 


28 






92.0 


108.3 


120.0 


130.6 


139.1 


146.8 


153.4 


158.9 


164.0 


169.3 


173.3 


177.9 






89.0 


103 


115 


124.5 


132.5 


140 


147 


153.5 


159 


163 


169.5 


173.5 


29 






88.9 


105.2 


117.0 


127.5 


136 


143.8 


150.3 


155.8 


161 


166.2 


170.2 


174.8 






86.0 


99.5 


111.5 


121 


129 


136.5 


143 


149.5 


155 


160.5 


165 


170 


29.60 


26 


.898 


87 


103.3 


114.7 


125.4 


133.9 


141.6 


148.2 


153.8 


159 


164.3 


168.1 


172.7 


83.5 


97.5 


109.5 


119 


127 


134.5 


141 


147 


152.5 


158 


163.5 


167.5 


30 






85.8 


102.1 


113.5 


124.2 


132.7 


140.4 


147 


152.6 


157.8 


163.1 


166.9 


171.6 






82.5 


96.5 


108 


117.5 


125.5 


133 


139.5 


146 


152 


157 


162 


166 


31.05 


27 


.891 


82.6 


98.8 


110.2 


120.9 


129.4 


137.1 


143.5 


149.2 


154.5 


159.8 


163.6 


168.3 


79.0 


93.0 


104.5 


114 


122 


129.5 


136 


142 


148 


153 


158.5 


162.5 


32 






80.1 


96.2 


107.6 


118.3 


126.8 


134.5 


140.9 


145.6 


151.9 


157.2 


161.0 


165.7 






76.0 


89.5 


101 


110.5 


US. 5 


126 


132.5 


138.5 


144 . 5 


149.5 


154.5 


159 


33 






77.4 


93.5 


104.9 


115.6 


124.1 


131.8 


138.7 


143.9 


149.2 


154.5 


158.3 


163.0 






73.0 


86.5 


98 


107 


115.0 


122.0 


129 


135 


140.5 


146 


151.5 


155.5 


33.25 


28 


.886 


76.5 


92.6 


103.9 


114.6 


123.1 


130.8 


137.8 


143 


148.3 


153.6 


157.4 


162.1 


72.0 


85.5 


97 


106.5 


114.5 


121.5 


128 


134 


140 


145 


150.0 


154.5 


34 






74.6 


90.7 


102 


112.7 


121.2 


128.9 


135.9 


141.1 


146.4 


151.7 


155.5 


160.2 






69.5 


83.0 


94.5 


104.0 


111.5 


119 


125.5 


131.5 


137.5 


142.5 


147.5 


152 


35 






72 


88.1 


99.4 


110.1 


118.6 


126.3 


133.3 


138.5 


143.8 


149.1 


152.9 


157.6 






67.5 


80.0 


91.5 


100.5 


108.5 


115.5 


122 


128 


134.0 


139 


144 


148.5 


35.60 


29 


.881 


70.4 


86.5 


97.8 


108.5 


117 


124.7 


131.7 


137.9 


142.2 


147.5 


151.3 


156.0 


64.5 


78.0 


89 


98.5 


106 


113.5 


120 


126 


132 


136.5 


142 


146 


36 






60.5 


85.6 


96.9 


107.5 


116.1 


123.8 


130.8 


137.0 


141.7 


147.2 


151.0 


155.7 






63.5 


77 


88 


97 


105 


112.5 


118.5 


124.5 


130 


135 


140 


145 


37 






67.2 


83.3 


94.6 


105.2 


113.8 


121.5 


128.5 


134.7 


140.7 


146.8 


150.2 


154.9 






60.5 


73.3 


85.0 


94 


101.5 


108.5 


115.0 


121.5 


127 


132 


137 


141 


38 






65.0 


81.0 


92.3 


104.9 


111.5 


119.2 


126.2 


132.5 


138.4 


143.9 


149.4 


154.0 






57.5 


70.5 


81.5 


90.5 


98.5 


105.5 


112 


117.5 


123.5 


138.5 


133.5 


137.5 


38.20 


30 


.875 


64.5 


80.5 


91.8 


102.5 


111.0 


118.7 


125.7 


132 


138.1 


143.6 


149.3 


153.9 


56.5 


70.0 


81.0 


90 


97.5 


105 


111.5 


117.0 


123.0 


127.5 


133 


137.0 



TABLES 



631 



LI — Continued 

AMMONIA IN WATER 

AND PER CENT NH 3 IN SOLUTION 



Above One Standard Atmosphere 


o >> 

Q, M 


09 . 

<PSU 




60 


65 


70 


75 


80 


85 


90 


95 


3 100 


105 


110 


115 




211.3 


215.6 


219.8 


223.8 


227.6 


231.2 


234.6 


237.9 


241.1 


244.2 


247.2 


250.1 


.928 


21 


19.87 


209.5 


214 


218 


221.5 


225 


229.5 


233 


236 


239 


242 


245.5 


248 


210.8 


215.2 


219.3 


223.4 


227.1 


230.7 


234.1 


237.4 


240.7 


243.8 


246.7 


249.6 






20 


200 


213.5 


217.5 


221 


224.5 


229 


232.5 


235.5 


238.5 


241.5 


245 


247.5 






207.1 


211.5 


215.6 


219.7 


223.3 


227 


230.4 


233.7 


237 


240.1 


243 


245.9 






21 


205 


209.5 


213.5 


217.5 


221 


224.5 


227.5 


231 


234.5 


237.5 


240.5 


243.5 






204.1 


208.4 


212.6 


216.6 


220.4 


224 


227.4 


230.7 


233.9 


237 


240 


242.9 


.921 


22 


21.75 


202 


206.5 


210.5 


214 


218 


221.5 


225.5 


228.5 


232 


234.5 


237.5 


240.5 


203.2 


207.5 


211.7 


215.7 


219.5 


223.1 


226.5 


229.8 


233 


236.1 


239.1 


242 






22 


201 


205.5 


209.5 


213 


215 


220.5 


224.5 


227 


230.5 


233 


236.5 


239.5 






199 


203.3 


207.5 


211.5 


215.3 


218.9 


222.3 


225.6 


228.8 


231.9 


234.9 


237.8 


.915 


23 


23.03 


196.5 


201.5 


205 


209 


211 


216.5 


220 


223 


226.5 


229 


232.5 


235 


195.8 


200.1 


204.2 


208.3 


212.1 


215.7 


219.1 


222.4 


225.6 


228.7 


231.7 


234.6 






24 


193 


197.5 


201.5 


205 


207 


212.5 


216 


219 


222.5 


225 


228.5 


231 






192.2 


196.5 


200.7 


204.7 


208.5 


212.1 


215.5 


218.8 


222 


225.1 


228.1 


231 


.909 


24 


24.99 


188.5 


193 


197.5 


201.5 


205 


208.5 


212 


215.0 


218.5 


221.5 


224.5 


227 


189.1 


193.3 


197.5 


201.6 


205.3 


208.9 


212.2 


215.6 


218.9 


221.9 


225 


237.8 






26 


185.5 


190 


194 


197.5 


201.5 


205 


208 


211.5 


214.5 


271.5 


220.5 


223.5 






185.9 


190.2 


194.3 


198.4 


202.2 


205.7 


209 


212.5 


215.8 


218.7 


221.8 


234.7 






27 


181.5 


186 


190 


194 


197.5 


201 


204.5 


207.5 


210.5 


213.5 


216.5 


219.5 






183.3 


187.6 


191.8 


195.8 


199.6 


203.2 


206.6 


209.9 


213.1 


216.2 


219.2 


222.1 


.904 


25 


27.66 


179 


183.5 


187.5 


191.5 


195 


198.5 


202 


205.5 


208.5 


211 


214.5 


217 


183.2 


186.6 


190.7 


194.8 


198.5 


202.2 


205.6 


208.8 


212.1 


215.1 


21S.2 


221.0 






28 


177.5 


182 


186.5 


190 


193.5 


197.5 


200.5 


204 


207 


210 


212.5 


215.5 






180.2 


183.5 


187.6 


191. S 


195.4 


199.1 


202.6 


205.7 


209.0 


212.1 


215.1 


217.9 






29 


174 


178 


182.5 


186 


190 


193.5 


196.5 


200 


203 


206 


209 


211.5 




• • • 


178.1 


181.4 


185.6 


189.6 


193.4 


197.0 


200.4 


203.7 


206.9 


210 


213.0 


215.9 


.898 


26 


29.60 


171.5 


176 


180 


184 


187.5 


191 


194.5 


198 


201 


203.5 


207 


209.5 


176.9 


180.2 


184.4 


188.4 


192.2 


195.8 


199.2 


202.5 


205.7 


208.8 


211.8 


214.7 






30 


170 


174.5 


179 


182.5 


186 


189.5 


192.5 


196.5 


199.5 


202.0 


205 


208 






173.5 


177.0 


181.2 


185.2 


189.0 


192.6 


196 


199.3 


202.5 


206.6 


209.6 


212.5 


.891 


27 


31.05 


166.5 


171 


174.5 


178.5 


182.5 


185.5 


189 


192.5 


195.0 


198.0 


201 


204.5 


170.9 


174.4 


178.6 


182.6 


186.4 


190 


193.4 


196.7 


199.9 


204 


207 


209.9 






32 


163 


167 


167.5 


175 


178.5 


182 


185.5 


188.5 


192 


194.5 


197.5 


200.5 






168.2 


171.7 


175.9 


179.9 


183.7 


187.3 


190.7 


194.0 


197.2 


201.3 


204.3 


207.2 






33 


159.5 


163.5 


163.5 


171.5 


175 


178.5 


181.5 


185 


188 


191.0 


194 


196.5 






157.3 


170.8 


175 


179 


182.8 


186.4 


189.8 


193.1 


196.3 


200.4 


203.4 


206.3 


.886 


28 


33.25 


169.0 


163 


162.5 


170.5 


174.5 


177.5 


180.5 


184 


187.5 


190 


193 


196.0 


165.4 


168.9 


173.1 


177.1 


180.9 


184.5 


187.9 


191.2 


195.4 


198.5 


201.5 


204.4 






34 


156 


160 


160 


168 


171.5 


175.5 


178 


181.5 


184.5 


187.5 


190 


193.0 






162.8 


166.3 


170.5 


174.5 


178.3 


181.9 


185.3 


188.6 


192.8 


195.9 


198.9 


201.8 






35 


152.5 


156.5 


156.5 


164 


168 


171.0 


174 


177.5 


180.5 


183.5 


187 


189.5 






161.2 


164.7 


168.9 


172.9 


176.7 


180.3 


183.7 


187.0 


191.2 


194.3 


197.3 


200.2 


.881 


29 


35.60 


150.5 


154.5 


154.5 


163 


165.5 


169 


172 


175.5 


178.5 


181.0 


184.5 


187 


160.8 


164.5 


168.7 


172.7 


176.5 


180.1 


183.5 


186.8 


191 


193.9 


196.9 


199.8 






36 


149.0 


153 


153.0 


160.5 


160.5 


167.5 


170.5 


174 


177.0 


179.5 


182.5 


185.5 






159.7 


163.7 


167.9 


171.9 


175.8 


179.3 


182.7 


186. Q 


190.2 


192.8 


195.8 


198.7 






37 


145.5 


149.5 


149.5 


157 


153 


164.0 


167 


170.5 


173 


176.0 


179.5 


182.0 






158.6 


162.9 


167.1 


171.1 


175 


178.5 


181.9 


185.2 


189.4 


191.7 


194.7 


197.6 






38 


142 


146 


146 


153.5 


150 


160.5 


163.5 


166.5 


170 


172.5 


175.5 


178.5 






158.3 


162.6 


167 


171.0 


174.8 


178.4 


181.8 


185.1 


188.3 


191.4 


194.4 


197.3 


.875 


30 


38.20 


141.5 


145.5 


145.5 


153 


149.5 


160 


163 


166 


169.5 


172 


175 


178 



632 



ENGINEERING THERMODYNAMICS 



Table LII 
AMMONIA— WATER SOLUTIONS 

VALUES OF PARTIAL PRESSURES OF AMMONIA AND WATER VAPOR FOR 
VARIOUS TEMPERATURES AND PER CENTS OF AMMONIA IN SOLUTION 



Per cent 
NH 3 



32. 
35.6 
39.2 
42.8 
46.4 
50.0 
53.6 
57.2 
60.8 
64.4 
68. 
71.6 
75.2 
78.8 
82.4 
86 

89.6 
93.2 
96.8 
100.4 
104.0 
107.6 
111.2 
114.8 
118.4 
122.0 
125.6 
129.2 
132.8 
136.4 
140 



2.5 



fn c3 

o 

5 B 

Ph 







o 


a 


<L> 




Pressu 
Vapor. 




~* I* 


p^"^ 








15 Ph 







Press. Inches Hg. 



.236 

.256 

.276 

.295 

.315 

.354 

.394 

.434 

.492 

.552 

.611 

.670 

.729 

.807 

.885 

.985 

.085 

.18 

.28 

.38 

1.455 

1.655 

1.811 

1.970 

2.15 

320 

520 

740 

955 

15 

,37 



.177 
.197 
.236 
.276 
.315 
.355 
.413 
.472 
.532 
.590 
.670 
.748 
.847 
.945 
1.06 
1.2 
1.36 



515 

69 

89 

125 

36 



2.62 
2.95 
3.21 
3.54 
3.88 
4.29 
4.73 
5.21 
5.77 



413 
453 
512 
571 
630 
709 
807 
906 
024 
142 
281 
318 
576 
752 
945 
185 
445 
695 
97 
27 
580 
015 
431 
920 
5.36 
5.860 
6.400 
7.030 
7.685 
8.36 
9.14 



a 

o . 



1.3 

1.5 

1.6 

1.8 

2. 

2.1 

2.5 

2.8 

3 

3.4 



3.8 
4 

4.6 
5 

5.2 
5.9 
6.4 
7 

7.8 
,2 
9 



5.0 



u & 

3 e3 

8> 

2 « 
Phg 

_ o 

.2 a 

tB 

03 <J 



"8 


a 




? 




OQ 


2 fa 


<D 


to O 
TO p. 

*^ 

Ph> 


s3 


^■2 


^ S3 




tfPH, 


3£ 
Ph 


g* 



Press. Inches Hg. 



.512 
.571 
.591 
.650 
.709 
.788 
.866 
.965 
1.062 
1.18 
1.319 
1.455 
1.592 
1.75 
1.925 
2.125 
30 
52 
725 
01 
29 
58 
3.90 
4.23 
4.58 
4.96 
5.35 
5.80 
6.25 
6.72 
7.2 



.158 


.670 


.197 


.768 


.236 


.827 


.276 


.926 


.315 


1.024 


.355 


1.343 


.394 


1.260 


.452 


1.417 


.511 


1.573 


.590 


1.770 


.649 


1.958 


.728 


2.183 


.826 


2.418 


.925 


2.675 


1.043 


2.968 


1.180 


3.305 


1.34 


3.64 


1.495 


4.015 


1.672 


4.397 


1.870 


4.880 


2.085 


5.375 


2.30 


5.88 


2.56 


6.46 


2.815 


7.045 


3.11 


7.69 


3.44 


8.40 


3.80 


9.15 


4.22 


10.02 


4.65 


10.90 


5.12 


11.84 


5.63 | 


12.83 



a 

o . 
J* » 






i 
i 

2 

2 

2 

2.8 

3. 

3.5" 

3.8 



4.1 

4.5 

5 

5.2 

6 

6.5 

7 

7.8 

8.5 

9 

10. 
11 
12 
12.9 



7.5 



3a 



*o 


a 




3 




W 


3 i- 


<o 






p> 


cu.S 


-3 S3 


*£ 


"43 o3 


-3Ph 


c3^ 


lo 


Ph 


H 



<n 
Phoq 



Press. Inches Hg. 



.788 

.867 

.945 

.041 

.16 

.28 



1.415 



1.575 
1.75 
1.925 
2.125 
2.34 
2.58 
2.835 
3.09 
3.49 
3.70 
4.06 
4.42 
4.82 
5.27 
5.72 
6.18 
6.78 
7.33 
7.89 
8.55 
9.25 
9.89 
10.06 
11.45 



.158 

.197 

.216 

.256 

.295 

.335 

.374 

.433 

.473 

.552 

.611 

.689 

.788 

.866 

.985 

.122 

.28 

.435 

.615 

.81 

.03 

.245 

2.50 

2.76 

3.05 

3.37 

3.70 

4.07 

4.5 

4.98 

5.49 



.946 
1.064 
1.161 
1.297 
1.455 
1.615 



789 
008 
223 
477 
736 
029 
368 
701 
075 
4.612 
4.98 
5.495 
6.035 
6.63 
7.30 
7.965 
8.68 
9.54 
10.38 
11.26 
12.25 
13.32 
14.39 
15.04 
16.94 



1.6 

1.8 
2. 
2.1 
2.5 
2.9 
3.1 
3.6 
3.9 
4.1 
4.8 
5.2 
5.8 
6 

6.7 
7.3 
8 

8.8 
9.4 
10.2 
11.3 
12. 
13.2 
14.4 
15.8 
16.9 





10 


12.5 


15 


32 


1.21 


.158 


1.368 


1 


1.58 


.138 


1.718 


1.5 


2.11 


.138 


2.248 


2 


35.6 


1.24 


.177 


L417 


1.5 


1.72 


.157 


1.877 


1.8 


2.3 


.157 


2.457 


2.5 


39.2 


1.36 


.197 


1.557 


1.5 


1.89 


.177 


2.067 


2.1 


2.54 


.177 


2.717 


2.8 


42.8 


1.495 


.236 


1.731 


1.7 


2.09 


.217 


2.307 


2.5 


2.79 


.217 


3.007 


3 


46.4 


1.67 


.276 


1.946 


1.9 


2.31 


.256 


2.566 


2.8 


3.07 


.256 


3.326 


3.2 


50 


1.87 


.315 


2.185 


2 


2.56 


.295 


2.855 


3 


3.41 


.295 


3.705 


3.8 


53.6 


2.05 


.355 


2.405 


2.4 


2.82 


.335 


3.155 


3.3 


3.76 


.335 


4.095 


4.1 


57.2 


2.28 


.413 


2.693 


2.9 


3.12 


.394 


3.514 


3.7 


4.14 


.374 


4.514 


4.7 


60.8 


2.52 


.472 


2.992 


3 


3.45 


.453 


3.903 


4 


4.55 


.433 


4.983 


5 


64.4 


2.79 


.532 


3.322 


3.4 


3.82 


.512 


4.332 


4.5 


5.02 


.492 


5.512 


5.5 



TABLES 



633 



Table LII — Continued 



Percent 
NHi 



10 



2.2 

fta 

o 

.2 a 



"o 


a 




3 


03 


CO 




0) 






<n o. 


3 ^ 


ft^ 


si 


13 ® 


^ 


'43 e3 


lift 


c3^ 


r° O 


ft 


H 



a 

8« 



ft 10 



Press. Inches Hg. 



3.09 


.590 


3.680 


3.4 


.670 


4.070 


3.74 


.767 


4.507 


4.09 


.847 


4.937 


4.49 


.965 


5.455 


4.9 


1.1 


6.0 


5.35 


1.24 


6.59 


5.86 


1.4 


7.26 


6.37 


1.555 


7.925 


6.94 


1.75 


8.69 


7.5 


1.95 


9.45 


8.19 


2.165 


10.355 


8.88 


2.42 


11.30 


9.6 


2.68 


12.28 


10.38 


2.97 


13.35 


11.22 


3.25 


14.47 


12.05 


3.58 


15.63 


12.95 


3.96 


16.91 


13.95 


4.37 


18.32 


15.0 


4.81 


19.81 


16.5 


5.29 


21.79 



3.8 
4 

4.6 
5 



9.5 
10.4 
11.4 
12.2 
13.3 
14.5 
15.5 
17 

18.2 
20 
21.2 



12.5 



fta 
. o 

* a 

ft 



*3 


a 




3 


2 . 


CO 


3 «■« 


<u 








N 


ft> 


(H *^ 


■a J 


*3 




13ft 


3£ 


r° O 


ft 


H 



ft M 

T5£ 






Press. Inches Hg. 



.22 
61 
04 
55 
08 
66 
26 
92 
63 
38 
18 
11.02 
11.9 
12.88 
13.85 
14.95 



.571 


4.791 


.65 


5.26 


.729 


5.769 


.827 


6.377 


.926 


7.006 


1.04 


7.70 


1.18 


8.44 


1.32 


9.24 


1.47 


10.10 


1.67 


11.05 


1.87 


12.05 


2.07 


13.09 


2.32 


14.22 


2.56 


15.44 


2.83 


16.68 


3.13 


18.08 



5 

5 

6 

6 

7 

7 

8 

9. 

10 

11 

12 

13 

14 

15. 

17 

18 



15 



fta 
_ o 

la 

o3<£ 



"o 


a 




3 




CO 




® . 






m a 


3 ™t 


2." 


00,2 

3.5 


ft!> 


13 <" 


W Cj 


'•J3 c3 


13ft 


Bjf£ 


r° O 


ft 


H 



a 

o . 



£co 



Press. Inches Hg. 



5.55 

6.1 

6.7 

7.33 

7.98 

8.66 

9.5 

10.35 

11.28 

12.25 

13.22 

14.30 

15.45 

16.62 

17.9 

19.3 



.552 


6.102 


.631 


6.731 


.71 


7.41 


.81 


8.14 


.906 


8.886 


1.005 


9.665 


1.12 


10.62 


1.26 


11.61 


1.42 


12.70 


1.59 


13.84 


1.77 


14.99 


1.98 


16.28 


2.2 . 


17.65 


2.44 


19.06 


2.69 


20.59 


2.97 


22.27 



6 

7.7 
7.6 
8 

8.9 
9.9 
10.7 
11.9 
12.8 
13.9 
15 

16.3 
17.8 
19 

20.6 
22.2 





17.5 


20 


22.5 


32 


2.72 


.138 


2.858 


2.8 


3.46 


.118 


3.578 


3.5 


4.37 


.118 


4.488 


4.6 


35.6 


3.0 


.157 


3.157 


3.1 


3.84 


.138 


3.978 


4 


4.85 


.138 


4.988 


5 


39.2 


3.29 


.177 


3.467 


3.5 


4.22 


.158 


4.378 


4.3 


5.33 


.158 


5.488 


5.9 


42.8 


3.62 


.217 


3.837 


3.9 


4.65 


.177 


4.827 


4.9 


5.86 


.177 


6.037 


7 


46.4 


4.02 


.256 


4.276 


4.2 


5.12 


.217 


5.337 


5.1 


6.43 


.197 


6.627 


6.7 


50 


4.41 


.295 


4.705 


4.8 


5.63 


.256 


5.886 


5.9 


7.07 


.236 


7.306 


7.3 


53.6 


4.87 


.335 


5.205 


5.2 


6.2 


.295 


6.495 


6.4 


7.74 


.275 


8.015 


8 


57.2 


5.36 


.374 


5.734 


5.9 


6.8 


.335 


7.135 


7.1 


8.48 


.315 


8.795 


9 


60.8 


5.92 


.433 


6.353 


6.5 


7.49 


.394 


7.884 


7.8 


9.3 


.354 


9.654 


9.7 


64.4 


6.5 


.492 


6.992 


7 


8.2 


.453 


8.653 


8.6 


10.18 


.394 


10.574 


10.8 


68 


7.13 


.552 


7.682 


7.8 


9.0 


.512 


9.512 


9.5 


11.12 


.453 


11.573 


12 


71.6 


7.8 


.631 


8.431 


8.5 


9.85 


.571 


10.421 


10.3 


12.15 


.512 


12.662 


12.9 


75.2 


8.55 


.71 


9.26 


9.3 


10.75 


.65 


11.40 


11.5 


13.25 


.571 


13.821 


14 


78.8 


9.33 


.788 


17.118 


10.3 


11.75 


.73 


12.48 


12.4 


14.45 


.65 


15.10 


15.2 


82.4 


10.2 


.866 


11.066 


11.4 


12.75 


.85 


13.60 


13.6 


15.85 


.729 


16.579 


17 


86 


11.1 


.966 


12.066 


12 


13.9 


.905 


14.805 


15 


17.40 


.807 


18.207 


18 


89.6 


12.1 


1.08 


13.18 


13.3 


15.05 


1.14 


16.19 


16.1 










93.2 


13.2 


1.22 


14.24 


14.5 


16.30 


1.26 


17.56 


17.9 










96.8 


14.35 


1.36 


15.71 


15.8 


17.75 


1.4 


19.15 


18.9 










100.4 


15.6 


1.5 


17.1 


17 


19.35 


1.55 


20.90 


20.6 










104.0 


16.95 


1.67 


18.62 


18.6 


21.05 


1.71 


22.76 


22.3 










107.6 


18.45 


1.85 


20.30 


19.9 



















634 



ENGINEERING THERMODYNAMICS 



Table LIII 

ABSORPTION OF GASES BY LIQUIDS 

Selected from Smithsonian Physical Tables. 

Values of x t = volume of gases referred to 32° F. and 29.92 ins. Hg which one volume of 
water can absorb at atmospheric pressure and temperature of first column. 



Temperature. 


C0 2 . 


CO. 


H. 


N. 


O. 


Air. 


NHs. 


H2S. 


Me- 
thane. 


Ethy- 


°C. 


°F. 


lene. 





32 


1.797 


.0354 


.02110 


.02399 


.04925 


.02471 


1174.6 


4.371 


.04573 


.2563 


5 


41 


1.450 


.0315 


.02022 


.02134 


.04335 


.02179 


971.5 


3.965 


.04889 


.2153 


10 


50 


1.185 


.0282 


.01944 


.01918 


.03852 


.01953 


840.2 


3.586 


.04367 


.1837 


15 


59 


1.002 


.0254 


.01875 


.01742 


.03456 


01795 


756.0 


3.233 


.03903 


.1615 


20 


68 


.901 


.0232 


.01809 


.01599 


.03137 


.01704 


683.1 


2.905 


.03499 


.1488 


25 


77 


.772 


.0214 


.01745 


.01481 


.02874 




610.8 


2.604 


.02542 




30 


86 




.0200 


.01690 


.01370 


.02646 











.... 


40 


104 


"506 


.0177 


.01644 


.01195 


.02316 













50 


122 




.0161 


.01608 


.01074 


.02080 













100 


212 


!244 


.0141 


.01600 


.01011 


.01690 













L TABLES 

Table LIV 

ABSORPTION OF AIR IN WATER (Winkler, 1904) 
Air free of CO* and NH 3 measured at 29.92 ins. and 32° F. 



635 



Temper- 
ature. 
°C. 


Cu.ft. 

Oxygen at 

29.92 ins. 

Hg per 1000 

cu.ft. water. 


Cu.ft. 

Nitrogen, 
Oxygen, etc., 

per 1000 
cu.ft. water. 


Sum of 
Oxygen 

and 
Nitrogen. 


Temper- 
ature. 
°C. 


Cu.ft. 

Oxygen at 

29.92 ins. 

Hg per 1000 

, cu.ft. water. 


Cu.ft. 

Nitrogen, 
Oxygen, etc., 

per 1000 
cu.ft. water. 


Sum of 
Oxygen 

and 
Nitrogen. 





10.19 


18.99 


29.18 


16 


6.89 


13.25 


20.14 


1 


9.91 


18.51 


28.42 


17 


6.75 


13.00 


19.75 


2 


9.64 


18.05 


27.69 


18 


6.61 


12.77 


19.38 


3 


9.39 


17.60 


26.99 


19 


6.48 


12.54 


19.02 


4 


9.14 


17.18 


26.32 


20 


6.36 


12.32 


18.68 


5 


8.91 


16.77 


25.68 


21 


6.23 


12.11 


18.34 


6 


8.68 


16.38 


25.06 


22 


6.11 


11.90 


18.01 


7 


8.47 


16.00 


24.47 


23 


6.00 


11.69 


17.69 


8 


8.26 


15.64 


23.90 


24 


5.89 


11.49 


17.38 


9 


8.06 


15.30 


23.36 


25 


5.78 


11.30 


17.08 


10 


7.87 


14.97 


22.84 


26 


5.67 


11.12 


16.79 


11 


7.68 


14.65 


22.33 


27 


5.56 


10.94 


16.50 


12 


7.52 


14.35 


21.87 


28 


5.46 


10.75 


16.21 


13 


7.35 


14.06 


21.41 


29 


5.36 


10.56 


15.92 


14 


7.19 


13.78 


20.97 


30 


5.26 


10.38 


15.64 


15 


7.04 


13.51 


20.55 











636 



ENGINEERING THERMODYNAMICS 



P 

o 

Q 

3 



s a 

9 9 






H 

O 

gq H 

1° 

as 

o <i 

3 9 
<3 



o 

Q 

m 
H 



w 
p 

o 
o 

H 
cc 

PQ 

O 
O 

o 

H 
W 



B.T.U. per 

Cu.ft. at 32° F. 

and 29.92 

ins. Hg. 


















cc 


CO <N 


HQOH 

^1 CO ^ 
CO CO CO 




O OS 


i> 

u 

o 

>> 

"C 
o 
A 


» 




bfl 

'53 


X 



CO 




PI 

CU 
02 

a 


H 


Cu.ft. per 

Lb. at 32° F. 

and 29.92 

ins. Hg. 






CO 
OS 

o 

OS 

t^ 

T-l 


2 

O 
00 

CM 

1—1 




C5i 
CO 


P 
H 

."go, 

o 

<! 


Favre and Silberman 

Berthelot 

Berthelot 

Favre and Silberman 

Berthelot 


Favre and Silberman 

Berthelot 

Calc. from Thomsen 


Thomsen 

Andrews 

Favre and Silberman 

Calc. from Thomsen 

Calc. from Thomsen 

Favre and Silberman 


9 

a 

u 

PI PI 02 

& * g 

a 22 
> tj 

r^H 03 PI 


fl-g p 

72 Cl> 02 

a^ a 

HPQH 


Thomsen 

Berthelot 

Calc. from Thomsen 

Calc. from Thomsen 




<* N M CO CO 
-* r*H CM CO rjH 

«5CD(NOH 

rH i— 1 rH rH i— 1 


rH O iH 
»OOCiO 


i-l O r- 1 O tH i— 1 


Ol W3 CO 
CDWN 
CO CO CO 


00 i> 

OS OS i-l 
OS GO 00 
CO CO lO 


HN©OS 
T*H ,-1 -^ CD 
00 O «© ^ 
COHOSH 
CM CM CM Ol 


m 

o 

o 

Ah 


6 

O" " - - 


o- - 


H 2 liquid 64° F. 
90° F. 
64° F. 
212° F. 
vapor 212° F. 
« 212° F. 


6. - 
0- - 


» "S. 

bC S 

6- d 


CM CM 

^ Tt< y~< rH 

CO CO CM CM 

s s 

S.- - p^ 

O 

w 

'TJ ^ ^ ^ 

a ^ ^ ^ 

o3 

d 

O 


o 

a 

03 

OQ 

as 


c 

c 
s 


c 

' p 
c 

1 


c 

a 
P 

E 

C 


c 

a 

-t- 

R 

c 


c 

p 

J 

8 


* 


C 

-r= 
«+. 

c 

V 

P 
C 
X 
f- 
K 

c 




p 

a 
fc 





6 


of 


PI 



a 

PI 


1 




m 

u 
pi 

ft 

m 




s 

-1-3 



TABLES 



G37 



HO OS 

iH i-i 


© US 

© © 

iH r* 


© 00 
CM t>. 

tH iH 


© © 

US WO 
© ** 
CM CM 


© © 
t>. CM 

rH 00 
CM CM 


CM "* 

U0 iffl 
00 CM 
CM CM 


«<* © 

CM 00 

-* iH 
CO 00 


00 CM 
© t>. 

CM © 

00 CO 


CO 

'o 

M 


0) 

a 

CO 

§ 


O 

^>- - 

CO 

PQ 


s 

TO 

"b ^ 

o3 
bD 
O 
> 
< 


: 


- 


-o 
d 

d 

o3 

tH 


i 

"b 

I 

bD 
O 

> 
<1 


rH 

5 

OS 


00 
CM 


iO 

CO 

CO 


00 

o 

00 


oo 


O 
OS 

00 


CO 


CO 

© 


Thomsen 

Berthelot 

Calc. from Thomsen 

Calc. from Thomsen 


a d 

TO TO 

a s 

o o 
rd x, 
HH 

d- § S 
S° 2 2 

TO CD <£ vG 

g rd • • 

rd co o3 o3 

HPQOO 


d d 

a co 

TO TO 

S B 

o o 

to <; t2 d 

offl^rS 

X! . o3 o3 

HJOO 


d d 

co co 

TO TO 

s a 

o o 

rd x 

§-2 2 2 

TO CO m-h «<-c 
S -d • • 

o-g » « 
d 0) cj d 

HPQOU 


d d 

CO CO 
TO TO 

a a 

o o 
rd -d 

HH 

d- § § 

S^ 2 2 

TO CO <£ <£ 
S -d . . 

Xi CO o3 o3 

HfflOO 


d d 

CO CO 
TO TO 

s s 

o o 

HH 

d- § § 
Sr2 2 2 

to co <£3 a 

g rd • . 
O "g ^ r2 

d ca cj cj 
HPQOO 


d d 

CO CO 
TO CO 

a a 

o o 

ja 

S 2 2 

co<i, tS 

a • • 

X o3 o3 


d d 

CO CO 
CO TO 

a a 

o o 

X rd 

o s a 

S 2 2 

CO Vl «+H 
rd 03 03 

boo 


CO 00 •>• o 

<N CO © CN 
<N CM CM O 
CM CM CM CM 


OS 1>- © 00 
CM CM © *0 

T* OS 00 © 

CM CM CM CM 


CO CM © CM 
tH t- 00 t^ 

rH rH i-l O 

CM CM CM CM 


O CO © CM 

IO rH © •>• 

CO CO "** 00 

HHHSS 

CM CM CM tH 


O 00 CM «* 

CM Oi C» CM 

rH CO l>- ■>• 

rH rH © © 

CM CM CM rH 


oo »o wo © 
CO oo 00 © 
O CM © © 

rH rH © © 

<M CM CM CM 


© r-l rJH 
CM rH OO 

CO rH © 
rH rH © 
(M CM rH 


CM OO © 
rH US rH 
OS !>• iO 

O © © 
CM CM rH 


CM CM 

T* Tt^ rH rH 

CO CO (N (M 

3 S 

rd > 

q 
w 

*0 „ ,. „ 


CM CM 

TjH tH rH rH 

CO CO CM CM 

El- - p^ 


* 

CM CM 

T* T* rH rH 
CO CO CM CM 

rd > 


* 

CM CM 

TJ<^HH 

CO CO CM CM 
CT 1 ^ ^ o3 

rd > 


* 

CM CM 

t^H r^ rH rH 

CO CO CM CM 

d Oh 

o^: - o3 

rd > 


* 

CM CM 

^ ^H rH rH 
CO CO CM CM 
, U 

12 ° 


* 

CM CM 

TH rH rH 
© CM CM 

11 ^ 

rd > 


* 

CM CM 

^ rH rH 

© CM CM 

3 s 
&i & 

rd > 


9 

d 
















o 

a 

rd 

w 


w 

Q 

co 
d 

rd 

-r3 


w 
d 

ca 
d 

CO 
CO 

eg 


W 
d 

co" 

9 

ft 

o 


d 

CO 

d 

CO 

">> 

Oh 

o 
Ph 


w 

d 

£ 

CO 

rS 

5 


d 

i 

+3 

d 

PQ 


d 

co 
d 

CO 

1 

PQ 



n3 co 

co •> 



^2 o3 
fag 



3§ 

H<5 



638 



ENGINEEKING THEKMODYNAMICS 



t3 

1 



> 

w 
m 



A 

<! 

H 
hh" 

O 
Q 






H 
so 

H 

w 

o 

! 

o 
h w 

O H 

co H 

~ CO 
CO 5! 

O <1 
° ,0 

3 g 
i-i <j 

I 5 

^ & 
Q O 

S§ 

^ S 

g§ 
fe w 

fe Pi 

°£ 

o S 

rH 

co s 

P 

pq 

O 



go 
«1 



P 

CO 

o 



pq 



B.P.U. per 

Cu.ft. at 32° F. 
and 29.92 
ins. Hg. 


OS 00 
00 OS 

© •>; 


© © 

i>- 00 

© © 


-* © 
© t>. 

00 00 


i>. © 

© © 

*© 4© 


!>• © 

tH tH 
© GO 


© © 
©1 l>- 

t>» I© 

rH rl 


6 

a 

o 
►> 

o 

>> 

o 
-a 

< 


03 

"b 

S-l 

bD 
O 
> 


- 


- 


: 


: 


- 


Cu.ft. per 

Lb. at32°F. 

and 29.92 

ins. Hg. 


00 

o 




CO 


CO 

CO 


CM 

i—i 

i—l 
i—i 


CO 


H 
pq 

°hh" 

■a ft 

o 

<1 


9 

s^go 

GO ^ f=H 

S o o 

03 ?H »H 

=+-< =-1-1 

03 . . 

03 03 03 


£ s3 

03 03 

a a 

03 ^0) 

o "o 

■+J += 
CO GO 

d B B 
§ 2 2 

GOOO 


<D ® 
CD 03 

pq m 

^ a a 

o o o 
^ . . 

03 03 03 

PQOO 


03 03 

a a 

^03 ^03 

"o "o 

-t-3 += 

CO GO 

| 2 2 

a^^ 
s ^ ^ 

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642 



ENGINEERING THERMODYNAMICS 



Table LVII 
RELATIVE THERMAL CONDUCTIVITY 

Conductivities Carbon Dioxide ) 

and |-=lat32 F. 

Resistances Silver J 



Substance. 



Conductivity Carbon 
Dioxide =1. 



Resistance = 



Conductivity 
Silver =1. 



Iron 

Iron (Wiederman and Franz) . . . 

Copper 

Copper (Wiederman and Franz). 

Steel 

Steel (Wiederman and Franz) . . 

Aluminum 

Lead 

Lead (Wiederman and Franz) . . 

Tin 

Tin (Wiederman and Franz) 

Zinc 

Zinc (Wiederman and Franz) . . . 
Silver 



5700 
4165 

23000 

25760 
3600 
4165 

11000 
2700 
2975 
5000 
5320 
5000 
9835 

35000 



5.23 
8.60 
1.52 
1.36 
9.74 
8.60 
3.18 
12.95 
11.75 
7 
6 
7 
3 
1 



58 



56 



Slate 

Granite and sandstone . 
Marble, limestone, etc . 

Portland cement 

Plaster of Paris 

Soil 

Sand, white dry 

Chalk .. 

Firebrick 

Carbon 

Glass 

Diatomic earth 

Paramne 



Ice 

Sawdust 

Snow, packed . 

Woods 

Strawboard. . . 
Pasteboard 
Asbestos paper , 
Blotting paper . 

Felt 

Cotton wool. . . 



117 

176 

153-182 

23.2 

22.8 

10.7 dry; 52.2 wet 

30.4 

6.52 

9.12 

13.2 

35.8 to 75 
4.24 
7.50atO°C. to 
55.0 at 100° C. 
72.7; 18.5 
3.92 
16.6 
9.8 w.g.; 2.94 a.g. 
9.8 
14.7 
14.0 
4.9 
2.84 
1.4 



3 
199 
228-192 
1511 
1531 
3270 dry; 6700 wet 
1150 
5370 
3840 
2650 
978 to 467 
8260 
4670 at 32° F. to 637 at 212° 

481; 189.0 

8940 

2110 
3570 with grain; 11900 ag.gr. 

3570 

2380 

2500 

7150 
12300 
25000 



TABLES 



643 



Table LVII — Continued 
RELATIVE THERMAL CONDUCTIVITY. 



Substance. 


Conductivity Carbon 
Dioxide =1. 


l 


Conductivity 
Silver =1. 


Cotton wool, pressed 

Flannel 

Haircloth 


1.08 
3.92 
1.37 
2.34 
13.7 


32400 

8930 

25600 


Cork 


1495 


Leather, cowhide 


2560 


Water 


39.09 
16.12 
10.70 
13.78 
12.07 
12.53 
21.22 
10.83 
11.25 
11.56 


896 


Methyl alcohol 


2170 


Methyl alcohol (De Heen) 


3270 


Ethyl alcohol 


2540 


Ethyl alcohol (Henneberg) 


2900 


Ethyl alcohol 90% (Henneberg) 

Ethyl alcohol (Henneberg) 


2990 
1650 


Benzole 


3240 


Benzole (Weber) 


3100 


Petroleum 


3030 






Air 


1.85 
1.27 
1.7 
1.28 
1.37 
10.65 
12.97 
13.14 
1.71 
1.83 
1.89 
2.30 
2.57 
1.62 
1.81 
1.00 
1.15 
1.09 
4.94 


18900 


Ammonia 


27600 


Ammonia (Plank) 


20600 


Ethylene 


27400 




2960 


Hydrogen 


3280 


Hydrogen (Stefan) 


2960 


Hydrogen (Kindt and Warberg) 

Nitrogen 


7100 
20450 


Oxygen 


19100 


Oxygen (Stefan) 


25500 


Methane 


15200 




18500 


Carbon monoxide 


21600 


Carbon monoxide (Kindt and Warberg) 
Carbon dioxide 


19300 
35000 




30400 


Carbon dioxide (Kindt and Warberg) . . 
Illuminating gas (Plank) 


32100 
13600 







CHAPTER V 

HEATING BY COMBUSTION. 
FUELS, FURNACES, GAS PRODUCERS AND STEAM BOILERS 

1. Origin of Heat and Transformation to Useful Form. Complexity of 
Fuels as Sources of Heat. General Classification. Solid, Liquid, Gaseous, 
Natural and Artificial. All heat for power purposes is derived from the 
combustion of fuels of some sort, whether natural or artificial, as also is nearly 
all the heat for the warming of things for domestic and industrial purposes, 
but some heat is derived directly from bodies in their natural state, as for 
example, when foods are cooled for safe storage or when ice is made. Thus, 
the source of heat with which engineering operations must deal is (a), that 
of an exothermic chemical reaction, or (6), that of bodies in their natural 
state, and this heat when taken into some system for the purpose of accom- 
plishing some industrial end must be traced through the whole series of 
processes necessary to that end and finally disposed of, as no energy can be 
destroyed. The ultimate disposition of the heat is a matter of just as much 
importance as its generation or acquirement at a point of origin, and the 
almost infinite variety of intermediate exchanges and transformations between 
its reception into the system and its final disposition. 

Heat received into a system is ultimately disposed of in one of three ways: 
first, it is most commonly discharged from the system in the form of hot 
water, hot gases or vapors, and from them communicated to the surrounding 
bodies or radiated into space; or second, it may be bound up by endothermic 
chemical reactions or changes of state as when certain chemical compounds 
or solutions are formed by heat absorption; or finally it may have been trans- 
formed into energy of another form, permanent or transient, as when a mine 
cage has been lifted or light generated in the electric lamp. Omitting such dis- 
positions as correspond to transformation of energy into another form as but 
temporary, and to be treated in the next chapter, the two final dispositions 
are the inverse of the two original sources, chemical or physical changes of 
state of bodies or temperature changes of the general surroundings, earth and 
water and air. 

When the heat is received into the system at high temperatures as is the case 
when its origin is the combustion of fuel, its progress is generally downward as 
to temperature, each exchange from substance to substance resulting in lower 
temperature but the final disposition by transfer to water or atmospheric 
air, in steam condensers, for example, must take place at a temperature of 
the heat carrier, in excess of that of the final heat receiver, that is, the final 

644 



HEATING BY COMBUSTION 645 

temperature of the steam, the heat of which must be disposed of by heating 
water or air must be higher than that of the available water that is to receive it. 
Between the reception and final disposition all sorts of things may be caused 
to happen; power may be generated and transformed into all the various 
forms of energy; metals be extracted from their ores or chemical compounds 
of value made from those of no value. 

When, however, the heat is received at low temperatures, as when ice 
is to be made or a quicksand to be frozen to permit of shaft sinking, or food 
is to be preserved, or natural water cooled for industrial or drinking purposes, 
or air to be dried of its moisture, then heat can be received into the system 
only by having in the system something colder than the body whose heat is 
to be taken. In general, these operations result in heat reception at a lower 
temperature than the surrounding air and water to which it must ultimately 
be returned and which can receive it only when it has been raised to a higher 
temperature, so that at least one step in the series of processes must be a raising 
of temperature by doing work on the working substance, usually by compression. 

The practical use of heat, therefore, is absolutely limited as to process, 
methods and quantities by the conditions of reception and those of ultimate 
disposition, but between these limits all operations or processes may be clas- 
sified as 

(a) Heat transformation into work or vice versa; 

(b) Change of substance form, chemically or physically, with heat absorption 
or liberation, 

(c) Heat transfer from regions of high to regions of low temperature. 
These may also be grouped into two broader divisions, somewhat more 

convenient than those above. 

I. Thermal processes in which no mechanical work is done and involving 
heat exchange or substance change with heat addition or abstraction. 

II. Thermal processes involving the doing of mechanical work on, or by, 
a working medium or heat carrier. These are all work-heat transformations. 

It is with the first class that this chapter deals in as much as the majority 
of fuels must be changed in character before the heat in them is in a form 
suitable for conversion into work. All work is produced by a change in the 
volume of a substance with or without a change in pressure. As before stated, 
those substances giving the greatest volume changes are the gases and vapors. 
For this reason then the heat in the fuel must be transferred to a gas or vapor 
unless the fuel be originally one or the other of these. 

The first type of heat-carrying substance, a gas, is generally prepared under 
pressure by the so-called explosive burning of a gas with air in a closed 
vessel. As few of the natural fuels are gases, the first step in the utilization 
of other forms of fuels for such cases as these, is to gasify them by means of 
gas producers, retorts or similar apparatus at the expense of some of their 
own heat or heat from another source. 

The second type of heat-carrying substance, used for securing work at the 
expense of heat, a vapor, is obtained by evaporating some volatile liquid, 



646 ENGINEERING THERMODYNAMICS 

usually water, in a boiler by means of the heat produced by the combustion 
of any natural or artificial fuel. 

Therefore, the problem of the fuels, their transformation, one form into 
another, more especially their gasification and complete combustion, is of the 
utmost practical importance and of considerable difficulty. Although the 
only combustible chemical elements of all fuels are carbon, hydrogen and 
to a small extent sulphur, and the complete combustion reactions few and 
simple, it must not be understood that fuels are simple in nature, their suitable 
and proper combustion easy, or that it is possible to predict their heat of 
combustion or its effects with precision, even when their chemical compo- 
sition is known. No problem of equal practical importance is more difficult 
or elusive than the prediction of fuel effects by calculation, or the determination 
of the most suitable fuel for a given purpose. 

Fuels may be regarded as mixtures of the chemical elements, carbon, 
hydrogen and sulphur, so far as their combustible parts are concerned, in 
the proportions indicated by their ultimate analysis, but it is known that these 
elements make a great variety of compounds, so that if these compounds are 
present in a fuel, its heat of combustion cannot be the sum of the heats of com- 
bustion of its chemical elements, but this sum less the amount necessary to 
break the bonds of molecular union. It would be more proper from the 
heat liberating standpoint, to say that the heat of combustion of a fuel is the 
sum of the heats of combustion of all those chemical compounds that as a 
mixture, make up the fuel. Unfortunately this precise statement is of little 
help because no one knows that a fuel except the gaseous varieties, does con- 
sist of a mixture of separate and distinct fuel compounds, or what they are. 
It is probably nearer the truth to say that solid fuels are themselves chemical 
compounds of great molecular complexity with other chemical compounds 
in solution or some equivalent state, and it is known that all liquid fuels are 
solutions of a lot of liquids in each other with perhaps a little of the simple 
mixture condition. If all the compounds that together constitute a fuel were 
known, if the heats of solution or absorption of such substances so held by 
the others were known and their quantities, then the heat of combustion 
of a fuel could be expressed as an algebraic sum of all these heats exactly. 
This, however, is quite impossible, and even if it were possible would be of 
comparatively little use because of its complexity and the fact that no two 
samples of the same fuel externally judged, give quite the same results, though 
all will average up within certain reasonably close limits. 

There is another possible assumption with regard to a solid fuel compo- 
sition that is, while approximate, quite useful, which considers the combustible 
to consist of fixed carbon or carbon that burns directly from the solid coke, 
mixed with a gaseous combustible, termed the volatile, and the heat of com- 
bustion as the sum of the heat of so much fixed carbon as is present and the 
heat of the volatile. This assumption leads to an estimate of the fuel value 
of one coal, from calorimeter determinations of others of the same class in 
terms of the heat of combustion per pound of the volatile which is different 



HEATING BY COMBUSTION 647 

for each class of coals, and that of the fixed carbon which is the same for all. 
In a precisely similar approximate way, may the gaseous and liquid fuels be 
treated, which treatment emphasizes the tremendously variable and complex 
nature of fuels which must be handled for the accomplishment of engineering 
ends and the necessity for some study of their nature that will throw light 
not only on their calorific power, but on the best way in which to burn them 
to develop it to the fullest extent and in the most efficient manner, that is, 
with least loss. 

The very chemical complexity that the study of calorific powers reveals, 
is also responsible for the remarkably different behaviors of the fuels even 
of the same class, in fires. Furthermore, each different use to which fuels 
are put, requires within certain limits special properties in the fuel to make 
it adaptable to the purpose. For example, internal combustion engines require 
gaseous fuel and where gaseous fuel is not available, naturally it must be made 
from whatever solid or liquid fuel is most available, availability depending 
partly on cost. Blast furnaces for the extraction of iron from the ore are 
best served with coke, and coke must be made for the purpose from coal, or 
where this is not available, a hard, natural coal may be substituted in some 
cases. Certain metallurgical and chemical manufacturing operations require 
a very steady, high temperature such as can be obtained only with gaseous 
and in some cases liquid fuel. Boilers are fired with all sorts of fuel, that 
which produces the cheapest steam being selected, and this may be a gas, 
an oil, or any one of the great number of different solid fuels that may be had 
in a given district, each of different quality and price and usually each giving 
a different boiler horse power and efficiency, so it may be, that either the 
most expensive coal per ton, or the lowest price coal per ton, or some inter- 
mediate one, may make the most steam per dollar's worth of fuel. Additional 
complexity arises from the fact that the cheapest steam-making coal may 
yield so low a boiler horse-power as to make the boiler installation investment 
charge per unit of steam so great, that greater horse-power with more expen- 
sive fuel might really be more economical. In cities the household use of 
gas for illumination is now combined with the use of the same gas for domestic 
heating, and a suitable gas must be manufactured to meet the demand for 
this double service. 

These examples will suffice to show first: that the use to which a fuel is 
put more or less dictates the type of fuel, second, that in apparatus capable 
of using more than one fuel the problem of selection is partly one of properties 
and partly one of price, factors that may or may not be related; third, that 
while natural fuels exist in great variety it is good engineering to manufacture 
from them other fuels to secure desired results. These facts make it necessary 
to enquire a little more closely than has been done into the properties of the 
natural and artificial fuels and their differences and the sort of process by 
which one class or kind of fuel may be manufactured from another. 

The first step in such a study is a comprehensive classification of all fuels 
as given in the following Table LVIII as a basis for the treatment: 



648 



ENGINEEKING THERMODYNAMICS 



Table LVIII 
GENERAL CLASSIFICATION OF FUELS 



Physical State. 


Name of Natural Fuel. 


Name of Manufactured Fuel. 


Solid 


Wood, peat, lignite, bituminous and 
anthracite coal 


Charcoal, coke, briquettes. 


Liquid 


Crude mineral oil 


Alcohol, mineral oil, distillates and resi~ 




, dues, oil, gasolene, kerosene, fuel coal, 
and oil tar. 


Gaseous. 


Natural gas 


^Coal gas, coke-oven gas, water gas, car- 
buretted water gas, producer gas, blast- 
furnace gas, oil gas, oil and alcohol 
vapors. 







The general processes by which the artificial fuels are made from the natural 
ones are divisible into groups. The first is a simple heating or roasting of 
natural solid fuels, which leaves a coke and produces a volatile mixture which 
on cooling leaves some permanent gases more or less saturated with vapors 
and some liquid generally termed tar or tar liquor. The second is analogous 
but as applied to liquid fuels is termed distillation or fractionation, but 'the 
third is quite different. It is based on the reactions of fixed carbon with the 
oxygen of the air and with steam to make carbon monoxide and hydrogen. 
Often these three type processes are mixed and all three take place at once 
and are associated with all sorts of dissociations and inter-reactions. It is, 
however, convenient to think of all processes being divisible into 

(a) Mere heating of the natural fuel; 

(6) Chemical reactions between the fuel constituents, air, and steam. 

In the succeeding sections there will be taken up in order the physical, 
chemical, and thermal characteristics of the natural fuels, their treatment 
for the creation of the common manufactured forms of fuel, the combustion 
of all sorts, solid, liquid, and gaseous, both explosively and non-explosively 
together with the conditions which surround and control each change. Finally, 
the gas producer and steam boiler as characteristic apparatus having certain 
properties will be analyzed and the laws of performance derived, both from 
experimental data and from purely rational considerations. 

Prob. 1. The simplest power-plant consists of a boiler and non-condensing engine. 
Trace the heat from its source in the fuel to its final disposition. 

Prob. 2. The above power-plant has had added to it various auxiliaries for increas- 
ing the efficiency. They consist of a feed- water heater in the exhaust of all auxiliaries, 
and an economizer in the flue, a superheater in the first pass of the boiler, an induced 
draft steam-driven fan, and a condenser. Trace the heat through this system from 
the coal to its ultimate disposition. 



HEATING BY COMBUSTION 649 

Prob. 3. Gasolene is burned in the cylinder of a water-jacketed engine, provided 
with an air-cooled radiator. Show where the original heat of the fuel goes. 

Prob. 4. Coke is burned in a blast furnace and some of the gaseous products, which are 
rich in CO, are washed and used in gas engines driving the blast compressors. Between 
the blower and furnace the air passes over brickwork previously heated by burning 
some of the gas. The gas-engine cylinder is, of course, water jacketed. Through 
what steps does that portion of the heat of the fuel not used up by the reduction of 
the iron ore pass and what is its ultimate disposition? 

Prob. 5. Heat is removed from a cold-storage room and delivered to the atmospheric 
air. The process in an ammonia-compression system is to allow ammonia to evaporate 
in coils at a low pressure, compress the vapor and condense it by allowing water and 
air to pass over tubes receiving the high pressure vapor. Through what transfer and 
transformation processes does the heat pass before reaching the atmosphere? Assume 
the compressor to be steam driven and trace the heat of the coal as well. 

Prob. 6. Air is compressed by a water-jacketed steam-driven compressor and dis- 
charged to a large pipe-coil receiver, then through a long pipe line, through a coal- 
fired preheater to an air engine. Show where the air receives heat and what becomes 
of it and also what becomes of the heat of the coal. 

Prob. 7. A coal-fired boiler supplies a steam-driven electric generator, the exhaust 
from which passes to radiators placed throughout a building lighted by the current, 
the excess steam passes to the roof exhaust. Trace the heat of coal to its final dis- 
position. 

Prob. 8. Air and steam are supplied to a gas producer. What becomes of the heat 
originally in the coal? 

2. Natural Solid Fuels, Wood, Peat, Lignite, Bituminous and Anthracite 
Coal. Chemical and Physical Properties. Classifications Based on Ultimate 
and Proximate Analysis and an Behavior on Heating. All solid fuels are of 
vegetable origin in spite of the variations in properties from the hard anthracites 
to the soft peats, and while the tracing of the process of transformation of 
each from its origin or one to another is more a problem of chemical geology 
or physical and organic chemistry than of the engineering of fuel manufacture 
or combustion, yet such an analysis throws so much light on the nature of 
the fuels and the methods of treatment in combustion or gasification as to 
render a brief review decidedly worth while. Such an analysis will not only 
show the relation between the different varieties of solid fuels and their 
chemical, physical, and calorific properties, but will also show the intimate 
relation between the solid, liquid, and gaseous form of the natural fuels and 
the corresponding artificial products made from the first two as raw materials. 

There are three ways of defining a fuel, first, by a general name, indicating 
geologic condition, behavior on heating, or some other generally understood 
but not very definite characteristic, and which is responsible for the names 
lignite, peat, bituminous and anthracite, long flame, short flame, cannel, coking, 
caking, and gas coal, together with many others, some indicating size like 
run-of-mine, slack, broken, furnace, egg, nut, pea, buckwheat, and rice, the 
latter names applying principally to sizes of anthracite. Also the name of 
the district or mine is used and any American engineer knows pretty well 



650 



ENGINEERING THERMODYNAMICS 



what is meant by Pocahontas coal or Pennsylvania anthracite or Illinois 
coal. 

The second way of defining or fixing a solid fuel is based on what is termed 
its proximate analysis, which has a meaning dependent solely on the methods 
employed, the following applying to the laboratory work of the U. S. Geo- 
logical Survey. One gram of pulverized coal selected with great care so 
that it represents as nearly as possible the average of the lot, the properties 
of which must be found, is heated in an oven at 221° F. for one hour and 
quickly weighed. The loss in weight is termed the moisture and the residue 
after burning the dried sample in a crucible termed the ash, the difference 
or combustible, being divided by other operations into two parts, the volatile 
and the fixed carbon, the former being the loss of weight by seven minutes 
heating of the dry sample, in closed crucibles, excluding the air, in the flame 
of a Bunsen burner and the latter the residue after subtracting the ash. 

By complete analytical methods the chemical elements of the coal may be 
found and reported, which define the coal by its ultimate analysis, the methods 
being those common to quantitative organic chemistry. Special constit- 
uents reported by analysts for fuels are each determined by special methods, 
out of place here. 

Wood is in all cases the primary organic substance from which solid fuels 
are derived and the term must be used to include not only the trunks of trees, 
but branches, leaves, and roots as well as small plants and mosses. It is 
composed chemically of cellulose, C6H10O5, as the fiber, and of sap, or sap 
deposits between the fiber. Cellulose consists according to its formula, of 
a definite weight proportion of carbon, hydrogen, and oxygen, but the pro- 
portions of these same elements will be different in real woods by reason of 
the sap properties or the materials deposited by sap, which are of both nitro- 
genous and non-nitrogenous character. Sap is really a very complex substance, 
consisting of protein, tannin and several vegetable acids, starch, sugar, essential 
oils and resins among its organic constituents, together with all sorts of inor- 
ganic salts derived from the soil. These are of some importance in explaining 
the differences found in the coals derived from the woods and are mentioned 
here to call attention to the relation. In the following Table LIX is compared 
the ultimate composition of pure cellulose with the average of maple, oak, 

Table LIX 
COMPARISON OF CELLULOSE AND AVERAGE WOOD (Dry and Ash Free) 



Constituent. 



Carbon 

Hydrogen 

Oxygen 

Oxygen and nitrogen 



Cellulose. 



44.44% 

6.17% 

49.39% 



Wood, Average of 

Maple, Oak, Pine, 

Willow. 



49.2% 
6.1% 



44.7% 



Spores of Club Moss. 



63.0% 
8.6% 



28.4% 



HEATING BY COMBUSTION 651 

pine and willow woods and with moss fiber, dried and free from ash and 
therefore, excluding much of the impregnating matter, as given by Chevandier, 
together with Percy's values for the spores of club moss, which forms peat. 

The table illustrates the increase of carbon content of wood over cellu- 
lose and still greater is the carbon content of the moss, but it does not indicate 
whether the carbon is free or combined with other elements. Of course, 
actual wood contains a great deal of water, even air-dried wood carrying 
from 15 per cent to 25 per cent of moisture, while the ash content will vary 
from one-tenth of one per cent up to perhaps 4 per cent, depending on the 
soil, the part of the plant, its age and a variety of similar circumstances, the 
ash of cellulose being, of course, zero. 

It is from such chemical origin that the solid fossil fuels were derived by 
a natural process, termed carbonification, because it is characterized by an 
increase of carbon content over the original vegetable substance just as it 
does itself show an increase over cellulose, its main constituent fiber. These 
fossil solid fuels are usually divisible by age into younger and older groups, 
though local conditions may disturb the division, the former including peat and 
lignite, and the latter bituminous and anthracite coals, which calls attention to the 
fact that time is the first element in increase of per cent carbon. The next 
factor is the action of oxygen. Wood will absorb oxygen with a resulting 
slow reaction to CO2 even at low temperatures, and so also, will all the coals 
of whatever grade. There is also a further reaction of organic nature between 
the hydrogen, whether free or combined and the carbon, to form hydrocarbons, 
among which is methane CH4 as the principal one. Mixing of the deposits 
with sand and clay increases the ash content of the coals over that of the 
woods, just as sap deposits give to the woods an ash content not existing in 
cellulose. 

It is certain that the processes of carbonification produce certain products 
consisting of carbon dioxide and hydrocarbons and perhaps many other com- 
plex compounds and leave a residue which is the coal. Increase of both 
pressure and temperature promotes the process, the coal residue containing 
a higher per cent of carbon and ash than the original plants from mixture 
and reaction with earths, water and air, while oxygen and nitrogen decrease, 
with time and the favorable nature of surrounding conditions. One very 
active agent in the changes is heat, promoting chemical reaction and expelling 
gases and the vapors of liquid resultant substances, which are compounds 
of carbon, hydrogen, and nitrogen, largely hydrocarbons and carbon dioxide. 
Vapors may condense in other colder places into which they are driven by 
the pressure, forming oil deposits, while gases may impregnate rocks and 
sands and remain in place when overlaid by impervious rock. Thus is the 
origin of natural gas and crude oils traced with reference to coal formation 
as products of successive decomposition and reactions beginning with wood, 
the coal constituting the residue, but , there is good reason to believe that 
hydrocarbons forming natural gases and oils may have been formed from 
carbides of iron, as the U. S. Geological Survey has recently shown that 



652 



ENGINEERING THERMODYNAMICS 



the regions of great magnetic needle deflection are those of known oil 
deposits. 

The various coals are divided into classes with names for convenience of 
discussion, according to their properties relating to the completeness and 
character of the changes they have suffered, but unfortunately there is a 
general lack of agreement as to the meaning of the names. Coal classification 
is generally based on chemical properties, Muck making the total carbon 
content of the ultimate analysis the basis, Frazer, the fixed carbon divided 
by the volatile combustible matter of the proximate analysis, while Campbell 
of the U. S. Geological Survey uses the ratio of the total carbon to the total 
hydrogen of the ultimate analysis. These three classifications with the cor- 
responding names are given in the following Table LX. 

Table LX 
CLASSIFICATION OF COALS BY COMPOSITION 



Class. 


Name. 


Campbell. 
Total C. 
Total H. 


Frazer. 

Fixed C. 

Vol. Comb. 


Muck. 

Per Cent 

Total C, 

Dry and Ash 

Free 


General 

Per cent Total C, 
Dry and Ash Free. 


A 


Graphite and graphitic coal 


GO to? 


Anthracite 

100 

to 

12 


Anthracite, 
95 




B 


Anthracite (1) 


?to30 


Anthracite 






97 to 92.5 


C 


Anthracite (2) 


30 to 26 








D 


Semi-anthracite 


26 (?) to 23 


12 to 8 


92 . 5 to 87 . 5 








E 


Semi-bituminous 


23 (?) to 20 


8 to 5 


Common 
coal 

82 


87 . 5 to 75 . 








F 


Bituminous (1) 


20 to 17 


Bitumi- 
nous 
5 

to 












G 


Bituminous (2) 


17.0 to 14.4 


Bituminous east- 






ern U. S. 
75.0 to 60 


H 


Bituminous (3) 


14.4 to 12.5 


I 


Bituminous (4) 


12.5 to 11.2 


Bituminous west- 






ern U. S. 
65 to 50 


J 


Lignite 


11.2 to 9.3 




70 


Under 50 








K 


Peat 


9.3 (?)to? 


59 










L 


Wood or cellulose 


7.2 


50 





The latest of these classifications is Campbell's, which was proposed by 
reason of the failure of previous classifications to suitably divide the fifty 
odd U. S. samples, thoroughly studied by the Geological Survey, and which 
were mainly poor coals, into the lignite and bituminous classes, to which they 



HEATING BY COMBUSTION 653 

obviously belonged. It is to be regretted that there was not enough data 
to positively fix the numbers marked (?), but it is a fact that will more clearly 
appear later that no sharp line of division exists between one class and the 
next, each merging into the other by almost imperceptible gradations, as 
might properly be expected, considering their common origin. 

Another basis of classification of considerable value depends on the 
behavior of coals on heating with respect to (a) the amount of gas produced, 
(6) the character of the coke residue, and (c) the changes in the mass before 
the coke sets. The U. S. Geological Survey divisions were based on the sort 
of residue left in a rectangular platinum box after heating a powdered sample 
of coal with ten parts of ground silica, and permits of division into coking or 
non-coking coals. This is similar to the practice of Muck except that he heats 
powdered coal alone and gives other names to the coal, according to the 
character of the coke residue. Similarly Griiner, Sexton and Hilt give names 
relating partly to the formation of coke and partly to the quantity of gas yield 
as indicated by the size of the flame of the burning volatile. These various 
conditions are compared in the following Table LXI, which indicates, first, 
that this sort of classification, while common in practice, is still not reduced 
to any accepted form; second, that as an indication of the use to which the 
different coals are well adapted, a classification based on behavior under 
heating, as to gas yield and character of coke formed may be more valuable 
than others based on composition; third, gas and coke property classification 
is not definitely related to composition, though there is some sort of relation. 

These classification properties are not of mere laboratory importance, but 
largely control the actual useful service of the coals. Coals that melt too 
completely cannot well be burnt under boilers without constantly breaking 
up the cake that forms, so as to let air pass through, but moderate caking 
tendencies are good as preventing fine material from passing through the 
grate, so anthracite rice or dust is often advantageously mixed with some 
caking bituminous coal. 

Peat is the least removed in character from its vegetable origin of all the 
solid fossil fuels and two general varieties are recognized with reference to 
origin, sometimes described as (1), high-bog peats, formed from heath and swamp 
moss and usually located in high altitudes, and (2), low-bog peats, formed 
largely from grasses about the borders of low bodies of water. 

In most peats the fibrous structure is still visible and they range in color 
from yellow through brown to black; some are soft and others hard, hard 
especially when the fibrous structure has almost or nearly disappeared, and in 
all cases there is a very large moisture content even in air-dried peat. In 
but few cases is peat strong enough to resist crushing when piled high during 
combustion as in gas producers. Probably its most distinguishing charac- 
teristic is the very large percentage of nitrogen and oxygen and the poorly 
combustible character of its volatile matter. There are given in Table LXII 
on page 655 a few analyses of peats, as this sort of fuel is widely distributed and 
of growing importance. 



654 



ENGINEERING THERMODYNAMICS 



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HEATING BY COMBUSTION 

Table LXII 
COMPOSITION OF PEATS (Juptner) 



655 



Origin. 



Composition Per Cent by Weight, 
Dry 



Hs 



Nj 



Ash. 



Moisture 

Per Cent 

Air 

Dried. 



Physical 
Properties. 



Authority. 



Cappoge, Ireland. . . . 
Kulbeggen, Ireland . . . 
Philipstown, Ireland. . 

Wood of Allen, Ireland 

Vauclaire, France. . . . 

Lony, France 

Fremont, France 



51.05 
61.04 
58.69 

61.02 



57.03 
58.09 
57.79 



6.58 
6.87 
6.97 

5.77 



39 

.30 
1.45 

.81 



55 

.46 

32.88 

32.4 



2.55 
1.83 
1.99 



10 
to 
25 



Pale, red 

brown 

Dark brown 

dense 



Kane 



5.63 
5.93 
6.11 



09 



29.67 

77 
.37 



5.58 
4.61 
3.33 



Dark brown 

a a 

Incompletely- 
decomposed 



Regnault 



Rammstein, Rheinfalls, Ger 
Steinweden, Rheinfalls, Ger. 
Niedermoor, Rheinfalls, Ger . 



61.15 
57.50 
47.90 



6.29 
6.90 
5. 



1.66 
1.75 
42 



27.20 
31.81 
80 



2.70 
2.04 
3.50 



16.7 
16.0 
17.0 



Solid & dense 
Lighter 
Light, felty 
mass 



Walz 



Griinwald, Germany . . 

Harz, Germany 

Harz, Germany 

Limm, Germany 

Hundsmiihl, Germany 

Switzerland 



49.88 
50.86 
62.54 
59.47 
59.70 



6.5 

5.80 

6.81 

5.52 

5.70 



1.16 
.77 
1.41 
2.51 
1.56 



42.42 
42.70 
29.24 
31.51 
33.04 



3.72 

.57 

1.09 

18.53 
2.92 



Websky 



40.10 



4.53 



2.84 



21.51 



7.87 



23.17 



Pressed peat 



Goppels- 
roder 



Average Calorific Power of Peat (Scheere) B.T.U. per Pound 



Zero per cent water, zero per cent ash = 9090 
Zero per cent water, 15 per cent ash = 7992 



25 per cent water and zero per cent ash = 6840 
30 per cent water and 10 per cent ash = 1462 



Lignite is only one step removed from peat, yet it is very different in 
character, having been formed largely, if not entirely, from plants rich in resin, 
including the coniferous trees, cypress and palms, but in some cases others 
are included. Its appearance is non-fibrous and is either brown or black in 
color. It readily absorbs oxygen and gives off carbon dioxide at all tempera- 
tures. The volatiles of lignites is so largely carbon dioxide that they 
are chemically in the same general class as the peats, and this distinguishes 
both as much as any property from all other coals, the volatiles of which are 
mainly hydrocarbons. This is a most interesting distinction but must not be 
interpreted as meaning that no hydrocarbons are found in lignites but rather 
that they are mixed with large amounts of non-combustible diluents. To 
the lignite analyses in the general table there are added the following Austrian 
lignite analyses, Table LXIII, for comparison from Juptner. 



656 



ENGINEERING THERMODYNAMICS 

Table LXIII 
COMPOSITION OF AUSTRIAN LIGNITES 



Origin. 


Composition Per Cent by Weight. 


B.T.U. 




C 


H 2 


o 2 


N 2 


H 2 


Ash. 


s 


per Lb. 


Trefail, Styria, Austria . . 
Teplitz, Bohemia, Aus- 
tria 

Dax, Bohemia, Austria. . 


44.95 

44.93 

50.12 


3.67 

3.12 
4.06 


16.93 

12.51 
13.14 


.97 

.64 
.65 


20.15 

34.28 
25.50 


8.43 

4.43 
6.53 


1.64 

.50 
.93 


7895 

7065 
8334 



He gives the following as a characteristic average analysis: 

Lignite (Brown Coal) General Average Composition 

Carbon 50-65% 

Combustible H 1-2% 

Chemically combined water 20-30% 

Moisture water (hygroscopic) 10-25% 

Ash 5-10% 

Many lignites contain water, as water of crystallization, so that, on drying 
even in the sun, loss of water will sometimes cause a freshly mined, hard, 
shiny piece to fall into a heap of sand-like particles. This is important in 
the problem of firing this fuel because varieties that behave like this, may fall 
through a boiler grate or pack a gas producer bed, so that the air blast cannot 
be forced through. When they contain bituminous matter that melts on 
heating this will act as a binder and may correct the disintegration evil in 
boiler fires or perhaps increase the packing difficulty in" gas producers. There 
are among lignites, as among peats, very great ranges of change and variety, 
but all varieties contain much water and yield much non-combustible 
matter in the volatile gases. The latter fact is the reason why class- 
ifications based on the per cent fixed carbon or its relation to the volatile of 
coals, often fails in distinguishing lignites from bituminous coals and why 
also even the total carbcn hydrogen ratio also fails, though, of course, it is 
better since the ratio is very different for two fuels with the same total carbon, 
one with mainly hydrocarbon volatile and the other with much carbon dioxide. 
In America lignites are found in Dakota, Texas, Arkansas, Louisiana, Mis- 
sissippi, and Alabama with other varieties more properly bituminous in 
Montana, Idaho, Washington, Oregon, California, Wyoming, Utah, Colorado, 
New Mexico, and Texas. 

Bituminous coal is next to lignite in order of age, but embraces in its own 
class so great a variety of coals as to lead to the different designations of 
coking, non-coking, long flame and short flame, and to the four sub-classes 
based on composition, proposed by Campbell, in addition to a semi-bituminous 



HEATING BY COMBUSTION 657 

class. It requires only a brief inspection of the general coal table at the end 
of this chapter, Table CIV, where are recorded the analysis of nearly two 
hundred coals of all classes and grades, to show how hopeless must be any 
attempt to draw a sharp line between bituminous coals of low grade and the 
lignites. In general practice perhaps no single thing distinguishes a bitu- 
minous coal more than the considerable quantities of very rich hydrocarbon 
volatile that characterizes the best varieties, such as come from the center 
of the Appalachian Mountains, from Pennsylvania to Ohio on the west, to 
Alabama on the south or from separate deposits in many Western States. 
Perhaps the best division of bituminous varieties for practical purposes, however 
indefinite it may be, is that descriptive of the flame or the coke. According 
to this, the lowest coal in the bituminous series is the non-caking long-flame 
coal. 

Non-caking long -flame bituminous coals, called by the Germans " sand 
coals," leave coal particles after heating, unchanged as to form, except as to 
the cracking of large masses, but give off combustible hydrocarbons gases very 
freely. They are black or brown in color, generally hard, and include the 
English and Scotch splint coals, which have been used for blast furnaces 
instead of coke and also for reverberatory furnaces. They occur in 
America but are most common in England and on the Continent. According 
to Sexton the English varieties contain about 40 per cent volatile and from 
45 per cent to 55 per cent fixed carbon and yield about 60 per cent coke, 
which indicates that in coking some of the hydrocarbon volatile is decom- 
posed and its carbon deposited with the fixed carbon. According to Juptner 
the total carbon of these coals is between 75 per cent and 80 per cent, hydrogen 
5.5 per cent to 4.5 per cent, oxygen and nitrogen 19.5 per cent to 15.5 per 
cent, indicating the presence of non-combustible in the volatile. The calorific 
power is placed by him, between 14,400 and 15,300 B.T.U. per pound, which 
is very high compared to lignites. 

Cannel coals belong more nearly to this class than to any other, but are 
quite different in many ways from the general run of coals, hardly fitting 
into such a series as is here described in any really logical place. They are 
dull black, hard but easily broken, give a very large amount of gas on heating 
and a long flame when burning, crack as they burn but do not melt or cake. 
They appear to be related to the shales that yield oil on distillation more 
than to other coals, a fact that is brought out by the very small fixed carbon 
content given by Sexton as only 7 per cent for English boghead cannel and 
the correspondingly large per cent of ash given as 22 per cent. This same 
coal yields, however, about 29 per cent of coke on roasting, proving how freely 
the volatile hydrocarbon is decomposed and in so doing deposits its carbon 
with the fixed carbon to make coke. Industrially cannel coals are used only 
for gas making by retort roasting processes. 

Bituminous Gas Coals come next in order and are distinguished from the 
previous class by the melting, caking or coking tendency during heating, 
and by the greater volatile content or gas yield, and its hydrocarbon character, 



658 



ENGINEERING THERMODYNAMICS 



though they contain nitrogen diluents and yield ammonia. Lumps will stick 
together but not puff up, nor melt completely with loss of original form, during 
heating. They are especially adapted to coal-gas making by the retort roasting 
process, to reverberatory furnaces because of their long flames, but may also 
be used to advantage in boiler furnaces. According to Juptner the total carbon 
content is greater than for the non-caking variety, being between 80 per 
cent and 85 per cent, the hydrogen between 5.8 and 5 per cent, so that this 
is also a little higher, and oxygen and nitrogen 14.2 per cent to 10 per cent, 
somewhat lower, indicating gases of more distinctly hydrocarbon character 
with less non-combustible diluents. This also is indicated by a higher calorific 
power placed between 15,300 and 15,840 B.T.U. per pound. 

Bituminous Furnace Coals is the term applied by Sexton to the next in the 
series and they have more strongly developed caking tendencies, softening 
and swelling during combustion and yielding long, luminous gas flames by 
the combustion of the hydrocarbon volatile. According to Sexton the volatile 
matter is between 25 per cent and 35 per cent, materially less in total quantity, 
than for the long flame non-coking coals. The coke yield is from 65 per cent 
to 75 per cent, and the fixed carbon from 1 to 7 per cent, less than the coke, 
indicating once more the decomposition of volatile in coking and deposit 
of soot with the fixed carbon. Juptner places the total carbon between 84 
per cent and 89 per cent, total hydrogen 5 per cent to 5.5 per cent, oxygen 
and nitrogen 11 per cent to 5.5 per cent. The rise of the hydrocarbons is 
indicated also by the rise of calorific power, which is placed between 15,840 
and 16,740 B.T.U. per pound. These coals are used in England for domestic 
purposes and everywhere for boiler fires, forges, reverberatory furnaces and 
coal-gas making. 

Bituminous Coking Coals are those next in order, distinguished not only 
by a strong melting tendency on heating, and by the very hard, large masses 
of coke they yield from slack coal, but equally important by a lesser gas yield 
which is responsible for the designation short flame applied to them. This is the 
most important industrial class of coals as it is applied to the making of blast 
furnace coke and is good for boiler work. The coke yield varies from 65 
per cent to 80 per cent, depending as much on the oven as on the coal itself, 
and the content of oxygen and nitrogen is less than for the preceding varieties. 
Lewes gives the following compositions, Table LXIV, for some good English 
coking coals which may be compared with others in the general coal table 
at the end of the chapter. 

Table LXIV 
COMPOSITION OF ENGLISH COKING COALS (Lewes) 



Source. 


C 


H 2 


o 2 


N 2 


s 


Ash. 


Coke. 


Durham 

South Wales . . 


83.47 
83.78 
79.69 
77.90 


6.68 
4.79 
4.94 
5.32 


8.17 

4.15 

10.28 

9.43 


1.42 

.93 

1.41 

1.30 


.60 
1.43 
1.01 
1.44 


.20 
4.41 
2.65 
4.88 


62.70 
72.60 


Derbyshire 

Lancashire 


59.32 
60.22 







HEATING BY COMBUSTION 



659 



Semi-bituminous, Semi-anthracite, and Anthracite coals are not distinguish- 
able on the basis of coke or flame properties but rather on the basis of hard- 
ness and composition and they merge very gradually from the last group in 
the order named to graphite. The volatile becomes of a simple character, being 
almost entirely methane in the anthracites and small in quantity, usually 
less than 4 per cent, whereas the more distinctly bituminous coals contain 
very large quantities of volatile which are complex hydrocarbons, some of 
them of the liquid form, termed tar, and which may be over 5 per cent by 
weight while the total volatile may be quite high. The total carbon 
content of anthracites may reach 98 per cent and as they are very hard, are 
not so easy to burn. Some varieties termed graphitic coals cannot be burned 
at all except by mixing with other varieties and only in the hottest fires. 
The calorific power of the anthracites approaches very closely to that of pure 
carbon, the small excess heating value of the volatile balancing the ash, but 
there is a most important exception found in small sizes. Anthracite generally 
occurs with slate streaks, which, as the coal is broken into commercial sizes, 
is separated out by hand and as sizes become smaller the slate cannot be 
distinguished from the coal so that the smaller sizes often contain as much as 
35 per cent of ash. This relation is shown approximately in the following 
Table LXV, applying to the Wilkes-Barre and Scranton, Pa., products. 

Table LXV 
WILKES-BARRE ANTHRACITE SIZES AND AVERAGE ASH CONTENTS 





Passes through 
Hole, Inches. 


Passes over 
Hole, Inches. 


Ash. 


Name. 


General 
Average. 


One Mine. 


Broken 

Egg 


3* 

(a) 2f 
(6) 2* 
(a) 2 
(&) H 

(a) If 

(b) 11 

3. 

(a) | 

(b) 1 
(a) f 
(6) \ 


2| 
2 
If 
If 
11 
f 

4 
1 
2 
3 
8 
1 
4 
_3_ 
16 
1 
8 


5.00 
5.5 

8 

12 

15 
19 

25 


5.66 


Stove 


10.17 


Chestnut 


12 67 


Pea 


14.66 


Buckwheat 


16.62 


Rice or buckwheat, No. 2 









Semi-anthracites are usually the border coals of the anthracite fields as 
the semi -bituminous are border coals of more distinctly bituminous fields, 
but the latter may more often separately occur. In America these semi- 
bituminous coals include those from Clearfield, Cambria, and Somerset counties, 
Pa., Cumberland, Md., Pocahontas, Va., and New River, West Va. They 
are perhaps most definitely fixed by the volatile content of 18 per cent to 



660 ENGINEERING THERMODYNAMICS 

22 per cent, low ash and sulphur, and are considered the best boiler coals in 
America. 

More than any single thing the character of the volatile characterizes the 
different coals and serves to group them naturally into classes, the bituminous 
coals, more especially those of the caking or coking class, having complex 
hydrocarbon mixtures with much of the liquid variety yielding tar; the non- 
coking semi-bituminous varieties yield a simpler set of hydrocarbons with 
anthracite at one end characterized by almost pure methane volatile, and 
lignites and peats at the other, characterized by much non-combustible matter 
in their volatile diluting the hydrocarbon constituents. It is not possible 
to say that the volatile gases or vapors that finally appear really existed in 
the coal as such. It is more likely, as has been pointed out, that they 
did not, but appear as products of decomposition of other oxygenated 
compounds. 

In Table CIV there is presented a collection of proximate and ultimate 
analyses selected from thousands available, by reason of the high authority 
and reputation for accuracy of the analysts reporting them and because 
also they fairly well represent, when grouped together, the whole range of solid 
fuels from the peats to the anthracites. For each fuel the calorific power, 
as determined by the bomb calorimeter, is given and will be used later as a 
basis for a general study of calorific power. The reports of Mahler are repre- 
sentative of French, those of Bunte, of German, and those of Lord, working first 
with Haas and later alone as chemist for the U. S. Geological Survey, of American 
coals and scientific authority, and it is remarkable how well their results fit 
together. These analyses together with a few others, perhaps less authoritative, 
constituting about 200 different coals, have been classified according to Camp- 
bell's carbon hydrogen ratio for convenience, but some little errors are 
introduced by the lack of consistency in moisture which is different for all and 
not reported by Mahler. 

The impurities in coal, notably the ash and sulphur, have an appreciable 
influence on the uses to which coals may be put, often not inferior, to the 
properties of the volatile or the calorific power of the coal. Sulphur generally 
occurs as iron pyrites, FeS2, in nuggets or veins and on combustion usually 
gives off SO2 as a gas, leaving iron oxide, Fe2C>3, as part of the ash if there 
is insufficient air, as is usually the case in the interior of the cinder, while in 
the region of high temperature there will be formed iron sulphide, FeS, and 
sulphur which escapes as vapor. Sulphur is also present in some coals as 
calcium sulphate or gypsum, which leaves calcium sulphide when heated in 
contact with carbon. Sulphur in almost any form will make compounds with 
other things present, which fact bears on the availability of the fuel for certain 
metallurgical operations; for example, the making of iron. Sulphides, especially 
of iron, are apparently important factors in the fusing of the ash and the 
making of clinker. Both in the solid combined form and as a gas, sulphur 
is actively corrosive and grate bars as well as boiler plates and the pipes and 
scrubbers of gas generators will suffer from its presence so that high sulphur 



HEATING BY COMBUSTION 661 

coal is considered generally undesirable though sometimes some sulphur may 
be removed by washing. 

Fusion of ash, and clinker formation is a serious matter, limiting the air 
supply, sticking to grate bars and when it contains sulphur actually corroding 
the bar by chemical reaction, while in gas producers clinker collection is often 
so severe as to stop the operation entirely. In fact certain coals cannot be used 
in producers at all because of this tendency, though otherwise desirable. 
The formation of clinker starts with the fusion of some one of the constituents 
of the ash or the fluxing of one constituent by another, the liquefied material 
flowing over and enclosing solid matter, even large pieces of unburned coal, 
and tending to stick to other pieces, to firebrick linings or to grate bars. The 
clinkering tendency of a given coal cannot be predicted even when the 
ash is completely analyzed and the melting-points of its constituents 
known, because the possible compounds of those constituents are not 
and cannot be known. The chemical compounds in ash are usually 
oxides of aluminum AI2O3, silica, Si02, iron FeO and Fe203, calcium CaO, 
potassium K2O, magnesium MgO, sodium Na20, sulphur SO2, with many 
more and all combined with each other in unknown ways. Those occurring 
in the largest quantities are usually the alumina and silica, the melting-points 
of which are 3400° F. and 2510° F. respectively, which are within range of 
the usual furnace temperature of 3000° F. or thereabouts. The melting- 
point of either of these constituents or a mixture of them is lowered by fluxing 
with the other compounds, making glasses and slags, some of which melt below 
2000° F., so that in practically every ash some fusion will occur and clinker 
form if the fire is pushed at all. Just what is to be done in any case to avoid 
trouble, in view of the certainty of some clinker and the possibility of a pro- 
hibitive amount cannot at present be settled on any scientific grounds, but 
is a matter of trial and judgment, first in the treatment of the most available 
coal and second in the selection of a substitute. 

Prob. 1. Into what class of coal, anthracite, bituminous, etc., would coals Nos. 1, 
12, 16, 25, 49, 64, 148, 188, and 194 of Table CIV fall? 

Prob. 2. From Table CIV, classify according to Muck, the following coals, Nos. 5, 
25, 90, 116, 141, 189, 191, 197. 

Prob. 3. A coal gave as an ultimate analysis of the combustible, H=9.5; C=91; 
(S+O2+N2) =5.5; and as proximate analysis, fixed carbon, 83 per cent; volatile, 
17 per cent. Classify it according to Frazer and Muck. 

Prob. 4. The analysis of a coal gave, C =87.7; H 2 =4.7; O2 =5; ash =2.6. From the 
gas and coke qualities of Table LXI, what names could be applied to this coal? 

Prob. 5. How do the classifications according to Muck, Frazer, and Campbell 
agree on the following coals? Nos. 9, 25, 50, 60, 184? 

Prob. 6. Taking Juptner's figures of, C =75-80 per cent, H2 =5.5^4.5 per cent, and 
(O2+N2) =19.5-15.5 per cent, as the limits for non-caking, long-flame coals, which of 
those given in the general table would fall in this class? 

Prob. 7. How do the analyses of the lignites given in the table compare with the 
average as given by Jtiptner? 



662 ENGINEEEING THERMODYNAMICS 

Prob. 8. Sexton gives 7 per cent as a fair value for the fixed carbon for English 
cannel coal. What coals in the general table substantiate this value? 

Prob. 9. To what coals in the general table could the name gas-coal be applied, 
using Juptner's values as a standard? 

3. Calorific Power of Coals and the Combustible of Coals. Calculation 
of Calorific Power from Ultimate and Proximate Analyses. Calorific Power 
of the Volatile. If the carbon, hydrogen and sulphur in coal were just mixed 
together, the calorific power would be the sum of the products of each frac- 
tional weight into the calorific power of the respective element, but no such 
simple relation exists, so that calculation of calorific power from the ultimate 
analysis must be based on some sort of assumptions as to molecular relations, 
expressed or implied. 

Most of the ordinary fuels are mixtures of chemical compounds of unknown 
molecular constitution and usually great complexity of molecular structure, 
not at all revealed by ultimate analysis and only partly so by proximate, 
analysis, and in the case of the liquids by fractional distillation. They are 
all compounds and mixtures of the elements carbon, hydrogen, nitrogen 
oxygen, sulphur, and with or without neutral ash. This fact lies at the basis 
of most of the empiric formulas for their calorific power, for if the fuel elements, 
C, H2, and S, shown to be present by an ultimate analysis be assumed to be 
just mixed together in the weights given, the calorific power is the sum of 
that of all the parts and can be expressed symbolically. 

Let Q.ff = B.T.U. per pound of hydrogen; 

" (?c = B.T.U. per pound of carbon; 

" Q S = B.T.U. per pound sulphur; 

" Q = B.T.U. per pound fuel ; 

" w H = weight of hydrogen per pound fuel; 

" w c ^weight of carbon per pound fuel; 

" w s = weight of sulphur per pound fuel. 



Then 



Q = QcWc+QhW H +QsW S - . (754) 



This, however, takes no account of the oxygen which may be assumed 
to be free and so ignored, or assumed to be combined as an incombustible. 
By weight, 1 lb. H+8 lbs. O2, make 9 lbs. water, so that each pound of O2 taht 
is combined with H2 as water takes from the total H 2 present J of the oxygen 
weight, leaving as burnable only what is left or (wh — \w ). Introducing this 
expression in the formula above 

Q = Qcw c +QH(w H -iwo)+Qsw s , . o . o . (755) 
or factoring, 

Q = Qc\wc+^(w H -iw )+~Ws\. . . . . . (756) 

L Qc He A 



. 



HEATING BY COMBUSTION 663 

This is the; form of one of the first empiric formulas as given by Dulong, the 
sulphur effect was neglected and 14,500 and 62,10Q in round numbers taken 
as the calorific power of carbon and hydrogen respectively, giving the form 

Q = U,500wc+62,100(w H -iwo) (757) 

In the American Society of Mechanical Engineers Code for Boiler Testing 
there is recommended a formula which retains the sulphur effect and assigns 
4000 as the calorific value, changing the others for carbon and hydrogen 
to give 

Q = Ufi00w c +e>2,000(w H -lwo) +4000ws .... (758) 

Another older formula given by Mahler neglects both sulphur and combined 
oxygen-hydrogen effects, but introduces an empiric constant as follows: 

= 20,100w c +67,600wtf-5540 (759) 

None of these expressions or others to be noted will agree with experimental 
determinations due to the very wide complexity of molecular structure of 
the fuels they are supposed to represent. Most of them have been developed 
for coals of a single class, thus, for example, the Dulong expression Eq. (757) 
was intended to nearly represent a class of bituminous coals characterized 
by much volatile hydrocarbon while Dulong gave another for lignites, the 
peculiarity of which is a high water constant as follows : 

Q = 14,500^c+5333(^-|^o)-1147(1>o^tf) . . . (760) 

where w w is hygroscopic water, l\w combined and fixed water, and the 
negative term is supposed to represent the heat to evaporate all the water 
present. 

As an example of an attempt to base the formula at least in part on 
the physical character of the coal that of Goutal, Eq. (761), is especially 
interesting and suggestive: 

Q = U,670wc'+Kw v , (761) 

where w v is the volatile hydrocarbon weight, while w c ' represents the fixed carbon 
or coke, K is a constant supposed to have values, according to Goutal, as 
follows : 

When volatile by weight is 2-15% then K is 23,400 

15-30% " " 18,000 
30-35% " " 17,100 
35-40% " " 16,200 



30-35% " " 

tt ' ' ok \c\crf ( ( it 



664 ENGINEERING THERMODYNAMICS 

Juptner quotes a formula of the International Union of Steam Boiler 
Inspection Societies based on ultimate analysis as given in Eq. (762). 

Q = U,mw c +52,200(w H -±Wo)+4:500ws-108()ww, . . . (762) 

where w w is the hygroscopic water. He also notes some differences between 
the results of this formula and calorimeter tests by L. C. Wolf as follows: 

For bituminous coal ± 2% 

For lignite ± 5% 

For peat d= 8% 

For cellulose - 7.9% 

For wood ±12% 

It has been pointed out that the solid fuels may be regarded as the sum 
of a combustible and a non-combustible part; and the combustible as the 
sum of a fixed carbon and a volatile part. On this basis it is possible to make 
approximations of calorific power if the nature of the volatile is known or the 
average calorific power of the volatile, without an ultimate analysis, using 
instead the simpler and easier proximate analysis which does not require 
elaborate chemical laboratory equipment for its determination. 

The principal combustible gases of the volatile are hydrocarbons and these 
are of fairly constant calorific power per pound over a very considerable range 
of character. The highest calorific power of all is that of methane, 23,646 
B.T.U. per pound, and few of those hydrocarbons that enter into coal 
volatiles so far as is known, have less than 75 per cent of this. If, therefore, 
it is found, as might be expected, that one group of coals is characterized 
by the same group of hydrocarbons in its volatile and in even approximately 
constant proportions the calorific power of that volatile would be almost 
exactly constant, at least as constant as the calorific power of all the combustible 
taken together. It is also reasonable to expect that each class of coal would 
have a different group of hydrocarbon combustibles in the volatiles, the 
calorific power of which would vary accordingly, being highest for anthracites 
with methane as the principal constituent, lower for the bituminous varieties 
and least for lignites, the volatile of which is much diluted with non-com- 
bustible gases and vapors, similar to those yielded by wood. Thus the character 
of heating power of the volatile of the coals furnishes a new basis of classification 
with direct reference to availability as fuels. 

To illustrate the peculiarities of the calorific power of coals and the depend- 
ence of calorific power on class of coal, a new table, Table CV, has been com- 
puted and some curves plotted, all derived from the data on the two hundred 
coal analyses reported in Table CIV. As a first step, the calorific powers as 
given in the latter table are plotted in Fig. 173, to horizontal coordinates of 
total carbon to total hydrogen, which is the basis of Campbell's classification. 
Vertically, there are laid off for each coal two points, one the calorific power 



HEATING BY COMBUSTION 



665 





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666 ENGINEERING THERMODYNAMICS 

per pound of coal as reported from the calorimeter test, and the other 
the heat of formation of the products. Vertical classification bands are 
drawn across the sheet to divide the classes. From this chart it appears 
that there is no very definite relation between calorific power and the Camp- 
bell class of the coal, though there is a distinct falling off toward the ratios 
15 and over. Great confusion exists in the bituminous region and from 
there to the peats; so great as to make any representative curve an 
impossibility. 

The failure to establish any definite relation between calorific power and 
the total carbon-hydrogen ratio, as illustrated in Fig. 173, naturally suggests 
a trial on another basis. Accordingly, the calorific powers, both calorimetric 
values and that by heat of formation of products, are replotted in Fig. 174 
to a new horizontal coordinate, suggested by Frazer's classification, the ratio 
of fixed carbon to volatile. Here the points are differently distributed, but 
with no more hope of a smooth curve being drawn, through them. Again 
there is confusion in the bituminous region, especially where the fixed carbon 
to volatile ratio is less than 2, as here almost any calorific power may be found 
for the same value of the ratio. 

The explanation of this situation must be found in the nature of the com- 
bustible and in its amount and kind, but especially in kind. As combustible is 
partly fixed carbon and partly volatile, and as fixed carbon has always the same 
calorific power, the real explanation must be sought in variations in the nature 
of the volatile itself, and, of course, its amount. In order that this possibility 
may be traced, the chemical and thermal properties of these coals are redeter- 
mined, as ash and moisture free, in Table CV at the end of the Chapter, 
which, therefore, gives the properties of the combustible matter only. In 
this table the calorific power of the combustible is reported, total and as 
divided between the fixed carbon and the volatile parts and finally the calorific 
power of the volatile itself per pound is found. To get this result, the calo- 
rific power of the coal as determined by the bomb calorimeter is divided by the 
sum of the partial weights of fixed carbon and volatile, the quotient being 
the calorific power per pound of total combustible. The product of the 
fractional weight of the fixed carbon and 14,544, its known calorific power, 
gives the heat due to the combustion of the fixed carbon part of the combus- 
tible, and this subtracted from the B.T.U. per pound of combustible gives 
the heat per pound of combustible derived from its volatile. The heat 
per pound of combustible derived from its volatile only, when divided 
by the fractional weight of volatile in the combustible gives the B.T.U. per 
pound of volatile itself. These values of the calorific power of the volatile 
matter of coals from Table CV are plotted in Fig. 175 and through the points 
a fair curve can be drawn at least as fair, as the determination of the values 
from the indirect experimental results of the calorimeter for the whole coal, 
are themselves accurate. It is reasonable to expect that direct determination 
of the calorific power of volatile alone, will lead to a much smoother curve, but 
even with the present data it is possible to make a really useful approximation 



HEATING BY COMBUSTION 



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668 



ENGINEERING THERMODYNAMICS 




5 10 15 

Per Cent Ratio of m*ed Carbon 

Volatile 

Fig. 175. — Calorific Power of the Volatile of Coals. 



Per Cent 






HEATING BY COMBUSTION 669 

to the calorific power of a coal from its proximate analysis by the formula 
Eq. (763) derived from the smooth solid curve of Fig. 175. 



Let Wc = fractional weight of fixed carbon in coal ; 
Wv = fractional weight of volatile in coal. 



Then 



Q = 14,544w ( /+27,000w 1 



1- 



(-V.5 

\Wv) 



(763) 



This equation is of the same general form as Goutal's Eq. (761), but with his 
constant C evaluated as a function of the fixed carbon to volatile ratio, instead 
of assigning it separate values for each class of coal by the volatile content 
alone. The differences due to this are indicated in Fig. 175 by the relation 
of the dotted to the solid line. 

These relations not only make it possible to calculate the calorific power 
of a coal with fair accuracy from its easily found proximate analysis, but more 
important in a broad sense it brings out the relation of the coals as fuels to the 
oils and natural gas fuels as will appear a little more clearly when these latter 
are examined. 

Example. A coal which gives a heating value of 13,961 B.T.U. when tested in 
the calorimeter, was found to have an ultimate analysis as follows: C =80.03, H 2 =4.13, 
N 2 = 1.40, O 2 =3.20, S = 1.90, ash =9.34. How do the heating values derived by means 
of the Dulong Eq. (757) and A.S.M.E. Eq. (758) compare with the experimental? 

Substituting the above values in Eq. (757), 

(0320\ 
.0413 -'—^-) =13,952 B.T.U. per lb. 

and 

/ 0320\ 

Q = 14,600 X .8003 +62,000 .0413 -'— — +4000 X .019 = 14,069. B.T.U. per lb. 



Prob. 1. Which of the following coals would be worth more per ton on the basis 
of the heating value found by the A.S.M.E. formula; by the others? 

For the first C =83.75; 
For the second C =74.39; 
For the third C=73.5; 

Prob. 2. A boiler horse-power is 33,486 B.T.U. received by the water in the boiler, 
per hour. If 70 per cent of the heat of the coal is available, what will be the horse-power 
of a boiler under which is being burned one ton per hour of the following coal, the 
correct calorific power of which is given by the Dulong expression? What is the 
weight of the coal per hour per boiler horse-power? C =79.20; H 2 =4.30; 2 =2.15. 



H 2 =4.13; 


S= .57; 


2 =2.65 


H 2 =4.98; 


S=3.44; 


2 =6.42 


H 2 =5.19; 


S=2.54; 


O 2 =8.05 



670 ENGINEERING THERMODYNAMICS 

Prob. 3. Natural gas of 1000 B.T.U. per cubic foot may be purchased in a certain 
locality for 10 cents a thousand cubic feet. What price could coal containing 75 per 
cent of C, 6 per cent of H 2 , 1.5 per cent of 2 , and 1.5 per cent of S, bring in the same 
locality on the basis of heat content only? 

Prob. 4. A gas engine is operated on producer gas. The producer is supplied with 
coal of the following composition: C =80.15 per cent; H 2 =3.35 per cent; N 2 = 1.32 per 
cent; 2 =4.28 per cent; S = .9 per cent; ash = 10 per cent. What is the efficiency 
for the system if the engine delivers one horse-power for each pound of coal fed to the 
producer? 

Prob. 5. In a boiler plant it has been found that of the entire heat in the coal 60 
per cent gets to the steam, 25 per cent escapes to the stack, 13 per cent to the ashpit, 
and the remainder is radiated from the boiler setting. For a coal of the following com- 
position, what would be the number of B.T.U. per pound of coal in each loss and what 
would be the number actually used? C = 70 ; 2 = 2.5 ; N 2 = 2 ; H 2 = 12 2 ; S = .5 ; ash = 13. 

Prob. 6. What will be the per cent error in determining the calorific power of the 
following coals by Eqs. (754)-(763), the value by the calorimeter being ^considered 
correct? Nos. 12, 45, 98, 126, 147 of Table CV. 

Prob. 7. Coal is bought by the calorific value as found by the formula of the A.S.M.E. 
code. What bonus or penalty would be involved if the true value is that of the 
calorimeter for the first five coals of the table? 

Prob. 8. For coal No. 20 which of the formulas gives a result closest to the calorimeter 
value? Which for coal No. 2? 

Prob. 9. What would be the relative error in assuming that Eq. (763) gave the 
true heating value of coal No. 9 compared to the value as found from the ultimate 
analysis, the calorimeter value being used as the true one? 

4. Mineral Oil and Natural Gas Fuel. Chemical and Physical Properties. 
Calorific Power Direct and as Calculated for Oils from Ultimate Analysis or 
from Density, and for Gas from Sum of Constituent Gases. Mineral oils, no 
matter from what part of the world they come, are very much alike in ultimate 
composition, but most amazingly different in detailed properties such as boiling- 
point. When clean, all are absolutely ash free, all yield vapors on heating 
and have calorific powers very close together, compared to the variations noted 
for coals. Analysis shows them to be hydrocarbon compounds or rather mix- 
tures and solutions of a large number of hydrocarbons of the paraffine, ethylene 
and naphthalene series, the former predominating in American and the latter 
in Russian oils, these two countries furnishing the bulk of the world's supply, 
having produced respectively 16 J million and 7 million tons in 1905. 

From Pennsylvania crude oil the hydrocarbons in Table CVI at the 
end of the Chapter, of the paraffine series have been separated as reported 
partly by Lewes, and from Russian oils the corresponding ethylenes, and two 
naphthalenes by Redwood. In the paraffines, there are noted four isomers, 
that is, compounds having the same ultimate analysis and fractional weight 
but different physical properties, while with the ethylenes, all of which 
have the same ultimate analysis but different molecular weights, are noted 
two naphthalenes which are isomeric modifications of the ethylenes, having 
the molecular arrangement indicated by the formula C ra H2«_6+H6 = C w H 2 „, 



HEATING BY COMBUSTION 671 

which have the same weight proportions but whose properties are more closely 
those of the paraffines than the ethylenes. As pointed out in Chapter IV, 
the calorific power of all hydrocarbons varies with the value of n or their 
position in the series and is somewhat different for the same value of n in differ- 
ent series. It is also true that both liquid and vapor densities vary with the 
value of n so it might be expected that the calorific power would vary with the 
density of the oils, if they did not contain excessively variable proportions of 
different series. This was foreshadowed by the Slaby formula for the hydro- 
carbon gases and vapors which expresses calorific power as a linear function of 
density of those gases and is confirmed by the work of Sherman and Kropff at 
Columbia University, for the oils. In Table CVII at the end of this Chapter 
are given the results of their calorimeter determinations on 64 oils, ranging 
from specific gravity of .7100 to .9644 or 67.7° Be. to 15.2° Be., together with 
the calculated value as a linear function of density by Eq. (764). 

B.T.U. per lb. oil = 18,650+40(Be.-10) (750) 

In only i of the cases is the error more than 1 per cent, in only -fa over 2 per cent 
and never exceeds 3 per cent, which is fairly satisfactory. This makes it pos- 
sible to estimate calorific power of oils by simply taking the density, about as 
closely, in some cases more so, as could be done by an ultimate analysis and the 
application of the calorific powers of carbon and hydrogen to the respective 
fractions of each. 

Inspection of the general table of properties of mineral oils, Table CVIII 
at the end of the Chapter shows that the carbon hydrogen ratio, which for the 
ethylenes and naphthalenes is constant and equal to 6 and for the paraffines 
always less than this, does sometimes exceed this value. In such cases it must 
be assumed that solid carbon is present, having been deposited by the decom- 
position of some of the oil under heat treatment before or after leaving the earth, 
and such oils are invariably heavy and black in color. No natural or crude oil 
is heavy or dense because it consists only of heavy hydrocarbons, nor is any 
one light because its hydrocarbons are exclusively light ones, but as all crudes 
contain both heavy and light constituents the density is chiefly an indication 
of which class of hydrocarbons, the light or the heavy, predominate. These 
relations, which are of fundamental importance, in fixing the industrial value 
of an oil and in prescribing the treatment it shall receive in applying it to oil 
engines, boiler furnaces or the making of oil gas, will be brought out more 
clearly in a succeeding section under heat treatment for fractional distillation. 
This fractionation of oils is the only known way of reducing the range of hydro- 
carbon constituents and is a sort of separation process. 

The calorific power of the oils cannot be as accurately predicted from their 
ultimate analysis by computation of the heat of formation of the products of 
combustion, as it can fromt he density by Sherman and KropfFs formula, 
though there seems to be in some cases, large differences between the experi- 
mental bomb values and those for the formula. In most cases if not all, these 



672 



ENGINEERING THERMODYNAMICS 



can be traced to inaccurate experimental work, which, with oils, especially those 
that have light and easily volatile constituents, is a very difficult procedure. 
It is for this reason that two oil tables giving calorific powers are given, one 
containing the Sherman and Kropff values, known to be most accurate, and the 
other such values as are commonly reported in general engineering literature. 
To still more clearly bring out this point, there are plotted in Fig. 176 all 
the tabular values of calorific powers, exclusive of the heats of formation of the 
products, and through them the S. and K. line is drawn according to its equation. 
It will be noted how much better their experimental values lie with respect to 
this line than do others of more doubtful accuracy. 

Natural gas like mineral oil is also chiefly a mixture of hydrocarbons though, 
of course, no such complexity of mixture can exist in the gases as in the liquids 
because so few of the hydrocarbons are gaseous at ordinary temperatures; most 





















































































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Fig. 176. — Calorific Power of Mineral Oils. 



of them being liquids and a few solids. Accordingly, natural gas is a pretty 
definite thing and analysis shows it to be largely and in some cases almost 
entirely methane CH4. One sample, No. 1, from the series of Table CIX at 
the end of the Chapter, that from West Virginia, is over 99 per cent methane, 
but another sample, No. 38, reported for the Pittsburgh district is only about 
half methane, and contains nearly 36 per cent by volume of free hydrogen. 
Practically all natural gases contain some CO, which is a product of oxygen 
reacting on carbon, probably in a coal and as might be expected there is corre- 
spondingly some nitrogen present, as would be the case when the oxygen that 
combined had come from the air, leaving the nitrogen free after combustion. 
Similarly in many cases some CO2 is present, indicating a more complete oxida- 
tion of the carbon and accounting for more nitrogen on the assumption that the 
oxygen had been derived from air. 

One important class of chemical constituents of some natural gases 
is the higher hydrocarbons such as ethylene, but more peculiarly those 



HEATING BY COMBUSTION 



673 



that at ordinary temperatures are liquid and are present in the gas as vapors 
of those liquids together with a group that liquefy under pressure. This has led 
to a process of compression of natural gas with subsequent cooling for the 
recovery of liquid forms of hydrocarbons most of which are easily volatile and 
all of which are high in illuminating value. A recent paper of the Bureau of 
Mines by Allen and Burrel discusses the recovery of liquid forms from natural 
gas by compression, to which reference may be made, but which is not analyzed 
here because of lack of space. 

The calorific power of natural gas, like any other gas, is a thing very 
much easier to predict than that for coals and oils with reasonable precision 
when its constituent gases are known. Each of these has a definite calorific 
power per cubic foot at standard conditions of 32° and 29.92" Hg, so that the 
calorific power is to be found by multiplying the volumetric proportion of each 
constituent by the proper calorific power. The sum will be the calorific power 
of the gas in B.T.U. per cubic feet (standard) and, of course, there will be 
both a high and a low value for every gas containing any hydrogen, free or com- 
bined and this includes all of the natural gases. For convenience, in calcula- 
tions on temperature rise, due to combustion, the B.T.U. per pound is usually 
added. When some of the constituents are the heavier hydrocarbons, which 
are not separated by the absorption methods of gas analysis, they are generally 
reported together as heavy hydrocarbons, as illuminants, or in some equivalent 
terms. In all such cases, heat calculations require that a chemical constitu- 
tion be assigned to these, usually ethylene and benzole in some proportion, 
as judgment based on experience may indicate. This is the only element of 
uncertainty as to calorific power that is worth noting, and at present there is 
no remedy available. 

To illustrate the method of determination of density and calorific power 
per cubic foot and per pound one case is set down in convenient tabular form, 
Table LXVI. 

Table LXVI 

DENSITY AND CALORIFIC POWER OF NATURAL GAS FROM CONSTITUENTS 

(32° F. and 29.92" Hg) 





One Cubic Foot 






Contains 


Yields B.T.U. 


Summary 




Cu.Ft. 


Lbs. 


High. 


Low. 




Methane, CH 4 

Heavy hydroc., C 6 H 6 . . . . 
Carbon monoxide, CO. . . . 

Oxygen, 2 


.9820 

.0010 
.0025 

.0025 


.043935 

.000240 
.000195 

.000223 


1090.5 

5.0 

.8 


944.7 

4.6 

.8 


B.T.U./cu.ft. gas high = 1046.3 
B.T.U./cu.ft. gas low = 950.1 

Cu.ft/lb. gas = 22.42 
Lbs./cu.ft. gas = .0446 

B.T.U./lb. gas high =23458. 


Total for gas 


1.0000 


.044593 


1046.3 


950.1 


B.T.U./lb. gas low =21301, 



674 ENGINEERING THERMODYNAMICS 

Prob. 1. The following oil was used in an engine and one horse-power hour (2545 
B.T.U.) was obtained for each pound of the oil supplied. What was the thermal 
efficiency based on the heating value, (a) as derived by calculation from the ultimate 
analyses, and (b) by the Sherman and Kropff formula? 

Sp.gr. at 60° F.=. 926. C=83.36; H 2 = 12.41; S = .5; (N 2 +0 2 ) =3.83. 

Prob. 2. At one time a melting device was used to clear away snow from city streets. 
The fuel used was oil and the heat generated was used to melt the snow placed a tank 
above the fire, the water running off to the sewer as soon as formed. Assuming 90 
per cent of the heat of the fuel to be available, that the snow is put in at a tempera- 
ture of 20° F. and that the water runs away at a temperature of 40° F., how many 
pounds of snow will be disposed of per pound of the following oil? 

C=85.5; H 2 = 14.2; (0 2 +N 2 )=.3. 

Prob. 3. To prevent frost damage in orchards, crude oil is burned in smudge pots 
and the temperature of the air in the orchard kept above 32° F. Taking the specific 
heat of air as .243, how many cubic feet can be heated from 30° to 40° F. by a gallon 
of California crude oil, the density of which is 16.85° Be. and the ultimate analysis 
by weight is H 2 = 11.3 C =85.75; S = .67? Compute the results by S. and K. formula 
and by heat of formation of products. 

| Prob. 4. The analyses given are for a West Virginia and an Ohio natural gas respect- 
ively by volume. On the basis of the heat of formation of the products of each, what 
would be their relative heating values? 

CH 4 H 2 CO C 2 H 4 N 2 C0 2 2 

West Virginia. .99.5 5 

Ohio... 93:35 1.64 .41 .35 3.41 .25 .39 

Prob. 5. The oil, the analysis of which is given below by weight, may be had 
for 5 cents per gallon. At what price will the gas, the analysis of which is also given, 
but by volumes, be an equally economical fuel, the economybeing based solely on the 
heat value? 

Oil: 2 = 82; H 2 = 14.8; 2 +N 2 =3.2. 

Gas: CH 4 = 75.99; H 2 = 6.1; C 2 H 4 = 18.12; C0 2 = .34. 

Prob. 6. From the ultimate analysis of the two compare the heating value of 
1 bbl. (50 gals.) of oil No. 11, Table CVIII and one ton (2000 lbs.) of coal No. 98 and 
191, Table CIV. 

Prob. 7. A ton of coal occupies roughly 40 cu.ft. Compare the space occupied 
by a million B.T.U. in the form of fuel oil (Table CVIII) and a bituminous coal 
(Table CIV). 

Prob. 8. Assuming a boiler efficiency of 70 per cent for both fuels, what will be 
the consumption of a crude oil (Table CVIII) and an anthracite coal (Table CIV) per 
boiler horse-power? 

Prob. 9. An oil-burning locomotive is developing 1000 boiler horse-power. What 
would be the necessary oil tank capacity for a two-hour run? The oil used is No. 37, 
Table CVIII and the boiler efficiency is 65 per cent. 



HEATING BY COMBUSTION 675 

Prob. 10. What would be the space required to carry an equivalent amount of coal 
No. 24, for the same conditions as in Prob. 9? 

Prob. 11. Oil and gas may both be procured in a given locality. Assuming the gas 
to be equivalent to No. 19, Table CIV, and the oil to be equivalent to No. 69 Table 
CVIII, what would the oil be worth per gallon to equal the gas at 7 cents per 1000 
cubic feet based on heating value alone? 

5. Charcoal, Coke, Coke Oven and Retort Coal Gas as Products of Heating 
Wood and Coal. Chemical, Physical, and Calorific Properties per Pound. 
Calorific Power of Gases per Cubic Foot in Terms of Constituent Gases. Yield 
of Gas and Coke per Pound of Coal. When wood is heated, a distillation process 
begins at about 400° F. before which the discharge is mainly water vapor, but 
after which a series of complex gases and vapors are liberated, some of which 
may condense. Some of these distillates were present as such in the wood, and 
are liberated unchanged, but most of them are compounds formed by the heat 
action in the fiber, sap and ash constituents by mutual chemical reactions. 
The products of heating wood as given by Juptner show the surprising complexity 
of the process and products which are divisible into five groups: 

\ 
Products of Wood Distillation (Juptner) 

(1) Hygroscopic water, 

(2) Gas, consisting mainly of 

(a) Acetylene, C2H2 (e) Carbon monoxide, CO 

(6) Ethylene, C 2 H 4 (/) Carbon dioxide, C0 2 

(c) Benzole, C 6 H 6 (g) Methane, CH 4 

(d) Napthalene, CioHg (h) Hydrogen, H2 

(3) Liquid tar, consisting of 

(a) Benzol, CeH 6 (g) Crysylic acid, CyHgO 

(b) Naphthalene, Ci H 8 (h) Phlorylic acid, C 7 H 8 

(c) Paraffine, C20H42 to C 22 H 46 f C 7 H 8 2 

(d) Retene, CigHis (i) Creosote A CgHioCb 

(e) Phenol, C 6 H 6 1 C 9 Hi 2 2 
(/) Oxyphenic acid, CeH 6 (j) Resins 

(4) Pyroligneous acid, consisting of 

(a) Acetic acid, C2H4O2 (c) Acetone, C3H6O 

(6) Propionic acid, C 3 H 6 2 (d) Wood alcohol, CH3OH 

(5) Wood charcoal, 

When the heating of wood takes place in closed chambers, the wood melts 
just as do the coking coals, but at temperatures in the neighborhood of 600° 
F. and it yields a sort of hard coke quite different from charcoal. The whole 
process is summed up in the following Table LXVII from Violette which also 
indicates the changes in the result brought about by gradually rising tem- 
peratures of distillation, which are briefly, an increased gas yield continuously, 
and a liquid yfeld first increasing and then decreasing, showing that at high tern- 



676 



ENGINEERING THERMODYNAMICS 



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HEATING BY COMBUSTION 



677 



peratures some of the liquid is decomposed into gaseous constituents. In all 
cases the yield of charcoal decreases but its quality changes. The charcoal 
at higher temperatures contains more carbon, about the same hydrogen and 
regularly lessening amounts of oxygen and nitrogen, the sum of which is, how- 
ever, always large and indicates why peats and lignites near the woods have 
volatiles so weak in combustible matter. Quick coking produces only half 
the charcoal that results from a slow process. 

Peat on heating also gives off an equally large variety of volatile products, 
the nature of which is intimately related to the fuel value and its treatment 
in boiler fires and producers. The following is the result of heating a Swiss 
peat obtained by Vohl which it is interesting to compare with that for wood, 
already given and for the coals to follow; 

Products of Swiss Peat Distillation (Vohl) 

1. Water, associated with liquids called tar water, 25 per cent by weight, 
consisting of 



Bases • 



a. Ammonia 

b. Methylamin 

c. Picolin 

d. Lutidin 

e. Anilin 

I /. Caespidin 



Acids 



g. Carbon dioxide 
h. Carbon disulphide 
i. Hydrocyanic 
j. Acetic 
k. Propionic 
I. Butyric 
m. Valieranic 
n. Phenol 



of 



2. Gas, 17.62 per cent by weight, consisting of 

(a) Heavy hydrocarbons, C re H2n (c) Hydrogen, H2 

(b) Methane, CH4 (d) Carbon monoxide, CO 

3. Liquid, 5.37 per cent by weight condensed from the gas and consisting 

(a) Tar, specific gravity, .820 

(b) Heavy mineral oil, specific gravity, .855 

(c) Paraffine 

4. Peat coal or coke, 25 per cent by weight. 



The most striking difference between the products of peat and wood distil- 
lation from the fuel standpoint is the appearance of heavy hydrocarbon in the 
former in the gas form and an increase of it in the liquid forms or a general rise 
of hydrocarbon combustible matter in the volatile, because the liquids appear 
first as vapors mixed with the permanent gases. A comparison of the products 
of different coking peats is given in the following Table LXVIII from Kane and 
Sullivan who roasted 100 pounds in retorts, 



678 



ENGINEERING THERMODYNAMICS 






Table LXVIII 
PRODUCTS OF PEAT DISTILLATION (Kane and Sullivan) 









Composition 




Per Cent by Weight of 


Composition of Tar Water. 


of Tar 




Products. 


Per cent, by Weight 


Per cent, by 
Weight 












Ammonia. 


Acetic Acid. 










Origin. 










M 








o 
O 


«5 

03 

a 




o 

GO 
W 


6 
w 
6 


< 




1 

o3 

(H 

03 
PM 


o 
1 


a 

•go 


Light and dense mixed Mt. 


























Lucas Philipstown 


23.60 


2.00 


37.50 


36.90 


.302 


1.171 


.076 


.111 


.092 


.024 


.684 


.469 


Light peat Wood of Allen. 


32.27 


3.58 


39.13 


25.02 


.187 


.725 


.202 


.302 


.171 


.179 


.721 


.760 


Dense peat Wood of Allen. 


38.10 


2.76 


32.65 


26.49 


.393 


1.524 


.286 


.419 


.197 


.075 


.571 


.565 


Upper layer of Ticknevin . 


38.63 


2.92 


31.11 


32.34 


.210 


.814 


.196 


.287 


.147 


.170 


.262 


.617 


Same distilled at red heat . 


32.10 


3.34 


23.44 


42.12 


.195 


.756 


.208 


.365 


.161 


.196 


.816 


.493 


Upper layer of Shannon . . 


38.13 


4.12 


21.87 


35.69 


.404 


1.576 


.205 


.299 


.132 


.181 


.829 


.680 


Dense peat 


21.19 


1.46 


18.97 


57.74 


.181 


.702 


.161 


.236 


.119 


.112 


.647 


?66 







Lignite yields on heating much the same sort of products as does peat but 
there are two groups, one in which the hydrocarbon or bituminous substances 
are larger than in the other, probably because of different origin. Thecoke 
and gas are not of much value and being little used there is but little data 
available. The coke from such of the lignites as do not disintegrate by reason 
of loss of water of crystallization will be a little less than half the coal by 
weight, the volatile from one-seventh to one-third and tar water from one- 
eighth to one-fifth. Thus, in the lignites that will coke, there is an increase in 
volatile combustible while those that disintegrate have combustible close to 
peat. The former might properly be called sub-bituminous coal by reason of 
the difference and the latter solidified peat or lignite proper. 

Bituminous coals are the great raw materials for roasting treatment for the 
manufacture of coke, (a) in beehive ovens where the volatile is burned and 
wasted, (b) in by-product coke ovens which yield not only valuable coke but 
also illuminating and fuel gases besides chemical by-products, and also for 
the primary manufacture of illuminating gas in retorts where the coke is a by- 
product. The study of the distillation or roasting of the bituminous coals 
is, therefore, not only important as throwing light on their fuel properties but 
also because of its relation to the illuminating gas and coke industries. Much 
is known of these properties and their use in plants of the various sorts, so much, 
that what can be said here will be the merest sketch in which the fuel properties 
will be dwelt upon rather than the chemical by-products recovery, the produc- 
tion of illumination or the metallurgical properties of coke, each of which is 
itself a big subject. 

While there is a very great difference between the products from the dif- 



HEATING BY COMBUSTION 679 

ferent bituminous coals, the following by Wagner may be taken as a basis of 
comparison of the results with those of wood, peat and lignite. 



Products of Bituminous Coal Distillation (Wagner) 

Coal: 78 per cent C, 4 per cent disposable H 2 , 1| per cent N 2 , .8 per cent S, 
5.7 per cent combined H 2 0, 5 per cent hygroscopic H 2 0, 5 per cent ash. 

1. Tar water or ammonia water containing, 

(a) Water, carbonate of ammonia and sulphide of ammonia. 
(6) Chloride, cyanide, sulphocyanide of ammonia. 

2. Gas containing, 

' gases — acetylene, ethylene, propylene, butylene; 

(a) Illuminants - J benzole, styrole, naphthalene, propyl, 

} butyl, acetylnaphthalene; 

(b) Combustible non-illuminants — hydrogen, methane, carbon monoxide; 

(c) Impurities — carbon dioxide, ammonia, cyanogin, rhoden, sulphuretted 

hydrogen, carbon disulphide, nitrogen. 

3. Liquid tar containing, 

(a) Liquid hydrocarbons — benzole, toluol, propyl, butyl, etc.; 
(6) Solid hydrocarbons — naphthalene, etc.; 

(c) Substances containing oxygen — phenol, creosote, aniline, etc.; 

(d) Asphaltic substances and resins. 

4. Coke, 70 to 75 per cent of the coal containing 90 to 95 per cent of combus- 

tible matter, and 10-5 per cent ash. 



Comparison of these results with previous ones shows most clearly a rise 
of hydrocarbons in both gaseous and liquid form, which together constitute 
the volatile of the coal, so that the volatile has combustible properties very 
close to that of vapor of oils. The diluent materials in the volatile are small 
in amount but may increase on excessive heating. The manner of heating, 
as to time and temperature, largely control the nature of the products, gases 
coke or by-products. Moderate temperatures are necessary to secure high 
yields of ammonia and for highly illuminating gas, high temperatures increase 
the coke at the expense of the carbon of the hydrocarbons, decomposing them 
to methane, hydrogen and carbon. The products differ, moreover, with the 
period of coking, as the temperature does also as a rule, fresh cold coal being 



680 



ENGINEERING THERMODYNAMICS 



charged and slowly heating up. During the first periods most of the richer 
hydrocarbons forming the tar and illuminants come off, constituting the high 
calorific power part of the volatile. Toward the end of the coking period and 
that of highest temperature the gases are free of tar and low in calorific and 
illuminating power. It has been suggested that the first period corresponds 
to the formation of coke and ends when the coke has set or taken its permanent 
size and shape, while the second is a period of coke decomposition or rather 
distillation of the freshly set coke and this is a useful distinction to keep in mind 
for boiler and gas producer application of fuels. Two sets of data will serve 
to illustrate the variation of products with coal, and the variation of gas with 
the time of coking, the former, Table LXIX, by Juptner, and the second, 
reported by Blauvelt for both a high and a low volatile American coal in the 
Solvay by-product coke oven. These latter results relate entirely to the gas 
composition its^ quantity and rate of production, and are given in. graphical 
form in Fig. 177, 



Table LXIX 

PRODUCTS OF BITUMINOUS GAS COAL DISTILLATION (Juptner) 

(Variation with coal composition) 



Coal from 


Pas De Calais. 


England. 


Commentry 


Blanzy. 




' Moisture 

Ash 


2.17 2.70 
9.04 7.06 


3.31 
7.21 


4.34 
8.80 


6.17 
10.73 


Coal composition, 


2 


5.56 6.66 
5.06 5.36 

88.38 86.97 
1 1 


7.71 
5.40 

85.89 
1 


10.10 
5.53 

83.37 
1 


11.70 


per cent by weight 


H 2 


5.64 




C 


81.66 




1 N 2 


1 










rGas 


13.70 15.08 
3.90 4.65 
4.59 5.57 

71.48 57.63 
6.33 7.07 


15.81 
5.08 
6.80 

64.90 
7.41 


16.95 

5.48 

8.61 

60.88 

8.08 


17.00 




Tar 


5.59 


Products of distilla- . 
tion, per cent by 


Ammonia water 
Coke 


9.86 
58.00 


weight 


. Coal dust 


9.36 


Gas produced per V ol. cubic meter 
kg coal 


30.13 31.01 


30.64 


29.73 


27.44 


Volumetric analysis 
of gas 


{ co 2 


1.47 1.58 
6.68 7.17 

54.21 52.79 

34.37 34.43 

.79 .99 

2.48 3.02 


1.72 

8.81 

50.10 

35.03 

.96 

3.98 


2.79 
9.86 
45.45 
36.42 
1.04 
4.44 


3.13 


CO 


11.93 


H 2 


42.26 


CH 4 '.. 


37.14 


C 6 H 6 


.88 


[ C 2 H 4 


4.76 







Probably no coals have as high a gas yield as the English cannel coals and 
related varieties, so the following figures, Table LXX, by Robinson, are of 
interest as representative of high yields. 



HEATING BY COMBUSTION 



681 













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Hour of Roasting 

Fig 177. — Variation in the Composition of By-product Coke Oven Gas with Time of Roasting. 



682 



ENGINEERING THERMODYNAMICS 



Table LXX 
GAS YIELD OF ENGLISH CANNEL COAL (Robinson) 



Coal. 



Cubic Feet Retort 


Pounds Gas. 


Cubic Feet per 


Gas per Pound Coal 


Pounds Coal 


Pound of Gas. 


5.13 


.1776 


28.87 


4.24 


.2080 


30.39 


6.70 


.3866 


17.32 


5.05 


.2848 


17.73 


5.09 


.2058 


24.73 


5.00 


1790 


27.93 


4.20 


.1375 


30.52 


3.20 


.1143 


27.99 


4.67 


• .1348 


34.66 



Newcastle 

Scotch parrot 

Boghead cannel .... 
Lesmahagon cannel 
Lucehall cannel. . . . 
Neath, South Wales 

Derbyshire 

Derbyshire soft .... 
Staffordshire 



Average 4£ cu.ft. gas per pound of coal, weighing .036 lb. cubic foot and giving about 
680 B.T.U. per cubic foot. 

Coke ovens, so far as the fuel processes are concerned, are exactly the same 
as gas-house retorts except for size, shape and temperature. The former 
yield more coke than the latter where the temperatures are higher, time of gas 
contact with coal shorter, and with walls longer. The hydrogen is higher 
and hydrocarbons and tar yield lower in the retorts, probably because of the 
temperature and contact conditions but just how or why these variations occur 
no one seems to know. 

In Table CX at the end of this Chapter are given a number of analyses 
of coal gas and by-product oven gases for comparison and taken from a variety 
of sources to show possible limits. The following, Table LXXI, gives a com- 
parison of two analyses considered by Juptner as typical of the two classes, 
with Blauvelt's typical coke oven gas and Guldner's typical retort coal gas. 

Table LXXI 
COMPARISON OF COKE OVEN AND RETORT COAL GASES 



Constituents. 



Coke Oven Gas Per Cent by Volume. 



Blauvelt. 



Guldner. 



Juptner. 



Retort Coal Gas Per Cent 
by Volume. 



Guldner. 



Juptner 



Benzole, C 6 H 6 

Ethylene, C 2 H 4 

Methane, CH 4 

Hydrogen, H 2 

Hydrogen disulphide, H 2 S . 

Carbonic acid, C0 2 

Carbon monoxide, CO . . . 

Oxygen, 2 

Nitrogen, N 2 



1.2 

4.2 

35.5 

48.0 



1.3 

5.1 

.5 

4.2 



.81 

1.52 

32.40 

55.60 

1.21 
7.08 

1.38 



.61 

1.63 

36.11 

35.32 

.43 

1.41 

6.49 



4.5 
35.0 

48.5 

2.0 
7.0 
.25 
2.75 



1.54 

1.19 

36.00 

55.00 

.87 
5.40 



HEATING BY COMBUSTION 



683 



The coke itself is a by-product in the gas retort and no attempt is made to 
get a good quality, the retort management and the selection of coal for it being 
both guided by the desire for low coke and high gas yield with high illuminating 
value. Beehive coke ovens are operated with regard to nothing but the largest 
production of good, hard coke of large pieces and good porosity, for use in 
foundry cupolas and in blast furnaces, while by-product coke ovens are designed 
with several ends in view, first good coke, second large yield, third good gas 
for heating, power, and illumination, and fourth chemical by-product recovery. 
It has already been shown how the gases of the retort and coke oven compare; 
they contain not the same quantities but nearly so and the same kinds of con- 
stituents, the differences being mainly in the hydrocarbons which for the retorts 
and high heats are simpler in character with more hydrogen and less carbon 
content. The beehive coke is generally regarded as the best for iron making 
but the difference compared to oven coke is not great. Retort coke is, however, 
very different because (a) it is made from a non-coking or semi-coking coal 
and (6) has much soot deposited from decomposed hydrocarbons and is very 
dense, close grained, hard, black and always in small pieces. In the beehive the 
yield from a given coal must always be less than from the oven because some is 
burnt with the gases and so is less than the fixed carbon, whereas in ovens or 
retorts the yield is always greater than the fixed carbon. Some coke analyses 
are given in Table CXI at the end of the Chapter from the U. S. Geological 
Survey reports of tests on American coals, given mainly to show its character- 
istic as a fuel; reference must be made to metallurgical works for analyses and 
properties of the coke adapted for treatment of the metals or ores. The fol- 
lowing three English cokes (Sexton) are given here for comparison: 

ANALYSIS OF THREE ENGLISH COKES 





Composition Per Cent by Weight. 


Origin. 


C 


H 2 


o 2 


N 2 


Ash. 


Dunkenfield (Percy) 

Best Durham (Kubale) 

Average Durham (Kubale) 


85.84 
93.15 
84.92 


.52 

.72 
4.53 


1.38 

.90 

6.66 


.86 

1.28 

.65 


11.40 
3.95 

2.28 



A most interesting observation with regard to the relation between the coal, 
coke and gas, was made by Deulle and quoted with the following Table LXXII 
by Lewes. It is, that the best gas coals should contain from 7.5 to 9 per cent 
of oxygen; less than this means a high coke yield but gas poor; more will yield 
bad coke, indicating that the illuminating hydrocarbons are originally oxygenated 
substances like fatty acids or alcohols decomposing to form ethylene, methane, 
hydrogen and oxides of carbon, which are not primarily present as such. 



684 



ENGINEERING THERMODYNAMICS 



Table LXXII 

RELATION BETWEEN OXYGEN IN COAL AND HYDROCARBONS IN GAS 

(RETORT) 



Per Cent O2 in Coal. 


5 to 6 


6.5 to 7.5 


7.5 to 9 


9 to 11 


11 to 12 


Volumetric compo- , 
sition of gas 


' C0 2 

CO 

H 2 

C 6 H 6 

CH 4 

k C2H4 


1.47 
6.68 

54.21 
.79 

34.37 
2.48 


1.58 
7.19 

52.79 
.99 

34.43 
3.02 


1.72 
8.21 

50.10 
.66 

35.03 
3.98 


2.79 

9.86 
45.45 

1.04 
36.42 

4.44 


3.13 
11.93 

42.26 

. .88 

37.14 

4.76 


Density of gas 


.352 


.376 


.399 


.441 


.482 







The calorific power and density of these coke oven and retort coal gases 
may be calculated from the constituent gases and their respective constants as 
was done for natural gas, the operations being most conveniently set down in 
tabular form, as follows in Table LXXIII. 



Table LXXIII 

DENSITY AND CALORIFIC POWER OF COKE OVEN GAS 

From Constituents (32° F. and 29.92" Hg) 





One Cubic Foot 




Constituents of Average 


Contains 


Yields B.T.U. 


Summary 




Cu.ft. 


Pounds. 


High. 


Low. 




Hydrogen, H2 


.4200 
.3430 
.0600 
.0200 
.0200 
.0250 
.0110 
.1010 


.0023607 
.0153470 
.0046840 
. 0015880 
.0043800 
.0030620 
.0009812 
.0079070 


143.220 

365.638 

20.460 

34.000 

78.840 

0.0 

0.0 

0.0 


122.640 

328.937 

20.460 

31.900 

75.900 

0.0 

0.0 

0.0 


B.T.U. per cu.ft. 


Methane, CH 4 

Carbon monoxide, CO. . 
Heavy ( C 2 H 4 . . 
Hydrocarbons \ C 6 H 6 . . 
Carbon dioxide, C0 2 . . . 

Oxygen, 2 

Nitrogen, N 2 


gas high 642.16 

B.T.U. per cu.ft. 

gas low 579.84 

Pounds per cu.ft. 

gas 04031 

Cu.ft. per lb. gas. 24.807 
B.T.U. per lb. gas 

high 15930 

B.T.U. per lb. gas 


Total for gas 


1.0000 


.0403099 


642.158 


579.837 




low 14384. 



HEATING BY COMBUSTION 



685 



Table LXXIII, Concluded 

DENSITY AND CALORIFIC POWER OF RETORT COAL GAS 

From Constituents (32° F. and 29.92" Hg) 





One Cubic Foot 




Constituents of Average 


Contains 


Yields B.T.U. 


Summary 




Cu.ft. 


Pounds. 


High. 


Low. 




Hydrogen, H 2 

Methane, CH 4 

Carbon monoxide, CO. . 
Heavy ( C 2 H 4 . . 
Hydrocarbons \ C 6 H 6 . . 
Carbon dioxide, C0 2 . . . 
Oxygen, Oo 


.5250 
.3135 
.0860 
.0110 
.0110 
.0150 
.0035 
.0350 


.0029509 
.0140310 
.0067140 
.0008740 
.0024100 
.0018400 
.0003121 
.0027400 


179.025 

334.191 

29.326 

18.700 

43.362 

0.0 

0.0 

0.0 


153.3000 

300.6400 

29.3260 

17.5450 

41 . 7450 

0.0 

0.0 

0.0 


B.T.U. per cu.ft. 

gas high 604.6 

B.T.U. per cu.ft. 

gas low 542 . 56 

Lbs. per cu.ft. gas .031872 
Cu.ft. per lb. gas 31.375 


Nitrogen, N 2 


B.T.U. per lb. gas 

high 18969.5 

B.T.U. per lb. gas 




Total for gas 


1.0000 


.0318720 


604.604 


542.5560 




low 17022.7 



Prob. 1. A natural gas which was practically 100 per cent CH 4 was procurable for 
7 cents per 1000 cu.ft. Compare the cost of coal gases Nos. 1 and 17 of Table CX at 
$1.00 per 1000 cu.ft., with this on a heat-unit basis. 

Prob. 2. Five cubic feet of coal gas No. 10, Table CX, were made per pound 
of coal, and the gas was sold at $1.10 per million B.T.U. What was the cost per 
ton (2000 lbs.) of coal if 50 per cent of the value of gas was charged to coal? 

Prob. 3. What will be the cubic feet of gas used per horse-power per hour by an 
engine running on coal gas No. 71 with a thermal efficiency of 20 per cent? 

Prob. 4. A hall is lighted by 50 lights each consuming 5 cu.ft. of gas per hour. If 
the hall be 100 ft. X50 ft. x20 ft. and the gas used be No. 43, how many times per 
hour must the air be changed to prevent a temperature rise of over 5° F.? 

Prob. 5. A thousand cubic feet of coal gas No. 12, are being forced through a 
pipe per minute. How many of the following units are being transmitted per minute? 

(a) Foot-pounds; (b) B.T.U.; (c) Horse-power hours. 

6. Distillate Oils, Kerosene, Gasolene, Residue Oils; and Oil Gas as 
Products of Heating Mineral Oils. Chemical, Physical, and Calorific Properties. 
Calorific Power of Fractionated Oils in Terms of (a) Carbon and Hydrogen; 

(b) Density per Pound, and Estimated Value per Cubic Foot of Vapor. Calorific 
Power of Oil Gas per Pound and per Cubic Foot in Terms of Constituent 
Gases. Yield of Distillates and Oil Gas. The discovery that crude 
mineral oil could be made to yield by simple heat treatment such valuable 
products as kerosene for lamps and engines, and gasolene, which latter has 
made possible the modern motor boat, automobile, and aeroplane, was one of 
great industrial importance and it is a little surprising to find so little scientific 
information available as to the precise nature, chemically and molecularly, of 
such important raw materials and products. It is, of course, known that all 
aire hydrocarbons and that any sample contains many, and these usually of more 



686 



ENGINEERING THERMODYNAMICS 



than one series, and it appears that each sample consists of a mixture of several, 
more or less soluble in each other. Each hydrocarbon having a different boiling- 
point from the others, it should be possible to separate them by distillation 
but not very well or completely because of mutual influences. In the first 
place, if one is dissolved in the other the boiling-point of the solvent will be raised 
by the substance dissolved and the effect will be different as the proportions 
vary and these will vary with continued boiling. Again, vapors of constituents 
having high boiling-points will escape with other vapors coming off at their 
own lower boiling-points in some proportion because each exerts its own vapor 
tension in a mixture. Therefore, while distillation will accomplish some sort 
of separation no distillate from a complex mixture of solutions can itself be simple 
but will be of the same general nature, except that the proportions of the constit- 
uents will be different. This explains why distillates obtained at low tempera- 
tures will leave residues when they are themselves heated even to a very much 
higher temperature than that at which they were obtained. i 

The separation and naming of the various oil products is a purely local 
procedure and though based on difference in boiling-points mainly, is not prac- 
ticed in the different refineries in the same manner. Therefore, a product of 
a given name, like gasolene, may be quite different, one sample from another. 
The following Table LXXIV is given by Robinson as the average practice 
in separating and naming the products of American and Russian crude oil, 



Table LXXIV 
AVERAGE DISTILLATION PRODUCTS OF CRUDE MINERAL OILS (Robinson) 



Class. 


Name of Product. 


Average 

Per Cent 

Yield. 


Specific Gr. 
60° F. 


Be. 


Boiling- 
Point, F. 




Petroleum ether. . . . 


' Cymogene 

\ Rhigolene 


small 
.1 

1 -1.5 

10 

2 - 2.5 
2- 2.5 

12 -20 
40 -55 


.590 

.625-. 631 
.635-. 658 
.680-. 700 
.717-. 72 
.742-. 745 
.780-. 785 
.800-. 810 
.85 

.885-. 920 
.980 


107 
94-92 
91-83 
76-70 

65 

58 

49 

44 

35 
28-22 

13 


32 

64 
86-158 
140-212 
175-250 
212-265 
300-575 
300-700 






Gasolene 




Petroleum spirit. . . . 


' C naphtha (benzene) . . 
• B naphtha 


__, 


k A naphtha (benzene) . 
j Water white 


u 

d 
a 
u 


Lamp kerosene 

Intermediate 


\ Ordinary kerosene .... 
Gas oil 


I 


Heavy oils 


' Lubricating oil 

Paraffine 


17.5 

2 
5 -10 




. Residue and loss 




Petrol 


Gasolene or benzene . . 
Kerosene 


5 -16 
30 -40 
10 -12 
12 -15 
25 -40 

3 -5 

10 -15 


.725-. 765 
.817-. 828 
.840-. 860 
.870-. 897 
.908-. 912 
.915-. 920 

.900-. 950 


63-53 
41-39 
37-33 
31-26 

24 
23-22 

25-17 






Lamp oils 




Intermediate 


Solar oil 


!5J 

o 




f Spindle oil 


Lubricating oils .... 


\ Engine oil 


a 

S3 




1 Cylinder oil 


00 


Fuel oil 


Residue, astatki or 
gondron 


« 













HEATING BY COMBUSTION 



687 



but must be accepted with caution as the differences mentioned above may be 
considerable,t hough for lamp oil kerosene, considerations of safety have led to 
legal restrictions as to boiling and flash points that keep it reasonably constant. 
A somewhat more common distribution of products with the American 
names and densities as now understood in the oil business is given in the follow- 
ing Table LXXV. 

Table LXXV 

AMERICAN MINERAL OIL PRODUCTS AT 60° FAHRENHEIT 



Kind of Oil. 


Degrees 
Gravity, Baurre 


Weight^of one gallon. 
231 cu.in. in ounces. 


Gasolene 


93.5 

91. 

90. 

88. 

86. 

78.5 

74. 

74. 

72. 

62.0 

49. 

47.5 

46.5 

42. 


83 5 


Gasolene 


85. 


Gasolene 


85 5 


Gasolene 


86. 


Gasolene 


86.5 


Gasolene 


90. 


Gasolene 


91.5 


Sumatra naphtha 




Gasolene 


92. 


Gasolene 


97.5 


Kerosene, 120°, water white 

Kerosene, 150°, water white 


104.5 
105.5 


Kerosene 


106.5 ' 


Celcius 




Fuel oil, Pratt's 


108. 


Fuel oil, Lima. . 




109. 


Limpid residue, W. P 


34.5 

34.5 to 35.0 

32.5 


114. 


Gas oil, E. W 

Paraffine, .865° sp.gr 


115.5 







Water, at 62° F 133.4 



The percentages and gravities given in the preceding tables are supposed to 
be average practice but there is a question whether there is any such thing as aver- 
age practice; certainly there is no such thing as an average crude oil. For 
example, California oils yield practically no burnable lamp oil, that is, clean, 
smokeless burning, but do yield much asphalt residue and no paraffine and 
practically all the lighter constituents are collected in one lot and called dis- 
tillate, which has some of the properties of gasolene and some of kerosene. Also 
the quantity of any one product usually sold by its specific gravity, or Baume, 
as a means of definition or specification can be varied greatly by mixing lighter 
and heavier constituents and this is a regular practice as can be shown by 
fractional distillation of the product. Densities are always given at 60° F. 
and are corrected for kerosene at other temperatures by adding or subtracting 
.0004 sp.gr. per degree F. 

To illustrate, first, the variations in the products having a given trade 
name, and second, the complexity of product, the several fraction- 



6S8 



ENGINEERING THERMODYNAMICS 



ation tests are given in Table CXII for crude petroleums and kerosenes and 
Table CXIII for gasolenes. These are plotted in curve form, Fig. 178 and 
Fig. 179 respectively, on which are indicated the boiling-points of known hydro- 
carbons and bands added for the class of distillate in accordance with the 
Robinson classification. Horizontal distances represent fractions distilled, 
a fraction being the per cent by volume that has been discharged between 
two given temperatures in a boiling mass, the temperature continually rising. 
Incidentally it may be noted that the temperature is different in the vapor 
than in the boiling liquid, though that of the liquid is usually taken. The rate 
of boiling or application of heat very seriously affects these curves, any one of 
which might easily be changed thereby. 

The distillate between any two temperatures is by no means a simple substance 
nor is one sample the same as another, so that the gasolene collected in original 
manufacture between the same temperatures from different oils will not be the 
same. This is a fundamental characteristic of the boiling of solutions of differ- 
ent substances in each other, some of the vapor of every one will come off at any 
temperature and while at low temperatures the amount of heavy vapors is 
small they are always present and in varying amounts for different oils and 
especially for different ways and rates of boiling that cannot be fully explained 
here. 

For the United States the following, Table LXXVI, gives an estimate of the 
oil characteristics from the different fields, divided into two groups. The 
first represents in output about 15 per cent of the total for the whole country 
and yields from 8 to 12 per cent of gasolene while the second represents about 
85 per cent of the total production, and yields from 1 to 4 per cent of gasolene, 
the calorific power of all varying not over 12 per cent. 

Table LXXVI 
U. S. GASOLENE AND KEROSENE BEARING CRUDE OILS 



Group I (8%- 


12% gasolene) 


GroupIII (1%- 


-4% gasolene) 


Pennsylvania 


. . 40-50 Be. 


Illinois 


. . 30-34 Be. 


Kentucky. . . 


.. 40-44 " 


Kansas 


.. 22-32 " 


Ohio 


.. 37-40 " 


Oklahoma. . . 


. . 22-32 ' ' 


Indiana 


.. 37-40 " 


Louisiana. . . 


.. 22-32 " 






Texas 


.. 16-26 " 






California. . . 


. .. 12-28 " 



The calorific power of distillates can, of course, be estimated from their 
carbon, hydrogen and sulphur contents by formulas of the Dulong type but 
very much closer results will be obtained from the density formula of the type 
derived by Sherman and Kropff. Samples of gasolene and kerosene, tested by 
the U. S. Geological Survey, yielded the following formulas (Eq. (765) and 



HEATING BY COMBUSTION 



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Fig. 179. — Fractional Distillation of Gasolenes. 



HEATING BY COMBUSTION 



691 



Eq. (766) ) under the manipulation of Stanton and Strong for the gasolenes 
and Allen and Strong for the kerosenes, both suggested by Eq. (764). 



B.T.U. per lb. gasolene = 18,320+40(66". -10), 
B.T.U. per lb. kerosene = 18, 440 +40 (Be. -10). 



(765) 
(766) 



The details of these samples, with the measured and calculated calorific powers 
are given in the following Table LXXVII. The calorific power is calculated by 
the general Sherman and Kropff formula and by the Strong modification, show- 
ing how very close the new constant fits the case, the error being always less 
than 1 per cent. 

Table LXXVII 
CALORIFIC POWER OF GASOLENES AND KEROSENES 





Density, Sp. 
Gr. at 60° F. 


B6. 


B.T.U. 


per lb. 


Per Cent 


B.T.U. 


per lb. 














By 

Calorimeter. 


By S. and K. 
Formula. 


Error. 


By Strong 
Constants. 


Error by 
Strong Con- 
stants. 




.7122 


66.6 


20581 


20910 


+ 1.61 


20580 


±.00 




.7165 


65.3 


20477 


20862 


+ 1.88 


20532 


+ .27 




.7168 


65.2 


' 20527 


20858 


+ 1.61 


20528 


±.00 


L 


.7175 


65.1 


20579 


20854 


+ 1.34 


20524 


-.27 


60 

a 


.7285 


62.2 


20401 


20738 


+1.65 


20408 


+ .03 


2$ 


.7285 


62.2 


20410 


20738 


+1.61 


20408 


-.01 


T3*3 


.7289 


62.1 


20407 


20734 


+ 1.60 


20404 


-.02 


3 8 
o 


.7289 


62.1 


20381 


20734 


+ 1.75 


20404 


+ .11 


o 


.7292 


62.0 


20390 


20730 


+ 1.67 


20400 


+ .05 


1 


.7292 


62.0 


20423 


20730 


+ 1.50 


20400 


-.11 


*3 


.7294 


62.0 


20389 


20730 


+ 1.67 


20400 


+ .05 




.7301 


61.8 


20407 


20722 


+ 1.54 


20392 


-.07 




.7912 


46.9 


19922 


20126 


+1.02 


19916 


+ .03 




.7925 


46.7 


19894 


20118 


+1.13 


19908 


+ .07 


a 2 

g 2 | 


.7930 


46.5 


19881 


20110 


+ 1.15 


19900 


+ .10 


=3 £.3 


.7999 


45.0 


19872 


20050 


+ .95 


19840 


-.16 



Residue oils always contain carbon in the free state as suspended solid matter 
and they are heavy and viscous in nature. However, the carbon may be removed 
by filtration, or the density reduced and fluidity increased by adding some 
light constituent, in which case they are very different from distillates or crudes 
of the same specific gravity, a fact that renders their use very difficult, as oils 
are generally bought on a specific gravity or Baume specification. When 
the free carbon is not removed, the oils will burn very much more smoky than 
otherwise and such oils cannot be used in some oil engines at all, because of carbon 



692 ENGINEERING THERMODYNAMICS 

deposits, when the same engine can successfully use another oil of the same 
density and perhaps same calorific power. Specifications based on temperature 
fractions would avoid these difficulties. 

Mineral oils when heated, not in a liquid mass but by slow admission to a 
hot chamber the temperature of which is above the boiling-point of some or 
all of their constituents, undergo a decomposition which may be called destructive 
distillation and is in many ways similar to the roasting of coal. It will yield 
a large quantity of permanent gases if the temperature be high enough and 
mixed with them a series of hydrocarbon vapors, mainly benzole if the tem- 
perature is that of good red heat, and there will be left behind some heavy liquid 
residue and coke. Such gas after washing and cooling is called oil gas and as 
made by the Pintsch process is the main illuminating medium of over 90 per 
cent of all American steam railroad cars. Usually the oil used for this purpose 
is the grade between the lamp oil or kerosene and the lubricating oil, but there 
is considerable variation in practice and price, the latter largely controlling 
practice in the choice of raw material. Too low a retort temperature gives poor 
gasification, and tar and condensible vapors, in excess; too high a temperature 
gives fixed carbon, reduced illuminants and increased hydrogen and each oil 
must be treated a little differently from the others. The gas consists mainly 
of hydrocarbons and hydrogen, the hydrocarbons being much the same as those 
from coal gas J)ut in different proportions, and not usually separated from . each 
other by analysts except into saturated and unsaturated groups beyond methane 
and ethylene, but even these as reported are often included with others by the 
methods of the analysis. The hydrocarbons called saturated are those absorbed 
by bromine in potassium bromide or concentrated sulphuric acid, and consist 
mainly of methane, CH4, and ethane, C2H6. The rest called unsaturated, include 
ethylene, C2H4, butylene, C4H8, acetylene, C2H2, benzole, C6H 6 and naphtha- 
lene, CioHs. Analysts frequently assume no ethane present and report only 
methane for the first absorption and call the rest heavy hydrocarbons or illu- 
minants. Sometimes they are assumed to be ethylene or benzole and reported as 
such, but only the most careful chemists will verify results or reject assumptions. 
For this reason it is likely that the hydrocarbon analyses reported in the Table 
LXXVIII for oil gas are quite uncertain, but less so, than the hydrocarbons 
equivalent to kerosene and gasolene. A recent investigation of Pintsch oil 
gas at Columbia University showed it to consist of 63.1 per cent CH4, 5.6 per 
cent H 2 , 27.4 per cent illuminants, 0.8 per cent 2 , 0.4 per cent CO, and 27 per 
cent ^2- Blau gas is the name applied to the condensible constituents of the 
distillation rejected in the Pintsch process, and as obtained by very high com- 
pression they include only those that freely revaporize on release of pressure, 
From calorimeter results the nature of the illuminating hydrocarbons in 
the above-mentioned Pintsch gas was found to be represented by carbon 89 
per cent and hydrogen 11 per cent by weight which does not correspond to any 
one, but might be made up by mixtures of several hydrocarbons. In a somewhat 
similar way, the general conclusion has been reached that gasolene is most 
nearly represented by the simple hydrocarbon hexane, and kerosene by decane, 



HEATING BY COMBUSTION 



693 



Table LXXVIII 
PROPERTIES OF OIL GAS 



No. 



Description. 



Volumetric Analysis. 



CH< 



§5 CO. 



COa. 



At 32° F. and 29.92" Hg Pressure. 



Lbs. 
Cu.Ft. 



Cu.Ft 
per Lb 



B.T.U. per 
Cu. Ft. 



High. 



Low. 



B.T.U. per 
Lb. 



High. 



Low. 



Thwaite oil gas. . . . 

Pintsch American oil 

Pintsch American oil 

Oil gas 

Pintsch gas from 
petroleum residue 

Pintsch gas from 
paraffine oil 

American petroleum 
oil gas 

Pintsch gas, Moore- 
head 

General 48 

Crude oil Retort gas,! 

England 35 

English shale oil cas, 
Young and Bell. . . 19 



63.19 
63.1 
(il .2 
58.3 

58 . 

54 . 9 
53.7 
52.5 



31.61 
5.6 
6.4 

24.3 

24.3 
5.6 

4.8 

18.5 
32. 

6.6 

16.85 



27.4 
28.3 
17.4 

17. 

28.9 

41.2 

23.5 
16.5 

49.4 

44.83 



8.9 

.2 

1.0 

1.5 
.63 



5.06 



3.5 
3.0 



1.15 



.03427 
.05142 
.05109 
.04313 

.04081 

.0591 

.05726 

.04777 
.04318 

.05972 

.04670 



29.18 
19.45 
19.6 
23.2 

24.5 

16.92 

17.46 

17.32 
23.16 

16. 750 ; 

21.41 



893.5 
1173. 
1260.7 

995.9 

990.: 

1126.8 

1294.8 

1157.5 
901.3 

1390.7 

1043.1 



818.0 
1074. 
1064. 

803 . 9 



1034.8 
1192.0 



966.5 
716 



1107 
966.0 



26072 
22815 
24710 
23096 

24260 

19065 

22607 

20060 
20874 

23282 

22333 



238()9 
20880 
20854 
18650 

22000 

17509 

20812 

16940 
16583 

18542 

20682 





































































































































































































































o 














1300 
























































o 


*x 




l/6 
















1200 




















i? 
































O 










X 
















to 

-j 1100 

o 

u 


























o 


*"* 




























X 
X 


- \<& 


i^^ 


















£>.100Q 

















1 


fi> 


sT^£- 






X 


















c 


) 






~5s£ 


























900 


























x Low Value 
o High Value 










o 






































F.nn 










■x 































.034 



,038 



,042 ,046 ,050 ,054 

Density, Lbs. per Cu. Ft. 



,058 ,062 

Fig. 180.— Calorific Power of Oil Gas. 



694 



ENGINEEEING THERMODYNAMICS 



but this is hardly better than the roughest sort of guess, only really needed 
when the volume of vapor per pound is to be estimated by the molecular weight 
method, the only available means. 

The calorific power of these oil gases is plotted to a density base in Fig. 
180, which shows a straight line relation with density somewhat similar to that 
found for liquid oils. There are very little data available on the production 
or yield of oil gas except that of a confidential character, because its manu- 
facture is not very general, but the following figures (Table LXXIX) quoted 
by Giildner from the Gas Journal are useful as comparative if not as absolute 
data: 

Table LXXIX 

YIELD OF RETORT OIL GAS 



Oil Used. 


B-naphtha 
Sp.gr. .730. 


Kerosene. 
Sp.gr. .807. 


Heavy Oil, Sp.gr. .847. 


Heavy Oil, Sp.gr. .884. 


Retort temp. F. . . . 


1112 


1562 


1112 


1562 


1112 


1472 


2012 


932 


1112 


1562 


Cu. inches gas per 






















cubic inch oil . . . 


451 


625 


469 


582 


401 


513 


594 


213 


368 


657 


Cu.ft. gas per lb. . . 


10.4 


13.8 


9.3 


11.0 


7.6 


9.7 


11.3 


3.7 


6.7 


12.3 


Residue % weight . 


11.4 


5.1 


21.4 


7.5 


28.5 


12.2 


18.0 


62.3 


41.5 


9.4 



The calculation of the calorific power of oil gas with the assumption of 
nature of heavy hydrocarbons as half ethylene and half benzene may be laid 
out conveniently in tabular form as for natural and coal gas, and for this a 
typical analysis is given in Table LXXX. 

Table LXXX 
DENSITY AND CALORIFIC POWER OF OIL GAS 

From Constituent (32°. F. and 29.92" Hg) 





One Cubic Foot 




Constituents of Average 
Pintsch Oil Gas. 


Contains 


Yields B.T.U. 


Summary 


Cu.Ft. 


Pounds. 


High. 


Low. 


Methane, CH 4 

Hvdroeen. H<>. . . 


.525 

.185 

.117 
.118 
.010 
.005 
.005 


.02347 
.00104 

.00930 

.02588 
.00078 
. 00061 
.00045 


559.6 

63.0 

198.9 

465.1 

3.4 


504.5 

54.2 

186.6 

447.8 

3.4 


B.T.U. per cu.ft. 

gas high 1290 

B.T.U. per cu.ft. 

gas low 1196.5 

Cu.ft. per lb. of gas 16.25 
Lbs. per cu.ft. of gas . 06153 

B.T.U. per lb. high 20,962 
B.T.U. per lb. low 19,443 


Heavy / C 2 F 4 . . 
Hydrocarbons I C 6 H 6 . . 
Carbon monoxide. CO. . 
Carbon dioxide, C0 2 . . . 
Oxygen, 2 


Total for gas 


.963 


.06153 


1290.0 


1196.5 





HEATING BY COMBUSTION 



695 



The cooling of the distilling gases from either coal or oil in retorts produces 
a liquid generally termed tar, the relative character of which for the two cases 
is shown by the following ultimate analyses : 

TAR ANALYSES 



Tar 


Per Cent by Weight 


from 


C 


H 2 


N 2 


o 2 


Ash 


S 


B.T.U. per Lb. 


Coal 


89.21 
92.70 


4.95 
6.13 


1.05 
.11 


4.20 
.11 


.06 
.05 


.53 
.33 


15400 


Oil 


17300 







7. Gasification of Fixed Carbon and Coke by Air Blast Reactions, Pro- 
ducing Air Gas, and Blast Furnace Gas. Comparative Yield per Pound of 
Coke and Air. Sensible Heat and Heat of Combustion of Gas. Relation of 
Constituents in Gas. Efficiency of Gasification. Carbon when hot enough 
may react with oxygen to form carbon monoxide or carbon dioxide and this 
is the basic principle for the gasification of carbon with the oxygen of air when 
the process is so controlled as to make a maximum of carbon monoxide, the 
product being called air gas. The principal source of air gas is the blast furnace 
in which the fuel is usually coke but may be a hard coal, though the process 
as there carried out is not controlled with a view to getting high carbon mon- 
oxide content primarily, but rather to best permit of the ore reduction to metal. 
However, the conditions are generally such^as to yield an air gas almost as good 
as if its production were the end in view. 

This gasification process is characterized by a greater precision of relation 
between conditions and results than any dealing with hydrocarbons whether 
derived from coal volatile or from oils, and, therefore, it yields better to pre- 
diction of results though not so well as might be expected at first glance. 
While from the fundamental chemical reactions the relative weights of carbon, 
oxygen and gas can be set down exactly for a complete reaction, there are certain 
gaps to be filled by estimation, in attempting to predict a gas made from a coal 
or a coke even when its composition is known. In the first place the three 
substances, CO, CO2 and C are known to be in equilibrium in all proportions at 
some temperature, the ratio of CO to CO2 in the presence of an excess of C depend- 
ing on the temperature. Therefore, in any fire where carbon is a fuel there will 
always be some CO2 and some CO but no one can say how much of each even 
though the equilibrium proportions are known for all temperatures, because 
the temperature is a resultant and not an imposed condition, and to attain equilib- 
rium requires sufficient time of contact and this is always unknown. No real 
fuel is all fixed carbon and seldom is fixed carbon in the same state, so that 
dense carbons like retort coke require longer time or higher temperatures for 
reaction than charcoals, while coals containing other constituents than carbon, 



696 ENGINEERING THERMODYNAMICS 

such as nitrogen, oxygen, hydrogen, both free and combined in all sorts of 
ways, further complicates the problem. Cokes even when they yield no gases 
on high heating will do so when by partial combustion the cell walls are broken 
down and in such cases hydrocarbons will be added to the air gas even in the 
late stages of coke combustion. 

Accordingly, air gas while consisting mainly of carbon monoxide will also 
contain some carbon dioxide, some hydrogen and possibly some methane, all 
besides the nitrogen derived from the air. Presence of oxygen in air gas is an 
indication of air addition after gas formation is complete and at a point where 
the gas has cooled below the ignition point of the combustible constituents. 
Air mixture with the gas in hot regions always results in the burning of some 
of the gas produced previously and is one cause of higher carbon dioxide content 
than the conditions of CO-CO2-C equilibrium warrant at the temperature. 

Air gas as usually made is merely the product of blasting air into a thick 
bed of coal, coke or charcoal, the depth of carbon being maintained at from 
2 to 6 ft. to insure sufficient time of contact between gases and carbon to com- 
plete the reaction. The necessary depth depends partly on the blast pressure 
which determines the flow velocity in a bed of given porosity and partly on the 
sort or condition of the fixed carbon fuel, being least for charcoal. Air gas is 
seldom made for itself, but as a part of other processes, for example, the blast 
furnace yields air gas as a by-product of iron making, called in this case blast- 
furnace gas, also water-gas making, to be described presently, requires a heated 
bed usually so prepared by a preliminary making of air gas especially in the 
Lowe system of carburetted water-gas manufacture, where the air gas serves 
a double purpose, as it is burned to heat the carburetor after, by its formation, 
it has heated the carbon bed to a temperature suitable for water-gas making. 
Probably the nearest approach to straight air-gas making with a view to secur- 
ing it for a gas engine fuel is that of the Tait producer, though in all gas producers 
some air gas is made to be mixed with water gas separately or simultaneously 
as made. 

To show the characteristics of the composition of some air gases Table 
CXIV at the end of the Chapter has been compiled. These analyses show 
that the composition varies considerably for reasons just given, the carbon 
monoxide ranging from about 18 to 35 per cent, carbon dioxide from 4 to 16 
per cent, hydrogen from less than 1 to over 12 per cent, methane, the only 
hydrocarbon, from zero to 5 per cent, being least for cokes and greatest for coals. 

It is of value to examine some of these results in the light of such fundamental 
relations as are available, to indicate what a perfect air gas should be, the 
corresponding yield per pound of fuel and the efficiency of the transformation, 
this latter being defined as the ratio of the heat of combustion of the gas formed, 
to that of the fuel from which it came, each burning to the same final products of 
C0 2 and H 2 0. 

The first step is to establish such relations as may be possible between the 
CO and CO2 existing together and two investigations are available for this 
purpose. First, that of Boudouard, who established the equilibrium ratio as a 



HEATING BY COMBUSTION 



697 



function of temperature and later determined the time factors in attaining equi- 
librium with carbon in three different states, which latter relations were also 
studied by the second authority, Clement. 

By leaving amorphous carbon in contact with CO2 for very long times 
Boudouard found on analyzing the gases resulting, that, as might be expected, 
the relation of CO to CO2 became finally constant for any one temperature 
and as temperature rose the per cent of CO increased. His numbers are tabulated 
below, Table LXXXI, as calculated from the derived formula by Dowson and 

CO 
Larter, together with the ratio ~~which may be termed the carbon oxide ratio 

CO2 

for want of a better name. These numbers are also plotted to coordinates of 

CO CO 

and temperatures, and the corresponding values of ~^ , ^^ by volume, 



C0 2 

are added as an extra scale in Fig. 181. 



CO+COs 



Table LXXXI 
BOUDOUARD'S EQUILIBRIUM RELATIONS CO AND C0 2 WITH TEMPERATURE 





Per Cent by Vol. 








Per Cent by Vol. 






Temp. 




CO 

CO2 


CO 


Temp. 
F. 




CO 

co* 


CO 


F. 




CO+C02 






CO+CO2 




CO2 


CO 








C0 2 


CO 






835 


98 


2 


.0204 


.02 


1407 


20.0 


80.0 


4.00 


.80 


923 


95 


5 


.052 


.05 


1447 


15 


85 


5.68 


.85 


1000 


90 


10 


.11 


.10 


1497 


10 


90 


9.00 


.90 


1090 


80 


20 


.25 


.20 


1533 


7.5 


92.5 


12.34 


.925 


1153 


70 


30 


.43 


.30 


1582 


5 


95 


19.00 


.95 


1204 


60 


40 


.67 


.49 


1610 


4 


96 


24.00 


.96 


1252 


50 


50 


1.00 


.50 


1646 


3 


97 


32.33 


.97 


1297 


40 


60 


1.50 


.60 


1697 


2 


98 


49.00 


.98 


1348 


30 


70 


2.33 


.70 


1790 


1 


99 


99.00 


.99 


1376 


25 


75 


3.00 


.75 


1886 


0.5 


99.5 


199.00 


.995 



To attain these per cents of CO at such temperatures, which must exceed 
1800° F. to exceed 99 per cent CO, requires a longtime, and longer for some 
forms of carbon than others and longer than is available in air-gas producers 
or boiler fires to a still less degree. Moreover, the process in these cases is some- 
what different for instead of making all CO2 to be reduced at a fixed temperature 
to CO, air is blasted into a bed, the temperatures are whatever result from the reac- 
tion and the reaction such that both CO and CO2 are produced at once, such 
CO2 as is formed later reducing, and the CO2 arising at first partly from direct 
burning of C with O and partly by burning of CO formed at one spot by O that 
escaped being used up at another. However, the time factor is a most important 
one, and operates in such a way as to prevent the attainment of as much CO 
as is indicated in the preceding table and curve at the given temperature, or if 



69S 



ENGINEERING THERMODYNAMICS 



1900 



1700 



1300 



1100 



20 

o 

O40 
!°60 



100 



















































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































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re 


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*«■ 


■ — 








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■ — 


4- 


























































































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20 






















































































































































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=L.0 



Fig. 181. — Boudouard's Equilibrium Relations for CO and C0 2 with Temperature. 



HEATING BY. COMBUSTION 



699 



a high amount is to be attained in a limited time the temperature must be higher. 
Therefore, if a curve be plotted for CO/CO2 with temperature, for less time 
of contact than is necessary for equilibrium it will be above the equilibrium 
curve and such data have been given by both Boudouard and Clement, repro- 
duced in Table CXV at the end of the Chapter. It must be observed that the 
time element depends on the mode of experiment, a fact that makes the problem 
more complex especially when comparing results obtained from gases at rest 
in a closed tube with another through which a current passes. This latter is the 
more favorable as the gases sweep the carbon, and fresh CO2 can reach it by 
mechanical pushing away of the CO next to it, while in the former case displace- 
ment can only take place by diffusion. Boudouard used the closed tube and 
Clement the open tube with current passing through; the former knew accu- 
rately the time of contact while the latter could only estimate it from the volume 
of flow and temperature, a very uncertain method. 

These results are plotted in two ways in Fig. 182 and Fig. 183, to bring 

CO . 
out clearly the relations. In Fig. 182 the ratio t^t is plotted as a function of 

time as derived, while in Fig. 183 as a function of temperature and in each case 
the equilibrium line is added as a reference. 

It is clear from these data how strong an influence the time and form of carbon 
exert, but in no case do the conditions correctly represent the gas producer as 
well as another series by Boudouard, who passed air through a tube packed with 
coke in 5 mm. pieces, and held at a temperature of 1472° F. while the rate of 
flow was varied with the following results, Table LXXXII, the units of which 
make them unfortunately not comparable with others. 

Table LXXXII 
CHANGE OF 2 IN AIR TO CO AND C0 2 AT 1472° F. (Boudouard) 



Vel. air, in 

Inches per 

Minute. 


Per 


Cent by Volume. 


CO 

C02 


CO 


Per Cent of Carbon Gasi- 
fied in 


C0 2 


CO 


CO2+CO 


CO+CO2 


C0 2 


CO 


6.10 


18.20 


5.20 


23.40 


.28 


.222 


77.8 


22.2 


16.47 


18.43 


3.80 


22.23 


.205 


.171 


82.9 


17.1 


79.31 


18.92 


1.88 


20.80 


.0995 


.090 


91.0 


9.0 


89.36 


19.90 


1.83 


21.73 


.0915 


.084 


91.6 


8.4 


195.20 


19.40 


.93 


20.33 


.0478 


.046 


95.4 


4.6 



These results show that decreasing the time of contact decreases the total 
carbon gasified and increases the CO2 content of the gases, indicating that a 
reduction to CO of some CO2 previously made is the order of reaction and that 
this is more complete the greater the time available but it is quite likely that 
higher temperatures would materially change the result. 



700 



ENGINEERING THERMODYNAMICS 



f^Oft 


-20p 






Equi 


ibjrium 






I 


1 








































— il— 




, 


-6 








































-3 






































60 


-2 












Cok 


e 
































T 


3mp. 


16 


52 




















A0 


-.8 






































-.0 








































-.:; 








































-2 




































6 






a— 





































> 








40 








80 








120 








don 


20 


) 




Ec 


ui 


lib 


ni 


m 


____ 






















80 


-10 

I s 


f 


l—l 








































































-2 

It 




















































CI 


mi 


X'C 


al 


5 


m 


m 












40 


-.8 
















T 


in 


P- 


18 


32 














-.5 






































J20 


.3 






































,2 




































5 












































Equilibrium 




Anthracite 



t£ 



mp. 



2192 




M 


o 




































igo 


20 
















Equi 

^-f — 


lib 


rii 


in 


-~s 


-\ 












10 

-c 


-J 


lift- 


ill 


bL 


1U1 


n- 


























o 
o 


O 
+80 

O 

o 


-G 








































-3 






































-3 








































-a 

i 


7 


/ 


































-2 










































































1 
1 

3 so 
o 




/ 




















































Cl 


ai 


cc 


al 


5 


m 


DJ 










^8 


/ 




















C 


ok 


e 












i 
















T 


sir 


P- 


16 


52 














f 


















T( 


nu 


P. 


21 


J2 




























































































































3 








































9 





















































































U00 


) 
=20,0 






20 

! 








40 








c 











100 
80 
60 
40 
.20 


3 20 





5 








1 









l 


5 




so 


-0 




Ecjjui 


libjrium 


? 
























10 
"6 


I 








UI 


1 — 




























-3 






























n 








-3 








































60 


:2 

1,. 

1 
.8 

1 


. o, 




K 
































-2 








































































/ 






































dO 


- 
















































C 


ok 


e 




























ci 


iai 


cc 


al 


5 


mm 
























T< 


>1; 


t. 


20 


12 




































"Temp. 


1562| 










I ! 










































-2 




































2 








































8 



















































































100 



--op— 




1 




1 1 1 


1 1 























E< 


uilibriu'm 


72 


















3 

1 • 






















































































Q 






























-& 


















; 














r, 




/) 










(Tin 


ireo; 


i 


b 


11! 


.1. 












7 
















Temp 1 . 1 


17 


2 



























































































i)0 100 

Tinie in Seconds 



100 


»200 






2 


"> 








E 











7 


5 












lol 






■ •'■" ui librium 




















- 




80 


-6 








































-3 






























j 


■ — 










60 


=2 

-l,i 


















i 






























~j ^ 


































40 


-.8 


















c 


Ols 


c 




















-5 




/ 












T( 


•11] 


P. 


1832 


















20 


-3 










































? 


/ 




































7 















































±0 80 120 

Time in Seconds 



Fia. 182.— Rate of Reduction of C0 2 to CO with Time of Contact with C at Various Fixed 

Temperaturei. 






HEATING BY COMBUSTION 



701 



5 
2200 



1800 
1000 
1100 









GO Seconds / 


> 


Coke ^^ 


^^--^" n 


__ — f— " Equilibrium y \ 


— cL,cji_^</ 


, ., , ., .. .^r- s jMtT T 



£ 
2200 
2000 



25 50 75 100 



1800 
1000 
1100 

























-p 




































































•> 


^Se 


■o 


nils 
















^ 






































x 
























e 


ok 


e - 












































































h'ljui 


ibiii 


in 




























Cbai(t 












71 






.8 




.5 




.s 






t 


^- 


-1 t 




s? 


L* 


•,o 




L 




; 




d 



1000 
1400 

2200 
2000 
1800 
1000 



( 


) 








25 








50 








75 








J 


X) 














































2200 
























































-\ 


)Se 


h'° 


ml? 




















2000 
































































C 


ok 


PL'^ 


















1800 








































/\ 




























E 1 


i" 


lit 


ru 


m 




/ 






1600 
























c 


lavcoal-L^ 




































_L, 








W 


-21 




1+00 




• L 
J 


" 


2 


'i' 




.0 




.» 


J 






jn-2~ 


_J_ 






1 



Jl— 1100 





1 








25 








50 








75 








100 










































7 










































/ 


















Hi 


> Se< 


■o 


ncls 














, 


y 






































■»" 


^" 


"1 






2000 


















c 


ok 


e- 










A 


ithraci 


e 




1800 
































1 
































r^i 


Ul 


ibi luinL 








1000 




















01 


larcoal-L-J^ 
















2 

L 


1" 




-j 












2- 












H00 










.. 


3" 


•■> 


! 










~ U 


° 













25 








50 








75 








100 
































1 








2200 




























A 


nt 


iracite 




/ 


















2( 


sec 


O 


n< 


S 












































































c 


oUo- 








































































































/ 


/ 






1000 
1100 






















c 


ia 


CO 




















7l 




2" 


r3 




■*f 










^ 


f 




-s 


t 




r 


Lj 


20 


° 



2200 
2000 
1800 
1G00 
1400 



U) Seconds 






pqMlibi 

PHP 



2200 
2000 
1800 
1000 
1400 

2200 
2000 
1800 
1000 



L— 1400 



2200 
2000 
1800 
1000 
1400 

2200 
2000 



120 Seconds 



Eq i ilibii|U 



25 50 75 



LLOSeconds 



Equilibrium, ' 



~Z. 




50 75 100) 



101! 



Seconds 



Charcoal 



l]q hi 



-^T 



— 6^f0 2( 



25 50 75 100 



90 Seconds 



Coke, 



Equilibrium 



\± 



50 75 100 



80 Seconds 



t=* 



25 50 



% 



25 50 75 100 

Upper Scale =-52_ Lower Scaler ^0 

CO2 CO-rCOs 



0f200 

hr- 1400 







TQSeconcs / 


/ 




Coke^ ^.~~'~ /J 


„ ■— -" Equyibyinm / / 


Chaj-co l al-^>^/ 


— rl — 2+3 — r5 — -8 -1= -1-JtT =2" -3 J> ffr "200 



25 50 75 100 

Upper Scale=-@g- Lower Scale — c ff C( ^ 



g. 183. — Rate of Reduction of C0 2 to CO in Contact with C with Temperature at Various 

Periods of Contact Time. 



702 



ENGINEERING THERMODYNAMICS 



In using data on equilibrium it must be assumed that sufficient time has 
elapsed to allow of equilibrium being established and this will always be the 
case in producers except when blasted too vigorously, that is, overloaded. Thus, 
in the case of the Brymbo blast furnace the gas of which is reported by Allen, 
the attainment of a carbon oxide ratio of 6.0 indicates that the bed temperature 
must have reached an average value of 1460° F. from the curve if, (a) no other 
CO2 was formed either by combustion or other chemical reaction beyond the 
highest temperature zone which is always the case with blast furnaces, (6) the 
flow through the bed and the bed itself were homogeneous, which is never the 
case, the bed being usually divisible into chimneys of high temperature and easy 
path of gases which always flow where the resistance is least, and if (c) the time 
were sufficient to attain equilibrium. 

That there should be any difference at all in the temperatures of beds of fuel 
blasted with air and all excessively thick is due to cooling influences, external and 
internal as follows : (a) dry air is seldom used or available and frequently steam 
is added which on decomposition absorbs heat, (6) air and steam blasts are sup- 
plied at different temperatures and as the reactions fix only the rise of tem- 
perature, the temperature in the bed rises directly with initial mixture tempera- 
ture, other things being equal; (c) walls absorb, transmit and discharge heat 
to water jackets or radiate it to the air; (d) in blast furnaces chemical reac- 
tions are absorbing heat at various points of the path, and giving off gases from 
ore and flux changes, otherwise disturbing the purely combustion reaction. 
Of these influences the strongest is the hydrogen decomposition and reaction 
with carbon, which produces what is termed a water gas, the characteristics of 
which will be examined after establishing the quantitative fundamental rela- 
tions for air gas. 

The two fundamental relations fixing the weight and volume proportion! 
and heats for air-gas making are given by Eqs. (767) and (768). 



C+0 2 = C0 2 + (14,544 X12)B.T.U. 
2C+0 2 = 200+2(435.1 X 12)B.T.U. 



(767) 
(768) 



Accordingly for oxygen reacting with carbon, and measuring volumes at 32° F. 
and 29.92 ins. Hg, the proportions being the same at any other pressure, 



1 lb. C+ 



1 lb. C+ 



5? = 2.66 lbs. 




r 44 1 
||=3.66 lbs. 




or 
5 = 29.8 cu.ft. 


►0 2 = 


or 
~ = 29.8 cu.ft. 


► C0 2 +14,544B.T.U. 

(769) 


i| = 1.33 lbs. 




j| = 2.33 lbs. 




or 

358 i . „ ,. 
-7^ = 14.9 cu.tt. 
24 


2 = < 


or 

^ = 29.8 cu.ft, 

I 24 j 


CO+4351 B.T.U. 

(770) 



HEATING BY COMBUSTION 703 

Wt., 2 = 23.1 per cent; N 2 = 76.9 per cent 



Vol., O 2 = 20.9 per cent; N 2 = 79.1 per cent 



For air reactions taking by 



,, , J, , . 3.33 Xwt. of oxygen 

the nitrogen to be added is i _ „_ v . _ J ° 

{ 3.78 X vol. of oxygen 



and the air involved is 



4.33 Xwt. of oyxgen 
4.78 X vol. of oxygen 



These values substituted in Eqs. (769) and (770) will give the weights, volumes 
and heats of reacting with carbon, Eqs. (771) and (772) applying when the 
product is C0 2 , Eqs. (773) and (774) when it is CO. 



1 lb. C+ 



1 lb. C+ 



1 lb. C+ 



1 lb. C+ 



+ 



+ 



2.66 lbs. 














3.66 lbs. 






or 


2 






or 


C0 2 




_ 29.8 cu.ft. . 

8.857 lbs. 
or 


N 2 


• = • 


+ 


_ 29.8 cu.ft. _ 

8.857 lbs. 
or 


N 2 


+ 14,544 B.T.U. 

(771) 


.112.64 cu.ft.. 


J 






.112.64 cu.ft.. 


- 








3.66 lbs. 






' 11.517 lbs. " 

or 
.142.44 cu.ft.. 


1 
Air | = ■ 

J 


+ 


or 
_ 29.8 cu.ft.. 

" 8.857 lbs. 
or 


C0 2 

N 2 


+ 14,544 B.T.U. 

(772) 






.112.64 cu.ft.. 






1.33 lbs. " 








" 2.331 lbs. " 






or 


o 2 






or 


CO 




. 14.9 cu.ft. _ 
4.429 lbs. " 




. = . 




_ 29.8 cu.ft. _ 
" 4.429 lbs. " 




+4351 B.T.U. (773) 


or 


N 2 




+ 


or 


N 2 




.56.32 cu.ft.. 








.56.32 cu.ft.. 










" 2.33 lbs. 






" 5.759 lbs. " 

or 
.71.22 cu.ft.. 


Air [ = < 
J 


+ 


or 
.29.8 cu.ft. _ 
" 4.429 lbs. " 

or 


CO 

N 2 


+4351 B.T.U. (774) 














.56 


.32 cu.ft.. 






1 



Assuming that 1 lb. of carbon reacting with air produces both C0 2 and CO 
at the same time, 

Let x = fraction of 1 lb. of carbon burning to C0 2 , 
(1 — x) = fraction of 1 lb. of carbon burning to CO. 



704 



ENGINEERING THERMODYNAMICS 



1 lb. C+ 



(775) 



Then if the products of reaction of 1 lb. of carbon mix, the reaction is denned by 
Eq. (775) 

3.66 lbs. 
x or CO2 

_ 29.8 cu.ft. 

] 2.33 lbs. 

or Air +(l — x) or 

142.44 cu.ft.J L 29.8 cu.ft. 

5.759 lbs. 8.857 lbs. 

-h(l-x) or Air + x or 

71.22 cu.ft. J L112.64 cu.ft.. 

" 4.429 lbs. 
+ (l-z) or 

_ 56.32 cu.ft. 
+ [14,544^+4351 (l-x)] 

B.T.U. 

This equation reduces to the two following forms, Eq. (776) giving the weight 
relations and Eq. (777) those for volumes each associated with the heats of 
reaction. 

3.66a; lbs. C0 2 
+ 2.33(1 -s)lbs. CO 
+ 4.428(1 +x) lbs. N 2 
I +(10,193^+4351) B.T.U. J 

29.8x cu.ft. C0 2 
+29.8(1 -x) cu.ft. CO 
+56.32(1 +z)cu.ft. N 2 
+ (10,193^+4351) B.T.U. J 

The final gas will have the following composition and carbon monoxid- 
dioxide ratio by volumes (Tables LXXXIII and LXXXIV). It should be 
noted with respect to the latter that the ratio of carbon monoxide to dioxide by 
volumes in the gas is the same as the ratio of fractional weights of carbon burning 
to each as products. 

Table LXXXIII 
COMPOSITION OF HYPOTHETICAL AIR GAS. GENERAL 



1 lb. C+5.759(l+z) lbs. air 



1 lb. 0+71.22(1+*) cu.ft. air = 



(776) 



(777) 



Constituent. 


By Weight. 


By Volume. 


g^byVol. 


coq^ byVoL 


co 2 = 


3.66X 


29.8s 


1-X 

X 




5.758x +6. 758' 


56. 32s +86. 12 




co = 


2.33(l-z) 
5. 758s +6. 758 


29.8(1 -s) 
56. 32s +86. 12 


1-s 


N 2 = 


4. 428(1 +x) 
5. 758s +6. 758 


56. 32(1 +s) 
56. 32s +86. 12 





HEATING BY COMBUSTION 



705 



If the carbon burns all to CO then 2 = 0; if all to C0 2 then x = l; and the 
composition of the gas for the two cases will be given by Table LXXXIV. 



Table LXXXIV 
COMPOSITION OF HYPOTHETICAL AIR GAS. NO C0 2 , AND NO CO. 





By Weight Per Cent. 


By Volume Per Cent. 


CO 


CO. 








C0 2 


CO +CO2 


Constituent. 


All CO 


All CO.. 


All CO 


All C0 a 


All 
CO 


All 
CO2 


All 
CO 


All 
C0 2 


co 2 





366 -99 2 
12.516 





^ 8 -=20.8 
142.44 


00 





1 




CO 


li??. =34.47 
6.758 





^.=34.6 
86.12 








N 2 


4428 A* *Q 

=65.53 

6.758 


8.856 


5 -^? = 65.4 
86.12 


11264_ 79 




12.516 • 


142.44 





The general relations between the carbon and air as raw materials and the 
gas produced from them are given as follows, Eq. (778) : 



1 lb. carbon makes (5. 758a: +6. 758) lbs. gas (a) 

.758^+6.758 



1 lb. air makes 



(i 



1 lb. gas requires ( 



1 lb. gas requires 



759(l+z) 

5.758z+6.758 
5.759(1+*) 



lbs. gas (b) 

lbs. carbon (c) 
lbs. air (d) 



5.758Z+6.758, 
1 lb. carbon makes (56.32.T+86.12) cu.ft. gas (e) 
56.32z+86.12 x 



/56 

1 lb. air makes ( 5 '.759(1+T i ^ *** (D 

1 cu.ft. gas requires L 6 32a . ■ g 6 12 ) lbs " carbon ^ 

i * • I 5.759(1+*) \ „ . m 

1 cu.ft. gas requires! 56 32x _|_ 86 12 j lbs * air W 



. (778) 



The heats of reaction can be expressed in terms of each of the volumes or 
weights entering, those for the raw materials or for the gas formed, and are set 
down in tabular equation form below, Eq. (779) first for any proportion of CO 
and CO2, next for no CO and finally no CO2. 



706 



ENGINEERING THERMODYNAMICS 
HEATS OF REACTION FOR AIR GAS, B.T.TJ. 



Heat of Reaction. 
General. 



10193a; +4351 



10193a; +4351 
5. 759(1 +x) 



10193a; +43 51 

5. 758a; +6. 758 



10193a; +43 51 

56. 32a; +86. 12 



10193a; +4351 
3.66a; 



10193a; +4351 
2.33(1 -a;) 
10193a; +4351 

4. 428(1 +x) 



10193a; +4351 

29.8a; 



10193a;+4351 

29.8(1 -a;) 



10193a;+4351 
56. 32(1 +x) 



When CO =0 
or x=l. 



14544 



14544 
11.518 1 



1263 



14544 
12.517 



= 1162 



14544 
142.44 



1021 



14544 
3.66 



3974 



14544 
8.856 



1642 



14544 

29.8 



= 488 



14544 
112.64 



= 129 



When C0 2 = 
or x =0. 



4351 



4351 
5.759 


= 755 


4351 
6.759 = 


= 644 


4351 

86.12 = 


= 51 






4351 
2.33" 


= 1867 


4351 

4.428 


= 983 






4351 

29. 8 = 


= 146 



Unit 



435J_ 
56 . 32 



77 



Per Lb. 


carbon 


u 


Lb. 


air 


" 


Lb. 


gas 


" Cu.ft. gas 


k 


Lb. 


C0 2 


u 


Lb. 


CO 


" 


Lb. 


N 2 


a 


Cu.ft. C0 2 


a 


Cu.ft. CO 



Cu.ft. N 2 



(a) 
(b) 

(r) 
(d) 
GO 

(/) 

(g) 

(h) 
(tj 

(./) 



(779) 



These heats of reaction are all positive and act to raise the temperature 
of the bed and the products of the reaction, and the temperature rise could be 
calculated if the specific heat of the products were known, or on any assumption 
for its value. The heat of combustion of the gases formed, which is due to the 
CO, is the useful effect in gas producers and a loss in boiler fires to which the 
preceding relations also apply. The amount of this heat of combustion is, of 
course, the difference between the heat of complete combustion per pound C and 
its heat of reaction when combustible gases are formed by it. Therefore, the 
heat of combustion of the gas will be 14,544— (10,193:r +4351) per pound of 
carbon, from which the value per cubic foot and pound of gas can be found as 
given below in tabular equation form, Eq. (780). 

HEAT OF COMBUSTION OF HYPOTHETICAL AIR GAS, B.T.U. 



General. 


When CO =0 
or x = 1. 


When C0 2 =0 
or x =0. 


Unit 


10193(1 -a;) 





10193 


Per lb. C 


10193(1 -a;) 
5.758s +6. 759 





10193 -1508 
6.759 ~ 1508 


Per lb. gas 


10193(1 -a;) 
56. 32a; +86. 12 





10193 
86.12- 118 


Per cu.ft. gas 



(a) 
00 

(c) 



. (780) 



Dividing the heat of combustion of the gas formed per pound carbon, by 
the heat of combustion of the carbon, will give the efficiency of the producer 
as a fuel transformer from solid carbon to combustible gases. 



HEATING BY COMBUSTION 



707 



( Efficiency of gasification 1 10,193(1 —z) 

I of carbon = J ~ 14 544 



(a) 



70% if no C0 2 is formed (6) 



(781) 



Therefore, the efficiency of this process cannot exceed 70 per cent, the other 30 
per cent of the carbon heat appearing as sensible heat of the gas, some of which 
can be used as will be shown later for the formation and dissociation of steam, 
adding hydrogen thereby to the gas, and raising the efficiency more and more 
above 70 per cent as the hydrogen increases, except as such reactions cool the 
bed and tend to make more CO2 with its corresponding lessening air-gas efficiency 
effect. When the CO2 content is not zero, the heat of combustion of the gas 
and efficiency of the process becomes less and the curves of Fig. 184 are plotted 
to show graphically the relation. On the same sheet are given the gas 
composition, calorific power of the gas and heats of reaction per pound of 
carbon, and per pound of gas. The heat of reaction per pound of carbon is a 
measure of efficiency, and per pound of gas a measure of the temperature of 
the gas when specific heats are known. 

Air-gas making involves high temperatures, which cannot be calculated, 
and these are really so high as to make it an impracticable process for most 
fuels because of the fusibility of the ash and formation of clinkers large enough 
to stop the flow of blast. The blast furnace, fluxing and melting everything 
it contains to slag and iron does not suffer in this way and it is not impossible 
that fluxing of ash and removal as liquid slag may be practical for gas making in 
producers some day. At the present time steam is almost universally introduced 
with the blast of air, and by its decomposition absorbs enough heat to prevent 
serious clinkering of most coals, though there are still some that give trouble. 

The calorific power and density of an air gas can, like others, be calculated 
from the properties of its constituents and such a calculation is arranged in 
tabular form below, Table LXXXV. 

Table LXXXV 

DENSITY AND CALORIFIC POWER OF BLAST FURNACE GAS 

From Constituents (32° F. and 29.92" Hg) 





One Cubic Foot 




Average Composition of 
Blast Furnace Gas. 


Contains 


Yeilds B.T.U. 


Summary. 




Cu. Ft. 


Pounds. 


High. 


Low. 




Carbon monoxide, CO. . 
Hydrogen, H2 


.2861 
.0274 
.0020 
.1139 
.5706 


.022336 
.000143 
.000089 
.013974 
.044672 


97.560 
9.343 
2.132 


97.560 

8.001 
1.918 


B.T.U. cu.ft. gas 
high 109.035 

B.T.U. cu.ft. gas 
low 107.479 

Cu.ft. per lb. gas.. 12.3132 

Lbs. per cu.ft. gas . 081214 

B.T.U. lb. gas 


Methane, CH 4 

Carbon dioxide, C0 2 . . . 
Nitrogen, N 2 




Total for gas 


1.0000 


.081214 


109.035 


107.479 




high 1342 . 57 

B.T.U. lb. gas 
low 1323.41 



708 



ENGINEERING THERMODYNAMICS 



P 

P3 
15000 



12000 



i9000 



1 

®6000 



3000 



,5- 

So" 
o 









































~D 






















































/ 


































































































/ 
































































































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' 
































































































/ 






b 


























































































< 




























































































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/ 






























































































/ 
































































































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u 


(X 




















































































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luu 


1C 


s. 




































































































s 




































































































s 


s 
































H 






H 




































































































































































































































































































































































































































































































































-8 


ot 
























































































































































































ou 


8 
































































































































































































































































































































































































































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per 


Ib.of 


Darbo 


n 














































































































































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of G-as.TB 


TlTI's' 




















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c o 






out 


If 


u 




















































































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Lq— 




































































































fH 






















































k, 




































Pi 


























































































h 


























































































o 
























































2 


































+J 


























































































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8 












































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v Efficiency of 
























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cj 










































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S 


S 




































































































































































































































































































































































































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/ 






























































































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y 






























































































































































































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T 




jr 


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lc 


GO 




























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l-Odo 
















^ 


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f 












S 


o- 


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,1 




•1 


A 







;8 


— ( 










1 


5 








-o 






3 






- 


1 






- 


30 


-2 

1. 




• 


f 


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fr30 


i 


r 


L 

ft: 


Lc 


T 


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0- 




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t 






.90- 


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80 
60 


































































































































































































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*2- 


































































































































































































































































































r 

















































80 



Fig. 184. — Characteristics of Air-gas. 



HEATING BY COMBUSTION 709 

Example 1. A sample of air gas showed a ratio of CO to C0 2 of 5.66. What was 
its composition by weight and volume, pounds of carbon and air supplied per cubic foot 
of gas made, and the heat of combustion per cubic foot of the gas? 

Since 7^7-= =5.66, x=.15. 

CU 2 x 



3.66X.15 
5.758X. 15+6.758 



t C0 2 by weight = g ?1 , Q w 1 g i a >ygo =7.2 per cent. 



2 33 X 85 
CO by weight = 5758xJ5+6758 =26 per eent. 



4 428 Xl 15 
N, by weight = 5758xl5+6758 =66.8 per eent. 

29 8 X 15 
CO, by volume = 56 32x 15+86 12 -4-7 per cent. 

CO by volume = 56 J^* f 8(U2 =26.8 per cent. 



■ . T 56.32xl.15 

N, by volume = 56 32>U5+8(U2 =68.5 per cent. 

1 cu.ft. gas requires ^-i^-^- = .0105 lb. C. 

5 759(1 15) 
1 cu.ft. gas requires 502xl5+86 . 12 =-0705 lb. air. 



The heat of combustion per cubic foot of gas is 



10 ' 193X - 85 -=92B.T.U. 



56.32 X. 15+86.12 

Prob. 1. If the gases Nos. 1 and 11 of Table CXIV were made with a bed 

CO 
temperature of 1400° F., what was the ratio of actual — — per cent to the equilibrium 

Gu 2 

per cent? What would it have been if the temperature had been 1600° F.? 

Prob. 2. In an air gas consisting only of CO and C0 2 , what would be the largest 
amount of CO per cubic foot attainable with a bed temperature of 1200° F.? 

Prob. 3. In a producer blasted with air alone 20 per cent of the carbon burns to 
C0 2 and 80 per cent to CO. How many cubic feet of CO, C0 2 , and N 2 will be formed 
per pound of carbon and how much air will be required? What heat will be developed 
by the reaction per pound C, per pound gases, and by the gas per pound and per cubic 
foot in burning? 

Prob. 4. Should the carbon all burn to CO what would be the composition of the 
gas by weight and volume and how much air would be needed? 



710 



ENGINEERING THERMODYNAMICS 



Prob. 5. How many cubic feet of air would be needed to make (a) 10 lbs. and (6) 
10 cu.ft. of air gas containing 25 per cent CO and 5 per cent C0 2 ? 

Prob. 6. A blast furnace yielded gas containing practically nothing but N 2 , CO, 
and C0 2 , the percentages being 30 per cent for CO and 10 for C0 2 . Assuming the 
blast to have been preheated to a temperature of 300° F. and that the specific heat of 
the products was .25, what was the final temperature? 

Prob. 7. What would be the heat of reaction in the above case per cubic foot and 
pound of gas made and per pound of carbon. What would be the heat of combustion 
of the gas calculated from constituents and from the general equation for the proper 
value of x. 

Prob. 8. Neglecting the H 2 and CH 4 in the gases what was the efficiency of the 
producer in the case of gases Nos. 1, 2, 3, and 9, Table CXIV? 

Prob. 9. What is the calorific power and density of gas No. 14 of Table CXIV? 

8. Gasification of Fixed Carbon, Coke and Coal, Previously Heated, by 
Steam Blast Reactions, Producing Water Gas. Composition and Relation 
of Constituents of Water Gas, Yield per Pound of Steam and Coal. Heat 
of Combustion of Gas and Limitation of Yield by Negative Heat of Reaction. 

Water gas is the term applied to the product obtained by blasting steam into a 
carbon, charcoal, coke or coal bed previously heated and producer gas generally, or 
Dowson gas in England and generator gas in Germany, where the heating of the 
bed by air-gas making, proceeds at the same time as the steam reaction which, 
of course, is endothermic. The fundamental relations of the temperature of 
the bed to the reaction characteristics were studied by Bunte with the results 
given in Table LXXXVI. 



Table LXXXVI. 
WATER GAS CHARACTERISTICS WITH BED TEMPERATURE (Bunte) 







i Composition by Vol. of Water Gas. 






Temp. F. 


% Steam H 2 
Decomposed. 






CO 

co 2 


CO 










CO+C02 






H 2 


CO 


C0 2 






1245 Dull red 


8.8 


65.2 


4.9 


29.8 


.16 


.141 


1396 


25.3 


65.2 


7.8 


27.0 


.29 


.224 


1540 


41.0 


61.9 


15.1 


22.9 


.65 


.397 


1749 


70.2 


53.3 


39.3 


6.8 


5.80 


.853 


1850 


94.0 


48.8 


49.7 


1.5 


33.1 


.972 


1940 


98.0 


50.7 


48.0 


1.3 


36.8 


.975 


2057 (White) 


99.4 


50.9 


48.5 


.6 


80.8 


.988 



These results are plotted in Fig. 185 to a base of carbon oxide ratio, together 
with the Boudouard equilibrium curve, and indicate that in no case was the 
equilibrium even approximately attained, as for example, at 1750° F. Bunte's 
ratio was less than 6, while Boudouard's equilibrium value is about 75 in round 



HEATING BY COMBUSTION 



711 



20 



Percentage Steam Decomposed 
40 60 



so 



loo 




Fig. 185. — Characteristics of Water-gas at Various Bed Temperatures 



712 ENGINEERING THERMODYNAMICS 

numbers. Assuming both experimenters' results to be accurate, this would 
be a measure of the importance of the time element, but there is considerable 
doubt as to the accuracy especially of the temperatures of the bed reported by 
Bunte. There is, however, another explanation based on the reactions of CO 
with steam to form CO2 and H2 according to 

CO+H 2 = C0 2 +H 2 

which would tend to oxidize the CO when formed in the presence of steam, 
and it is reported by Bunte that at the temperature named, 1750, there was 
30 per cent of steam free and still undecomposed. That this is the most prob- 
able explanation is indicated by the lesser discrepancy at low temperatures at 
which the above steam reaction takes place feebly or not at all. For example, 
at Bunte's lowest temperature 1245° F. his ratio is .16, while for equilibrium it 
would be about five times as much as against twelve times at 1750°F. In this 
work of Bunte the bed, previously heated to the temperature desired, must 
have begun to cool at once under the steam blast influence, which is precisely 
what happens in straight water-gas manufacture, an essentially intermittent 
process. Moreover, the cooling cannot be uniform because the resistance to 
flow is not uniform through a fuel bed, so that where the blast can pass most 
easily, there most of it will pass, and a non-homogeneous condition of tempera- 
ture and extent of reactions result. 

Another interesting fact shown by these data is the rapidity with which 
hydrogen forms in considerable volume even at low temperature and with only 
a small fraction of steam decomposed. There is a retardation of CO formation, 
until the temperatures have become considerable, at about 1400° F. the curve 
becomes more nearly horizontal and, for a very considerable temperature 
range the per cent hydrogen remains substantially constant between 65 per 
cent for 1250° F. where there is practically no CO, and 51 per cent at 2060° F. 
where there is nearly 49 per cent of CO. 

Straight water-gas making, which these experiments illustrate, proceeds 
with decreasing temperature, but producer-gas making is a continuous process 
with steady temperatures, enough air oxygen combining with carbon exotherm- 
ally to balance the endothermic heat of the water-gas making, to permit of the 
maintenance of a steady state. 

In Table CXVI at the end of the Chapter are given some analyses of water 
gases which show carbon oxide ratios as high as 17.5 indicating the attainment 
of very high bed temperatures before steam blasting, a moderately slow flow 
through the bed to give the CO2 a chance to reduce, and not an excessive amount 
of free H2O to react with the CO formed. In some cases CH4 is reported, indi- 
cating the use of a coal or an incomplete coke, some oxygen and nitrogen are 
also reported, indicating influx of air, but it is difficult to see how oxygen could 
exist unless its contact with combustible H2, CO or CH4 were delayed till cooling 
occurred. The hydrogen content of these water gases ranges a little above and 



HEATING BY COMBUSTION 



713 



below 50 per cent which is about the same as found by Bunte for all temperatures 
above that at which the formation of CO was appreciable, 1800° F., in which 
case, the volumetric amount is substantially equal for the hydrogen and carbon 
monoxide in Bunte's tests and in some of the analyses reported. When the 
analyses depart from this equality the CO is always less than H2 and as this 
was found by Bunte to be so for too low a temperature, it may be assumed that 
when such analyses are reported, the process had been carried on for too 
long a time, or started with too cold a bed. Usually five minutes is a fair 
average time for water-gas making in a previously heated coke or anthracite 
fuel bed. 

The volumes, weights and heats involved in water gas making are given 
by two fundamental equations of chemical reactions, Eqs. (782) and (783), 
using the low calorific value for hydrogen, and derived from, 

C+2H 2 = C0 2 + 2H 2 + [(14,544X 12)- (51,892X4)] B.T.U., 

C+ H 2 = CO + H 2 + [(4351 X 12)- (51,892X2)] B.T.U. 
Therefore, 

C+2H 2 O = CO 2 +2H 2 -33,040 B.T.U., .... (782) 



C+ H 2 = CO + H 2 -51,572 B.T.U. 
These may be interpreted as follows: 



(783 



1 lb. C+ 



1 lb. C+ 









~ 3.66 lbs. ~ 


' 


3 lbs. 

or 

_59.6 cu.ft._ 


1 
H 2 | = • 


+ 


or 
.29.8 cu.ft.. 

.333 lb. 

or 


C0 2 
H 2 






_59.6 cu.ft._ 


• > 






~ 2.33 lbs. ~ 


\ 


1.5 lbs. 

or 

_29.8 cu.ft.. 


' 1 
H 2 \ = < 

J 


+ 


or 
_29.8 cu.ft._ 

" .167 1b. 

or 
_29.8 cu.ft.. 


CO 
H 2 



2753 B.T.U. . (784) 



[. -4298 B.T.U. . (785) 



Assuming that 1 lb. of carbon yields both reactions at the same time and 
that x represents the fractional part burning to C0 2 , or (1—x) the correspond- 
ing part that burns to CO, then the double reaction will be represented by Eq. 

(786). 



714 



ENGINEEEING THERMODYNAMICS 



1 lb. C+ 



+ (l-x) 



3 lbs. 


' 


or 


H 2 


.59.6 cu.ft.j 




1.5 lbs. 




or 


H 2 


29.8 cu.ft.. 


k 



3.66 lbs. 

or 
29.8 cu.ft. 

' 2.33 lbs. 
+ (1 — x) or 

29.8 cu.ft. 

.333 1b. 
-{-x or 

.59.6 cu.ft. 

.167 1b. 
+ (1 — x) or 

_29.8 cu.ft. 
-[(2753x+4298(l-:r)] 
B.T.U. 



COs 



CO 



H 2 



H 2 



(786) 



This reduces to the two following forms, Eqs. (787) and (788), the first for 
weights and the second for volumes, all volumes being for standard gas con- 
ditions. 



lib. C+1.5(l+x)lbs. H 2 = 



3.66x lbs. C0 2 
+2.33(1 -x) lbs. CO 
+ .167(l+x) lbs. H 2 . 



+ (1545x-4298) B.T.U. 
(787) 



29.8 x cu.ft. C0 2 
+29.8(1 -x) cu.ft. CO 
+29.8(1 +z) cu.ft. H 2 . 



+ (1545a; -4298) B.T.U. 
(788) 



From these equations the weight, volume, heat and temperature changes can 
be set down in each of the various units. 

The composition of the gas will be for any proportion of CO to C0 2 as given 
below in Table LXXXVII. 

Table LXXXVII 
COMPOSITION OF HYPOTHETICAL WATER GAS. GENERAL 



Constituents. 


By Weight. 


By Volume. 


c^ (Vo1 ^ 


CO 

co+co 2 


co 2 


3.66a: 
1.5x+2.5 


29.8a; 


1-x 

X 




29.8^+59.6 
29.8(1 -x) 

29.8x+59.6 
29.8(l+z) 

29.8z+59.6 




CO 


2.33(1 -x) 
1.5Z+2.5 


1-x 


H 2 


. 167(1 +x) 

1.5x4-2.5 





HEATING BY COMBUSTION 



715 



It appears from this that, like the case of air-gas, the carbon monoxide- 
dioxide ratio by volume is the same as the fractional weight of carbon burning 
to these two constituents respectively. 

When the carbon burns all to CO, * = 0, or all to CO2, 35 = 1, the composition 
of the gas will be that given in Table LXXXVIII. 

Table LXXXVIII 
COMPOSITION OF HYPOTHETICAL WATER GAS. NO C0 2 , AND NO CO 





By Weight 


, Per Cent. 


By Volume 


, Per Cent. 


■a f CO 
Ratio ^ 


Ratio 
CO 


Constituents 






CO+CO2 


All CO 


All CO2 


All CO . 


All CO2 


All 
CO 


All 
C0 2 


All 
CO 


All 
CO2 


co 2 





3_66 = 917 
4 





^-33 
89. 4 "^ 


oc 





1 




CO 


2 -^-93 3 





59.6 _5U 








H 2 


167 

OT fl - 7 


-F-8.3 


59.6" 50 


59.6 

Wa =67 





The equality of volume of CO and H2 when no CO2 is formed is pretty well 
borne out by the test results quoted, at least as perfectly as the equilibrium 
ratio for the temperature permits, when the temperature is high and free 
steam is not in contact with CO in the hot zone, while the 67 per cent by volume 
of hydrogen for no CO formed, is pretty closely approached by Bunte's 65.2 
per cent when there was 4.9 per cent CO at the lowest temperature of 1245° F. 

The general relations between the carbon and steam as raw materials and 
the gas produced are given by Eq. (789). 

1 lb. carbon makes (2.5+1.5*) 

2.5+1.5* 



1 lb. steam makes 



lbs. of gas 
lbs. of gas 



lbs. of carbon (c) 



1.5(1+*) 

lib. gas requires ( 2 5+ \ bx 

1 ik • / 1.5(1+*) \ 

lib. gas requires ( 2 5+1 ^ ) 

1 lb. carbon makes (29.8x+59.6) cu.ft. gas 
.8x+59.6 x 



(a) 
<J>) 



lbs. of steam 



1 lb. steam makes ( * ,., , — f- ) cu.ft. 

V 1.5(1+*) / 



gas 



1 cu.ft. gas requires 
1 cu.ft. gas requires 



1 



29.8*+59.6 

1.5(1+*) ' 
29.82+59.6 



) lbs. of carbon 



(d) 

to 

(/) 

(9) 



lbs. of steam (h) 



(789) 



716 



ENGINEERING THERMODYNAMICS 



The heats of reaction per unit of each quantity entering are obtainable 
by dividing the total heat of reaction by the volume or weight of each substance 
entering, and these are set down in the tabular equation below, Eq. (790). 



HEATS OF REACTION FOR WATER GAS, B.T.U. LOW. 



General Heat of 
Reaction. 


When CO = 


=0, 3=1. 


When CO2 = 


= 0, x=0. 


Unit 


1545s -4298 


-2753 


-4298 


Per Lb. carbon 


1545 -4298s 
1.-5(1+2;) 


2753 
3 


-918 


4298 
1.5 ~ 


-2865 


"Lb. steam 


1545a: -4298 
1.53+2.5 


2753 
4 


-688 


4298 
2.5 ~ 


-1719 


"Lb. gas' made 


1545s -4298 

29.8^+59.6 


2753 

89.4 


-31 


4298 
59.6" 


-72 


" Cu.ft. gas made 


1545s -4298 


2753 
3.66 


-752 




" Lb. C0 2 


3.66s 




1545s -4298 




4298 
2.33" 


-1842 


" Lb. CO 


2.33(1 -s) 




1545s -4298 
.167(1+3) 


2753 
.33 


-8259 


4298 
.167" 


-25788 


" Lb. H 2 


1545s -4298 


2753 

29.8 


-92 




" Cu.ft. C0 2 


29.8s 




1545s -4298 




4298 
29.8" 


-144 


" Cu.ft. CO 


29.8(1 -s) 




1545s -4298 
29.8.(1+3) 


2753 
59.6 


-46 


4298 
29.8" 


-144 


" Cu.ft. H 2 



(a) 
(b) 

(c) 

id) 

(«) 

(/) 

(9) 

(h) 

(i) 

(J) 



(790) 



These heats all go to the lowering of temperature, being negative, and the 
temperature drop can be determined only by assuming an average specific 
heat for the products which are 93 per cent CO and 7 per cent H2 approximately 
by weight in one, and 92 per cent CO2 and 8 per cent H 2 approximately by weight 
in the other limiting case, and at the same time assuming the sensible heat 
capacity and the weight of the large fuel bed, and the weight of gas flow through 
it per minute. This would give an approximation to the rate of temperature 
lowering, which would be of doubtful value. In practical operation the process 
is controlled by observation and usually the bed drops in temperature enough 
in five minutes, from the highest starting temperature practicable to avoid 
serious clinkering, to that at which the process is no longer feasible, and this last 
temperature may be put roughly at 2000° F. and 1700° F. if excessive CO2 
is to be avoided in the gas. 

The pound of carbon for which the total heat of reaction is (1545^—4298) 
B.T.U. could give on combustion 14,544 B.T.U. so that the gas which results 
must yield the difference (14,544+4298- 1545x) B.T.U. (low) per pound of 
carbon with corresponding values in other units, as in tabular Eq. (791). 



HEATING BY COMBUSTION 



717 



HEAT OF COMBUSTION OF HYPOTHETICAL WATER GAS, B,T.U, LOW 



General 


When CO =0. x=l 


When CO 2 =0, x =0 


Unit 


18842 - 1545x 


17297 


18842 


Per lb. C 


18842 -1545* 
2. 5+1. 5s 


iI5i 7 =4342 
4 


S?-™ 


Per lb. gas 


18842-1545* 
29.8z+59.6 


i 7 ^ 7 = 174 
89.4 


S£-w 


Per cu.ft. gas 



(a) 
(&) 
(c) 



• (791) 



Dividing the heat of combustion of the gas per pound of carbon by the heat 
of its complete combustion per pound will give the apparent efficiency of the 
process by Eq. (792), 



E 



/ 18,842- 
V 14,1 



1545a: 



544 / 



(a) 



18,842 
14,544 



= 130 per cent if no CO2 is formed (b) 



. . (792) 



This is, of course, impossible continuously, and results from ignoring the heat 
that must be put into the bed before admitting steam to make the steam reaction 
which is so strongly endothermic, a possibility. Neglecting the heat necessary for 
the making of steam from water as it is possible sometimes to secure it from waste 
sources, the previous making of air gas can be computed in terms of the quantity 
that must be made, or in terms of the amount of carbon used in doing it, to permit 
of a given amount of water-gas production subsequently. 

Thus, for air-gas making, the heat of reaction per pound C is (10, 193a; +4351) 
while for water-gas making it is (1545a: — 4298) For these to equalize 

(Heat of reaction per pound C) X (lbs. C) used for air gas = (heat of reaction 
per pound C) X (lbs. C) used for water gas. 

Therefore 

Heat of air gas react, per lb. C _ Lbs. C gasified to make water gas 
Heat of water gas react, per lb. C Lbs. C gasified to make air gas 



10,193s+4351 \ 

1545a: -4298 / 

4351 



(a) 



4298 



= 1.01 if CO 2 = (b) 



(793) 



Similarly, 



(Heat of reaction per lb. of air gas) X (lbs. air gas) 

= (heat of reaction per lb. water gas) X (lbs. water gas). 



718 



ENGINEEEING THERMODYNAMICS 



Therefore 

Heat of reaction per lb. air gas 
Heat of reaction per lb. water gas 

Or in terms of volumes 

Heat of reaction per cu.ft. air gas 
Heat of reaction per cu.ft. water gas 



Lbs. water gas made _ J544 
Lbs. air gas made 1719 

= .37ifCO 2 = 0. (794) 

_ Cu.ft. water gas made _ 51 
Cu.ft. air gas made 72 

= .70ifCO 2 = 0(795) 



Therefore, assuming no CO2 to be formed, which is the limiting case, 2 cu.ft. 

of water gas can be made for each 3 cu.ft. of air gas previously made, in round 

numbers, or about a third of a pound of water gas for each pound of air gas, or 

about equal weights of carbon must be used for each. If, as in many cases, 

the process is intermittent and the air gas is thrown away, the efficiency would 

130 
be about -^" = 65 per cent for water-gas making alone, and proportionately 

higher as some of the air gas or its heat was put to useful duty. 

The density and calorific power of a water gas can, of course, be calculated 
from its constituents by the general method, and the results for a typical water 
gas are tabulated below, Table LXXXIX. 

Table LXXXIX 

DENSITY AND CALORIFIC POWER OF WATER GAS 

From Constituents (32° F. and 29.92" Hg) 





One Cubic Foot 




Constituents of Average 
Water Gas. 


Contains 


Yields B.T.U. 


Summary 




Cu. Ft. 


Lbs. 


High. 


Low. 




Hydrogen, H2 


.4557 

.4485 

.0441 

.0010 
.0445 
.0050 
.0012 


. 002561 

.035014 

.001971 

.000080 
.005460 
.000446 
.000094 


155.39 

152 . 94 

47.01 

1.7 


133.06 

152.94 

42.29 

1.57 


B.T.U. per cu.ft. 

gas high 357 . 04 

B.T.U. per cu.ft. 

gas low 329 . 80 

Cu.ft. per lb. gas. 21. 917 
Lbs. per cu.ft. gas. . .04562;, 

B.T.U. per lb. high.. 7825 
B.T.U. per lb. low 7228 


Carbon monoxide, CO. . 

Methane, CH 4 

Heavy Hydrocarbons 


Carbon dioxide, C0 2 . . 
Oxygen, O2 


Nitrogen, N2 




Totals 


1.0000 


.045626 


357.04 


329.86 









Example 1. A sample of water gas showed a ratio of CO to (CO+C0 2 ) of .9. What 
was its composition by weight and volume; pounds of carbon and of steam supplied 
per cubic foot of gas made, and the heat of combustion per cubic foot of the gas? Since 
COh-(CO+C0 2 )=1-z, (l-x)=.9, and x=.l. 



C0 2 by weight 



3.66 X.l 



1.5X. 1+2.5 



= 13.8 per cent. 



HEATING BY COMBUSTION 719 

2 33 X 9 
CO by weight x 5x 1+25 = 79.3 per cent. 

1 pfj \y ~l 1 

H 2 by weight 15xl < 2 5 = 6 - 9 P er cent - 

9Q Q y 1 

CO, by volume = 298xl+596 = 4.7 per cent. 

99 8x 9 
CO by volume = g^-l-HHM = 42 ' 8 P er cent - 

29 8 x 1 1 
H, by volume = 298xl+596 = 52 ' 5 P er cent 

1 cuit. gas requires n&xl+BiS = -16 lb. C. 

1 cuit. gas requires -—-■ — _ ' „ - .264 lb. steam. 
29.8 X. 1+59.6 

The heat of combustion per cubic foot of gas is 

Prob. 1. In a water-gas producer the bed temperature was found to be 1800° F. 
According to the results of Bunte's experiments, what would be the composition of 
the gas and the per cent of steam decomposed? How would the CO to C0 2 ratio, com- 
pare with that for the air-gas producer? 

Prob. 2. Lewes gives as an average analysis of water gas the following: H 2 =51.9; 
CO =40.08; CH 4 =.l; C0 2 =4.8; N 2 =3.13. How does the percentage of hydrogen 
as given compare with that found by the general equation in terms of x. 

Prob. 3. How much carbon and how much steam will be required to make 1000 
cu.ft. of gas which is 50 per cent H 2 , 45 per cent CO, and 5 per cent C0 2 by volume, 
and what will be the heat of reaction per pound of carbon and per pound of gas? 

Prob. 4. A producer gives a gas in which the ratio of CO to (CO+C0 2 ) is .8. How 
many cubic feet of gas will be made per pound of C and per pound of steam? 

Prob. 5. What will be the heat of combustion of the gas No. 10 in Table CXVI 
per pound and per cubic foot? Calculate this by the general equation in terms of x, 
and from the analysis. 

Prob. 6. For the same gas what would be the efficiency of the producer? 

9. Gasification of Coals by Steam and Air Blasts, Resulting in Producer 
Gas. Composition and Relation of Constituents of Producer Gas, Yield per 
Pound of Fixed Carbon, Air, and Steam. Modification of Composition by 
Addition of Volatile of Coal. Heat of Combustion of Gas, Sensible Heat, 
and Efficiency of Gasification, Horse-power of Gas Producers. The most 
important of these gasifying processes is the most complex, and as carried out 
in the power-gas producer is as important in gas-power systems using coal fuel 
as is steam generation in the boilers of steam-power systems. The gas producer 
is a continuously operated apparatus, having a brick-lined casing to hold its 
thick fuel bed, and provided with means for blasting the bed with a mixture 



720 ENGINEERING THERMODYNAMICS 

of steam and air in more or less closely regulated proportions. This gives rise 
to a condition of steady state, as to temperature and reactions, which are main- 
tained as long as working conditions permit and the most important of these 
is the physical condition of the bed as to porosity, ash content, homogeneity 
and clinker. Neglecting these physical conditions, it may be assumed, that the 
blast enters all points of the supply part of the bed uniformly, and passes through 
all parts at an equal rate, all parts of the blast remaining in contact with the 
carbon for the same length of time, and that finally that all points in any 
cross-section of the bed at right angles to the path are at an equal temperature. 
These things are not really as assumed, but it is difficult enough to fix any 
relations even with these assumptions and impossible without them except in a 
qualitative manner. 

In such producers there is made a combination of water gas and air gas, 
the ratio depending on the amounts of carbon reacting respectively with the 
oxygen of air and with the oxygen of steam. In all cases the exothermic heat 
of the air-gas process supports and more or less balances the endothermic heat 
of the water-gas process. The gas will thus contain as combustibles from the 
fixed carbon and steam, carbon monoxide, hydrogen, nitrogen, carbon dioxide, 
and some uncombined oxygen, usually very small in amount. Mixed with this 
gas will be the products of the coal volatile, which are most complex in charac- 
ter and amount as might be inferred from the discussion of retort-coal gas and 
coke-oven gas. The nature and amounts of these additional products of the 
volatile of the coal depends, of course, on the coal itself but also to a very 
considerable degree on the manner of treatment in producers and it is in this 
respect only, that different gas producer processes differ one from the other, 
however different they may look or vary in structure. 

Producers may be divided into the following classes with respect to the treat- 
ment of the volatile of the coal. 

1. Up draft. Coal is fed at top and blast at the bottom. Volatile and moist- 
ure roast off at the top layer in the presence of gases from the coke-bed reactions 
and in a temperature equal to that of the gas leaving. 

2. Down draft. Coal and blast are supplied to the top. Volatile and moisture 
distilled at the point of supply are mixed with the air and steam of the blast 
and pass down through the entire bed. Volatile is partly burned and partly 
decomposed in the hot zone. 

3. Combined up and down draft. Blast enters both top and bottom, coal 
is fed at the top and the gas leaves at the bottom. Volatile passes through 
hot zones, not so hot as (2) but hotter than (1) and in the presence of some 
air but less than (2). Volatile is partly burned, partly decomposed and partly 
unchanged. 

4. Combined up and down draft, blast entering only at bottom,. Volatile and 
moisture are roasted off at the top, and rising, are brought around the bed to 
its bottom, entering there with the blast gases, the finished gas leaving at a 
mid point. Usually, this is imperfectly executed, some volatile passing straight 
down to the gas outlet. Volatile is all burned as it enters the bed. 



HEATING BY COMBUSTION 721 

In the first class are to be found all standard forms of anthracite and coke 
producers, while the others are intended for bituminous coals, yielding much 
tar when supplied to the first class, and which cannot be completely separated 
from the gas, clogging mains and interfering with the operation of the engine. 
In no case can a caking or coking coal be properly handled in any standard 
producer because of draft interference due to melting of the coal without 
prohibitively costly attendance to keep the cake broken up and in some cases 
not even then. Even non-caking bituminous coals give some trouble, not only 
because of the tar they yield, but because also of the variability of the gas 
which is a mixture of water gas, air gas and more or less decomposed rich 
hydrocarbon volatile. Even when the ratio of air and water gas is kept con- 
stant, the volatile of the coal distilling off at a variable rate, will seriously 
impair the constancy of the gas quality, because the volatile of bituminous 
coals has about six times the heating power of the gasified carbon. With 
lignites the high moisture is mixed with the volatile, absorbing heat in its 
liberation and in the steam reactions with the other constituents like CO 
and so introducing a further cause of interference in a manner depending 
on its path. 

In the tables at the end of this Chapter, Tables CXVII and CXVIII, are 
collected some producer gas analyses, the first from a wide range of sources to 
illustrate the ranges of composition, while the second are from Fernald, report- 
ing for the U. S. Geological Survey the results of trials of bituminous coals 
and lignites, ranging from high to low grade in a common updraft anthracite 
or coke type of producer to which was added as an auxiliary a tar extractor of 
the centrifugal fan order. In these latter tests the volatile is all reported as 
methane and the calorific power calculated on that basis. For each case the 
carbon oxide ratio has been added, to indicate more clearly the relations of the 
CO, CO2 and H2, derived mainly from the coke bed reaction, the hydro- 
carbons being purely roasting products. Of course, in down draft and mixed 
draft producers some of the volatile has also reacted, part of it in a manner sim- 
ilar to the bed. 

Inspection of the general table, Table CXVII, shows a pretty steady value of 
the carbon-oxide ratio between 4 and 6, except for certain special conditions indi- 
cating that the temperatures and rates of blast feed are about the same in all, 
or that where high rates of combustion are used and time of contact small, the 
temperatures are allowed to rise to compensate. When the ratio departs 
from this, a reason is always available, for example, increase of steam in the blast, 

CO 
produced a regular decrease in the — — ratio according to the Bone and Wheeler 

CO2 

tests on air blasts saturated with water vapor at 60° F. to 80° F. in which the 

ratio changed from 5.2 to 1.2, and the hydrogen, as might be expected, rose 

from 16.60 to 22.65 per cent by volume. The increase of hydrogen with increase 

CO 
of steam and the corresponding reduction of temperature and ™— - ratio, is well 

CO2 



722 ENGINEERING THERMODYNAMICS 

confirmed by all results on Mond gas plants, operated primarily for ammonia 
recovery which requires low temperatures. In these the ratio falls between 
6 and 7, indicating according to Boudouard, temperatures above 1200° F. 
On the other hand reduction of steam permits the bed to attain high temperatures 
and the gas will contain little hydrogen which is indicated by the Thwaite test 
on charcoal where the hydrogen is .2 per cent by volume, which could easily 
have been derived from the moisture in even pretty dry air alone, and the 

CO 

— — - ratio = 42.5, corresponding to a temperature somewhere above 1500° F. 

CO2 

These results, therefore, show effectively how the steam supply controls the bed 

temperature and the gas quality as to CO and H2 content and, therefore, why 

for the constant gas quality needed in power gas, accurate control of bed and 

blast conditions is so necessary. 

In the second table, Table CXVIII, reporting the Geological Survey tests, 

the first remarkable thing noticeable is the regularity with which the carbon 

CO 
oxide ratio — — — decreases with decrease of CO2 and the small value of the ratio 
CO2 

compared to that obtained in other operating tests. This is an indication of very 
low temperature working, for the ratio is as small as .87 and never rises above 
2.4, possibly due to high rates of blasting. It is too much to expect equilib- 
rium to be established in any producer, yet departures from it give rise to inter- 
pretations as to what was happening in one case that did not take place in the 
other. The general average of the ratio of CO to CO2 being so low in the 
U.S.G.S. tests leads to the conclusion that; (a) the producer was driven to higher 
capacity than the average of others leaving less time for establishment of equi- 
librium; (b) more steam was used, resulting in lower temperatures and leaving a 
residue of undecomposed steam to react on the CO that was formed reducing it to 
CO2. There is no relat on, whatever, between the efficiencies reported for these 
tests and any of the fundamental factors that control efficiency so that these 
efficiencies must be considered as in error probably because of the shortness of the 
runs and the irregularity of working reported. Producer efficiencies cannot be 
obtained except by absolutely steady rate of output, which is necessary to a 
steady thermal state in the producers and for long periods of time because of 
the slowness of gasification compared to the body of coal in the producer. In 
these tests the producer could hold about 5000 lbs. of coal in all stages from 
green coal to ash and it was burnedt a rates ranging from 270 lbs. per hour 
for West Virginia No. 12 to 590 lbs. per hour for Texas No. 1, the lengths of 
run variyng from 4| hours for Missouri No. 2 to 43 hours for Alabama No. 2. 
The rate of combustion per hour, therefore, varied from a little over 5 per cent 
to 12 per cent of the weight in the full producer, making possible a very large 
error in judging equality of conditions of the bed before and after the run, for 
such short periods of time. For with the 5 per cent rate a particle of coal 
would be roughly 20 hours in passing through, as it would take this time to 
consume what was under it when it was fired. 



HEATING BY COMBUSTION 



723 



It is desirable to establish such standards of relation between the quantities 
involved as were found by analysis for air gas and water gas, to be applied to 
the producer gas case for the purpose of comparison of results, with possibilities, 
and this may be done by similar methods. From what has been said it is evident 
that when air gas and water gas are made simultaneously from the fixed carbon 
or coke, and mixed with the volatile more or less decomposed, that the final gas 
will depend on (a) the proportion of the air gas to water gas; (b) the volume of com- 
bined air and water gas to the volatile of the coal; (c) the degree to which the volatile 
is decomposed or suffers reactions. The first step, therefore, in a fundamental 
analysis is the establishment of formulas for the mixed gasified carbon gases 
in terms of the quantities of constituents, carbon, air and steam. 

From Eq. (776) of Section 7 for air gas in terms of weights the following 
Eq. (796) is derived by dividing by 5.759(1+2) and rearranging. 



1 lb. air+ 



|_5.759(1+2)J 



lbs. C = 



3.662 



5.759(1 +2) 
2.33(1 -x) 



lbs. C0 2 



5.759(1 +x) 
4.4281 „ AT 
5759J lbs ' N2 
10^193^+4351] 

5.759(1+2) 



lbs. CO 



B.T.U. 



. . (796) 



The corresponding volume relation follows from Eq. (777) and is given by 

Eq. (797). 

29.82 



Ub - air+ [ 5.759(1 +x) ] lbs - C 



5.759(l+z) 
29.8(1-^) 



5.759(1+2) 
56.3 



cu.ft. C0 2 
cu.ft. CO 



!2] 

r 9 J cu - 



ft. Ns 



5.759 
10,1932+4351 ] 
5.759(1+2) J 13 ' 1 -^ 



(797) 



Two similar equations are given below, Eqs. (798) and (799) derived from 
the water-gas relations Eqs. (787) and (788) of Section 8, for S lbs. of steam 
reacting with carbon. 



S lbs. H 2 0+ 



l_1.5(l+2)J 



lbs. C= • 



3.662£ 1 „ 
L5(T+2)J lbs ' 

2.33(1-2)S 

1.5(1+2) 

lbs. H 2 



C0 2 
CO 



7S1 



libs. 



1.5 J 
[ 15452-4298 ] 
^|_ 1-5(1+2) J i5 ' lAJ ' 



(798) 



724 



ENGINEERING THERMODYNAMICS 



S lbs. H 2 0+ 



|_1.5(1+3?)J 



lbs. C = 



+ 



+ 



29.SxS 



1.5(1+3?) 
29.8(1 



cu.ft. C0 2 



1.5(14 
29.8S 



x)Sl 
x) J 



cu.ft. CO 



+S 



, cu.ft. H2 

1.0 J 

1545a?-4298] 



B.T.U. 



(799) 



1.5(1+3?) J 

If 1 lb. of air and S lbs. of steam together react on carbon and the oxygen of 
the air and that of the steam act the same, or in symbols x is the same for both 
reactions, then together they will give a result expressed by Eqs. (800) and (801), 
one in terms of gas volumes and the other weights, both giving the heats of reac- 
tion and both obtained by addition. 
1 lb. air 



r T 1 lb. air 
L+Slbs.HsOj 



+ 



|_\5.759^1.5/l+a?J 



lbs.C 



+ 



+ 



+ 



3.66a; / 1 JT 
(l+3?)\5.759 + 1.5 
2.33(l- a;)/ 1 

(1-NO 

4.428 



lbs. C0 2 



5.759 
.167£ 



5.759 

lbs. N 2 



1.5 



lbs. CO 



f [ 1 lb. air 
L+£lbs. H2OJ 



+ 



lb. air 
lbs. H 2 

5.759^1.5 



-1-1 



lbs. H2 
1.5 

10,193^+4351 

5.759(1 +3?) 

1 

lbs.C 



+ 



+ 



+ 



+ 



+ 



29.83? 



\x_ I 
x)\l 



1 



(l+3?)\5.759 
"29.8(l-a;) 
. (1+*) 
56.32 



1.5 



\5.7l 



759 
cu.ft. N 2 



£(15453?-4298)" 

1.5(1+3?) " 



1 cu.ft. C0 2 

m 



B.T.U. 



(800) 



cu.ft. CO 



5.759 

29.85 
., , cu.ft. Hs 
1.5 

10,1933?+4351 



£(15453?-4298)1 



B.T.U. 



[ ■ (801) 



5.759(l+x) ' 1.5(1+3?) 

These two equations are fundamental to the gasification of the fixed carbon 
part of a coal in a gas producer fed by steam and air together. 

The composition of the producer gas is shown in Table XC, below for i 
general case and for the two limiting cases when CO = and when CO2 = 0. 



HEATING BY COMBUSTION 



725 









CO 

os 


CO 

00 
OS 














co 


CO 

00 

Oi 








00 
OS 




CO 












"o 


OS 
+ 


CM 




b^ 


CM 




CM 


i-H 








> 
>> 


+ 


O 


00 
Os 


+ 


b; 

oi 


+ 


CO 

q 


o 


o 


^ 


» 


OS 


CO 




1—1 


CO 




CO 


CM 






II 






cm' 






CM 




CM* 








a 




cm 












i— 1 










&5 

CM 
CM 
CM 

+ 


CO 






CO 




CO 








A 




CO 






CO 




CO 








* 


-t-> 
>> 


CO 
CO 

+ 


o 


2 


CO 
CO 

+ 


OS 

CO 
b; 


CO 
CO 

+ 


CO 


o 


o 




« 


00 


t^ 




^j 


I> 




l> 












00 






00 




00 












CO 


o 






q 




q 














CO 
b- 

00 

oi 
+ 


CO 




CO 




CO 














CO 




CO 




CO 














t^ 




t^ 




b; 










> 


o 


oi 

CO 

+ 


CO 

b- 

oc 
oi 


oi 

CO 

+ 


oo 


oi 

CO 

+ 


CO 

CO 

q 


8 


- 


© 


PQ 




i> 


Oi 




iO 
OS 




IO 

OS 


CM 










id 


Tt< 




«$i 




Tj" 








a 










1—1 




,-H 












co 


b» 




CO 




CO 








£ 






CO 
IO 
IO 

+ 

iO 

o 


CO 




CO 




CO 










>> 


o 


q 
+ 


CO 


cq 
+ 


OS 

CO 

b; 


q 


CO 

i-H 


8 


1— 1 




« 




TJH 


*; 


^ 




j*i 














b. 




t^ 




b- 




































Tlj 


^ 




rH* 




rl 














OQ 






QQ 






C0I 




CO 
















X' 


IO 




00 


iO 




oo|io 




00 


iO 














os 


fj) 




os 


,-* 




oi\^ 




OS 


1— i 














CM 






CM 






cmI 




CM 
















+ 

CM IOS 




+ 




+ 




+ 


CO 






1 1 


02 r. 


CM 


OS 


CM 


OS 


CM 


OS 






^12 


CO »o 


CO 


io 




CO 


IO 




co 


iO 


1—1 










c£|^ 


^ IrH 

+ 


CO 


b; 




CO 


b- 




CO 


b; 


CO 

q 










_1_ 


10I1O 


IO 


o 




IO 


id 




IO 


id 






0) 

s 






OS 

IO 


+ 




Oi 

iO 


+ 




+ 




+ 


cm" 

II 






^3 




IH 


^13 


1 




^12 


col 

00|iC 

OilrH 


^12 


CM 
CO 

CO 


Oi 
iQ 


^12 


OS 
IO 


2 
CM 


^1 

l u 


1 


>> 

« 






+ 


1 




— 


cmI 


+ 

OS 


iO 


id 


+ 


b. 

id 


CO 

CO 

IO 


1- ' 1 


^-1 


H w 




Oi 




Oi 






OS 






oo 

OS 
OI 


+ 


i 


to 
b. 

»d 
1 


1 

00 

oi 


+ 


■-■ 


to 

b; 

id 




i 


iO 

id 




1 


iO 

»d 


X 

oo 
Oi 
CM 


X 

iO 
1— 1 
















oc 


« 


CM 


00 


H 


00 


H 


1 

00 


H 










os 


+ 




Os 


+ 




OS 


+ 




oi 


+ 














CM 


T " H , 




(N 


i—i 
i 




CM 

i 


i—i 




OI 


i—i 










































CO 






*E3 


*I3 


^12 


^12 


°2|2 


H 






^p 


+ 


+ 

OS 
-J if" 5 


+ 




+ 




+ 


1 


H 












OS 


IOS 




IOS 




IOS 


"* 


,_( 






_^ 




+ 

IOS 


*■* 


IO 
b- 


_ r? 








^ if? 


y 


! 1 


( w 


CO 


J3 

"53 




|iO 


>o 

' 1 


u 


|iO 

i 1 


s 

CO 


IO 


lip 


oo 

CM 


OS 
IO 

id 


liO 

1 J 


CO 

OS 


CO 

q 
ll" 


1 

1—1 


CM 
b- 

iO 

+ 




1 


r^ — i 


1 


H 




— 1 


1 


— 1 


iO 

b. 


00 
CM 

X 


>> 

n 




CO 
q 


+ 




+ 


1? 
1 


+ 




+ 




^ 


+ 

i 




1 


+ 

i 


id 
X 


1 


CO 


















CO 


'— 1 


+ 


CO 




+ 




+ 




+ 


CO 


•o 


CO 
CO 


q 










CO 


CO 




SQ 




CO 




CO 


"~i 


th 


CO 










+ 


cm' 


+ 

1— 1 




+ 




+ 




CM 


































O 


a 
3 






















O 
O 


+ 


§ 
o 


















wl£ 


old 




c 
o 


















.o 


.o 


.o 


O 
























o 


O 


^ 






c3 


c3 


ta 








e. 


) 






C 


) 






pr 


i 






% 






P 


J 


P 


^ 


P 


J 



726 



ENGINEERING THERMODYNAMICS 



The general relations between the various quantities of raw materials and 
products are given by Eq. (802) below. 



1 lb. carbon makes 

1+5+ 



\l+x) ^5.759^1.5/ \l+s/ \5.759^1.5/ ^5.7 



^ + A\ (J_) 

759 1.5/ \1+*/ 



5.759 
1 lb. steam makes 



.32 29.8S-] 
.759 + 1.5 



\5.759^1.5/ \l+aj/ 



cu.ft gas. (a) 



r / 1 \ (_L ,_S_\ "I r / 29.8 \ /_1 _5_\ 56.32 29.85 ~| 

1 + ,S+ Vl+^/ U-759 + 1.5/ 1 = \l+x) ^.759+1.5/ +5.759+ 1.5 I ffc 



1 lb. air makes 



r lol / 1 \/ 1 . S\l, r/ 29 - 8 \ I l i S \ ,56.32,29.851 Jx 

L 1+fl+ tl+s) (5T759 + i^)J l0S - = L(l+^) (5^59 + i^) + 5^59-+lxJ CUft - gaS - 

1 lb. gas requires 

\ 5.759 1.5/ \l+a;/ 



1+5+ 



\l+z/ \5.759^1.5/ 

requires 



1 lb. gas requires 
S 



lbs. carbon. 



lbs. steam. 



1 lb. gas requires 

r— 

[l+5+ 

1 cu.ft. gas requires 



(— ) (--+-) 

\l+zj \5.759 1.5/ J 

is requires 

(_i_+A\ (_L_\ 

\5.759 1.5/ \l+x/ 



56.32 .,29.85 
5.759 1.5 J 



\l+x) \5.759^1.5/ 
1 cu.ft. gas requires 

S 1 

/ 29.8 \ / 1 5 \ 56.32 29.85 

\l+x) \5.759~ l ~1.5/~ i ~5.759 + 1.5 J 



lbs. carbon. 



lbs. steam. 



1 cu.ft. gas requires 



Ilbs. ail 



(&) 



(c) 



(d) 



w 



(/) 



(o) 



(h) 



(0 



(80:) 



In order to show the composition of the gas by volume for various propor- 
tions of steam to air in the blast assuming all the steam to react, the four curves 

CO 

of Fig. 186 have been plotted. The left-hand curve is for a ratio of = 2, 

CO2 

the other three for ratios of 6 15, and infinity. The first three cover the ranges 
obtained in practice, extending from a value that is lower, to one that is higher 
than usually obtained, while the last is the limiting case. The composition 
by volume is plotted vertically and the value of S, which is the weight of steam 



HEATING BY COMBUSTION 



727 











































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1 


















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1 

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CO 

II 














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► 

£ 






























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1 








£ 








s, 
















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z }f 








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£ : 8 8 



**-^ 



728 



ENGINEERING THERMODYNAMICS 



per pound of air reacting, horizontally. These are separately plotted because 
it is not convenient to show on one sheet the variations in each of the four gas 

CO 

constituents as a function of the two prime variables S and — — -'. 

GO2 

Heats of reaction are plotted as a function of S alone, laid off 

CO 
horizontally in Fig. 187, and a separate curve drawn for each value of the — — - 

CO2 

ratio, 2, 6, 15 and infinity. The vertical distances are heats of reaction, first, 

per pound of gases produced and second, per pound of carbon, the former being 

a measure of temperature rise, and the latter of efficiency of reaction. These 

two heats are given below as derived from Eq. (800), in the two Eqs. (803) 

and (804). 



Heat of reac- " 




tionB.T.U. 




per pound of 


■ = - 


gas pro- 




duced. 





710,1931+4351 , e /1545z-4298Y 
|A 5.759 lS \ 1.5 j 


V+x) 


H+GiX.™^). 





(803) 



Heat of reac- 
tion B.T.U, 
per pound 
of carbon 
gasified. 



10,193s+4 351\ 
5.759 



}+ 8 { L5 ) 



V5.759^1.5/ 



. . (804) 



Air saturated with water vapor carries a definite weight of vapor per pound 
of air and as the air blast is to carry water vapor in some proportion it is con- 
venient to define the amount in terms of the temperature of moisture-saturated 
air that carries the definite quantity in question. This is most conveniently 
done in curve form, Fig. 188, where the temperature of water-saturated air is 
plotted vertically to horizontals of pounds of steam per pound of air. 

As already pointed out, the greater the quantity of steam in the blast the 
more endothermic heat will be taken up from the exothermic heat of the air- 
gas reaction and when these two are equal there will be no temperature rise of 
the products of gasification over that of the blast. This is a limiting case, 
for which of course no reactions could take place. In order that it may, some 
temperature rise must be permitted or, in other terms, the net heat of reaction 
must be exothermic and sufficient in quantity to allow the final gas to be as hot 
as necessary. This final gas should not be less than 1100° F., corresponding 
roughly to 250 B.T.U. exothermic heat, per pound of gases, which fixes a limit- 

CO 
ing value of S on the chart, Fig. 187, depending on the — — ratio to be allowed 



in the gas. For 



CO 

C0 2 



C0 2 
6, the corresponding value of S is .215 which would be 



HEATING BY COMBUSTION 



729 





































































































































































\ 






















■ 
































\ 






















































\ 






















































\ 
























































\ 






















































\ 






















































\ 
























































\ 


















































\ 




\ 


















































\ 






\ 
















4 Upper Curves -Heat of Reaction 
per lb. of Carbon. 
4 Lower Curves -Heat of Reaction 
per lb. of Gas produced. 












A 






\ 


























\ N 


\ 






\ 
























\ 


\ 






\ 














































A 


\ 




















































\ 




\ 


















































\ 


\ 


\ 


V 
















































\ 


A 




\ 


















































\ 


\ 


\ 


\ 
















































\ 


\ 




\ 










CO - 

£_COo 


2 




































\ 


\ 




\ 
















































\ 


\ 


\ 




\ 
















































\ 


s 




\ 


V 






















































\ 


























































c 

Y C 


i=< 










































































































^ 






















































%\ 














X 


"s 


















— 


CO 

CO; 


4^0 
6& 


^? 






CO 
CO- 


=-0 








































=15 


co- 

COi 
























































































= 00 

1 
















s 




























• L 








.2 










r* 




















2 




















LTj 














\ 


> 


















































CO 


^oc 
























































































































































































































1 



















Fig. 187. — Heats of Reaction for Hypothetical Producer Gas from Fixed Carbon,B. T. U. 



730 



carried by air saturated at 150° F., for 



ENGINEEEING THERMODYNAMICS 
CO 



C0 2 



15 the value of S that gives 250 



B.T.U. per pound of gases exothermic heat, is .164, which would be carried by 

CO 

air saturated at 142° F., while for no CO2 in the gas or = oc, £=.15. 

CO2 



J3 

o 

I 



150 



125 



|100 
I 



50 



25 







































































































































































































































































































































































y 






























y 






























y 


/ 






























f 
































f 
































t 
































1 
































t 
































1 
































t 
































_r 
































1 
































1 
































j: 
































































_i 
































7 
































1 

































































































































































































































































































































































































































.1 .2 .3 .1 .5 

Lbs. of Steam per Lb. of Air 

Fig. 188.— Temperature of Air Blast at which Saturated Air Carries a Given Weight of Steam 

per Pound of Air. 



The heat of combustion of the gas produced from 1 lb. of carbon <- of course 
14,544 less the heat of reaction per pound of carbon or, Eq. (805) 



Heat of combus- 
tion of gas from 
one pound of 
carbon B. T. U, 
low. 



■ = 14,544- 



10,193z+435l\ . C /I545:r-429SY 
5759 r S \ U 1 



\5.759 ^1.5/ 



805) 



HEATING BY COMBUSTION 
Therefore the efficiency of the reaction is Eq. (806), 



731 



E=l- 



7l0,193z+435l\ . „/1545s-4298 
{- 5.759 -) +S {~~Ur 



"^(ot+ES 



(806) 



From this equation it appears that the efficiency of the reaction is primarily 

, CO 
a question of the quantity of steam used for any given value of — — , so the 

efficiency as given by this equation is plotted vertically to horizontal values 

CO 
of S, one curve each, for values of — — , 2, 6, 15 and oo in Fig. 189. 

CO2 

The heat of combustion of the gas itself can be found by dividing the heat 

of combustion of the gas produced from 1 lb. of carbon by its volume and is 

given by Eq. (807), 



B.T.U. cu.ft. gas 

(standard, low) 



14,544- 



/10,193x+4351\ . c ,/1545x-4298\ 
\ 5759 ) +S \ 1.5 ) 

\5.759^1.5/ 



1 



29.8 

l+z/V5.759 ' 1.5/ ' 5.759 ' 1.5 



56.32 29.8£ 



1 
1+x 



14 ' 544 l5T759 + 



\5.759^1.5/\l+;r 

S \ / 10,193x+4351 \ / 1545a; 
1.5/ \ 5.759 / \ 1 



(a) 



4298 N 



29. 



1+x 



\5.759^1.5/ 



56.32 29.8^ 
i_ 5.759 + 1.5 



(b) 



(807) 



If CO2 is absent x = and this becomes 



B.T.U. cu.ft. gas (standard, low) 



1770+6831£ 
14.95+39.73S * * 



(808) 



If no steam is supplied S = and B.T.U. per cubic foot gas=118, which checks 
the value found for air gas with 70 per cent efficiency. With .10 lb. steam 
per pound air, B.T.U. per cubic foot gas = 130, for which condition the reac- 
tion efficiency is 86.6 per cent. These figures show that steam in the blast 
increases both calorific power of gas and the thermal efficiency of its produc- 
tion, the steam-efficiency relations have been shown in Fig. 189, so, to correspond 
the calorific power of gas is plotted in Fig. 190 to the same horizontals, pounds 
of steam per pound of air. 



732 



ENGINEEKING THERMODYNAMICS 



To complete the series of graphic results, the gas production in cubic feet 
standard, per pound of fixed carbon is plotted in Fig. 191, to horizontals of S, 

CO 
one curve each for 7^- values of oc, 2, 6, and 15, Eq. (802). 
UO2 

So far the results all relate to the pure process of gasifying carbon with 

steam and air, but in real producers there are additional factors to be accounted 

for, but this cannot be done quantitatively. There must be considered the heat 

losses of radiation, steam making, distillation of volatile, escape of unchanged 







































































































CO 

cog 


= 15 




CO 

co- 


;0O 




















































100 
















































co 


^6 




















































go" 
co 2 


-2- 






















































































































S\ 


s 














































75 




/ 


1 


y 
















































/ 


'/ 


/ 




















































/ 






































































































































































50 






















































































































































































































































































26 















































































































































































































































































































































I .1 .2 .3 .4 ,5 

Lbs. of Steam per Lb, of Air 

Fig. 189. — Efficiency of Hypothetical Gas Producer Gasifying Fixed Carbon. 



steam and heats of reaction of volatile parts that suffer chemical change after 
being liberated. The production of the volatile always consumes less heat per 
B.T.U. in the gas formed, than the gasification of carbon, so that the coals which 
yield much volatile in the producer should give a higher efficiency than those 
that do not, unless some of the volatile is burned or its heat used up in other 
endothermic reactions. In fact, this gain by the volatile generally serves to 
pretty well balance the other heat losses just mentioned so that the real producer 
can have efficiencies pretty close to those calculated for the gasification of carbon 
if the volatile is 10 per cent and over. The addition of the volatile or its prod- 
ucts to the gasified carbon will not change the relation of the CO, CO2, H2 and 



HEATING BY COMBUSTION 



733 



N2 constituents very much, if at all, but will reduce the per cent of each. The 
following general relation must hold. 



Cubic feet gasl 
per pound coal [ 



[ 

, Tcu.ft. gasified] fWt. fixed carbon] 
> L C per lb. C J X per lb. coal 



cu.ft. volatile ] [~ Weight volatile ] 
per lb. volatile] ' per lb. coal 

;. fixed carb< 
per lb. coal 



(809) 



This Eq. (809) must be more or less exactly evaluated before the effects of vola- 
tile addition on each constituent can be estimated and to do this it is necessary 















































































































H 




























































































i_ioLs -co 




































































































SH^J 












H 






























































































U 




===■ 


\5 








pq 






























































































c 










































































































i. 






















































































































































































































g 




































































U- 
































cS£ 


■" 










•3 






























































































C^ 












s 












































































































^^00 












































































































a 
























































































































































































































CD 




















































































































































/ 


y 




















































rC 












a 






































y 


























































ftp'- 






































y 






/ 


y 








































































r y 
























/ 








s 


































































































/ 








s 


































































































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/ 


* 












o 






















































































/ 






/ 




















































































3 
























/ 








































































































/ 











































































































/ 


























































































^140 
























































































































































y 








































































































y 




































































H 












































































































ffl 
















































































































































































































































































































































































































































& 










































































































3 




































































































































































































































































































































O 












































































































«w 
























































































































































































































ts 












































































































S 80 












































































































W 

















































































































































































































































































































































































































































a .2 .3 .4. .5 

Lbs. of Steam per lib. of-Ate. 
Fig. 190.— Calorific Power of Hypothetical Producer Gas from Fixed Carbon. Low Value. 

to have some data on the cubic feet of gas per pound of volatile and the pro- 
portions of constituents making it up. This can be done by considering it to 
behave as coal gas from retorts or as coke oven gas. It was shown for the 
English cannel coals which yield the largest amount of volatile that they give 
on the average about 4.5 cu.ft. gas per pound coal, and are about 60 per cent by 
weight of volatile, so that the cubic foot of gas per pound of volatile = 7 app. 
Gas coals in retorts or coke ovens will yield about 8 or 9 cu.ft. gas per pound 
volatile, and about the same for quite a range of volatile per cent. The gasified 



734 



ENGINEEKING THERMODYNAMICS 



fixed oarbon yield is somewhere near 90 <;u.ft. per pound. Therefore, a coal 
with 30 per cent volatile and 60 per cent fixed carbon would give in round 
numbers 56 cu.ft. of mixed volatile and gasified carbon, of which the volatile 
makes up 4 per cent by volume, which has small influence on the proportion. 
The real influence may, however, be less, as in an up-draft producer, some of the 
volatile will condense to tar, several of the U.S.G.S. tests showing about .2 lb. 
tar per pound volatile of the proximate analysis, so that one-fifth of the volatile 
condensed. With down-draft producers the volatile may completely change 
and the gas show no trace except a small per cent of CH4. 











































































































































































100 






















































































































































































































































































d 
§90 














































































































O 
























































s 
























































§3 
e. 
•d 
§80 

1 


















































GO. 

ooj 


=2 
































































































































































CO 


= 6 




8 


















































i*Jl 






Si 


















































CO 


= 11 




^70 
S 


















































CO. 

CO. 
SQi 


^CO" 


























































































































































































































fin 

























































2 .3 

Lbs. of Steam per Lb, of Air 



Fig. 191.— Quantity of Hypothetical Producer Gas (Standard) per Pound of Fixed Carbon. 

Lignite coal will yield a volatile quite different in character and in up-draft 
producers will add something like the following gas, reported by Guldner for the 
yield by roasting brown lignite in retorts, weighing .0633 lb. cu.ft. or of specific 
volume 15.8 cu.ft. per pound. 





COMPOSITION 


OF LIGNITE VOLATILE (Guldner) 




H 2 . 


CH 4 . 


CnH 2 n. 


CO. 


H 2 S. 


C0 2 . 


o 2 . 


N 2 . 


24.3 


16.5 


1.4 


8.1 


1.1 


17.0 


3.1 


28.5 



HEATING BY COMBUSTION 



735 



Attention has already been called to two most important factors in the 
gasifying process which have only recently come within the scope of experimental 
measurement and these are the physical condition of the bed as to porosity, 
homogeneity, uniformity of gas flow, and the temperature of the bed with 
its homogenity .of distribution. As part of the general investigations on fuels 
in producers, Clement and Grine of the U.S.G.S. measured the temperature 




Fig. 192. — Section of Producer showing Temperature Variation. 



in the bed of a producer 7 ft. internal diameter, up-draft, steam jet blower, 
delivering the blast under a hood in the center of the producer bottom at a point 
normally covered with ash. The results are plotted in the diagram, Fig. 192, 
in the form of constant temperature lines. The highest temperature is found 
in the bottom and the same as would be found in a boiler fire when supplied with 
steam in the air. The temperature falls slowly and then more rapidly toward 



736 ENGINEEEING THERMODYNAMICS 

the top of the fuel bed, kept at 5J ft. above the blast entrance. The curvature 
of these lines shows that the blast rises toward the sides more than in the center 
and the difference is very marked indeed. The explanations and discussion of gas 
analyses which were made at the same time from the bed interior cannot be 
taken up here, the main facts being, first, that a non-homogeneous condition 
exists, partly because the blast is impelled outward by the hood but also because 
the resistance at the walls is less than anywhere else, second, that the tem- 
peratures actually attained are very much higher than correspond to equilib- 

CO 

rium of the — — 4 ratio reported. This ratio for the gases leaving, which were, 
CO2 

of course, the mixture of all sorts of different gases from different parts of the 

CO 
bed, was ~;^~ = 3, for which the equilibrium temperature is 1376° F., while in 
CO2 

this bed the temperature was 2200° F. at which value the equilibrium value of 

CO J 

the ratio — — - is over 200. 
CO2 

It has been found possible to generate a usable power gas from a jet of coal 
dust or oil blasted into a combustion chamber with air insufficient for complete 
combustion, if the walls and blast be so arranged as to keep the hot products in 
contact with glowing free carbon liberated from the fuel by its decomposition. 
Some analyses of such gas are given in Table CXIX at the end of the Chapter for 
Western heavy oils, which show a comparatively low CO content, though not so 
high a per cent CO2 as might be expected from the bed type of solid carbon pro- 
ducer. As the temperatures in these chambers are quite high, it must be con- 
cluded that equilibrium between CO and CO2 has not been established and the 
reason may be either too high a velocity of the gases, which is the same as 
insufficient time, or what is more likely, not enough free carbon present at the 
right time and place for CO2 reduction. These gases if passed through a coke bed 
would be very much improved, in fact would become almost perfect as to 
combustibles. Another characteristic of these analyses is the large CH4 content 
and with it an also large amount of heavy hydrocarbons. These two things 
are characteristic of the oil fuel as raw material, the methane having come 
from decomposition of higher hydrocarbons, yielding free carbon, some of which 
went into CO and some was left behind as coke in the producer. 

An almost identical operation with pulverized coal that has recently been 
tried gave gas analyses (Latta) as in Table CXX at the end of the chapter. The 
powder and air are blown into a large combustion chamber without steam, 
all hydrogen coming from coal volatile, or from moisture in air and coal. As 
the coal was bituminous, ranging from 60 to 77 per cent fixed carbon and 18 
to 33 per cent volatile, the small quantity of illuminants or heavy hydrocarbons 
shows that these were attacked by simple decomposition or by oxygen, more 
likely the latter, as the methane is also low while in coal gas it would be much 
higher and even higher also in up-draft producer gas with the same coal. The 
carbon monoxide-dioxide ratios are very much the same as for the oil-gas pro- 



HEATING BY COMBUSTION 



737 



ducer and probably for the same reason, insufficient amounts of free carbon 
for the more complete reduction of CO2 to CO, because the temperatures 
developed are quite high enough. 



Table XCI 
DENSITY AND CALORIFIC POWER OF PRODUCER GAS 







One Cubic Foot 






£ Constituent. 


Contains 


B.T.U. 






Cu.Ft. 


Lbs. 


High. 


Low. 




a 

< 

C3 
O 
> 


Carbon monoxide, CO. . 

Hydrogen, H2 

Methane, CH 4 

Carbon dioxide, CO2 . . . 

Oxygen, O2 

Nitrogen, N2 


.261 

.1500 

.0020 

.053 

.002 

.532 


.020376 
. 000843 
.000089 
. 000052 
. 000178 
.041650 


89.0 

51.2 

2.1 


89.0 

43.8 

1.9 


B.T.U. per cu.ft. gas high 142.3 
B.T.U. per cu.ft. gas low 134 . 7 

Cu.ft. per lb. gas 14.36 

Lbs. per cu.ft. gas .0696 

B.T.U. per lb. gas high . 2043 
B.T.U. per lb. gas low. . 1934 




Total 


1.00 


.069638 


143.3 


134.7 


+3 

ID 

Ph 


Carbon monoxide, CO. . 

Hydrogen, H2 

Methane, CH 4 

Carbon dioxide, CO2 . . . 
Nitrogen, N2 


.272 
.009 
.031 
.121 
.567 


.021235 
.000051 
.001385 
.014845 
.044390 


92.8 

3.1 

33.0 


92.8 

2.6 

29.7 


B.T.U. per cu. ft. gas high 128 . 9 
B.T.U. cu.ft. gas low . . . 125 . 1 

Cu.ft. per lb. gas 12.21 

Lbs. per cu.ft. gas .0819 

B T.U. lb. gas high 1574 

B.T.U. lb. gas low 1627 




Total 


1.000 


.081906 


128.9 


125.1 


1 
C, 

a 

c 

c 


Carbon monoxide, CO. . 

Hydrogen, H2 

Methane, CH 4 

Heavy Hydro., C2H4. . . 
Carbon dioxide, CO2 . . . 
Nitrogen, N 2 


.253 
.092 
.031 

.008 
.034 

.582 


.019751 
.000517 
.001385 
.000636 
.004171 
. 045565 


86.3 
31.4 
33.0 
13.6 


86.3 
26.9 
29.7 
12.6 


B.T.U. per cu.ft. gas high 164 . 3 
B.T.U. per cu.ft. gas low 155.5 

Cu.ft. per lbs. gas 13.88 

Lbs. per cu. ft. gas .0720 

B.T.U. per lb. gas high . . 2280 
B.T.U. per lb. gas low. . 2158 


3 
.■a 

PQ 


Total 




. 072025 


164.3 


155.5 









Horse-power, as a term applied to a gas producer, is just as much out of place 
but just [as necessary as when applied to a steam boiler. It is out of place 
because no mechanical work is done by either piece of apparatus, both being 
concerned with, and used for supplying the means for doing work in steam and 
gas engines. This gives the reason for the necessity of a name as engines are 
rated at, and develop a given horse-power, so it is natural to associate the horse- 
power name and number with the producer or boiler that makes it possible, 
however indirect or scientifically unsound it may be. There is, therefore 
necessary, some standard of producer as of boiler horse-power and it should be 
based on the capacity of either to supply respectively the requisite amount 
of gas or steam per hour that an engine may need to develop and maintain one 
horse-power of output. No such standard has, however, been generally accepted, 
but one can be easily created. 



738 



ENGINEERING THERMODYNAMICS 



It will be not far wrong to say the average good gas engine of almost any 
size above 100 H.P. can develope a horse-power on 10,000 B.T.U. per hour in the 
gas, low value, and many can do better, as low as 8500 having been proved 
possible, though a few two-cycle engines may use as high as 12,500. The figure 
10,000 B.T.U. per hour being an easily remembered round number and well within 
the range of practical performance with a fair margin of capacity for very good 
engines, may well be adopted as the equivalent of a gas-producer horse-power. 
Therefore, 



H.P. of gas producer 



B.T.U. per hr. (low) in gas produced 

10,000 — . . . 

(Cu.ft. gas per hr.) X (B.T.U. cu.ft. gas) (low) 
10,000 



(a) 
(6) 



(810) 



In the following tabular form, Table XCI, are given some average density and 
calorific properties for the producer gases, illustrating their calculation from 
the volumetric constituents in the usual way. 



Example 1. A producer is blasted with air and steam in the ratio of 5 to 1 by 
weight. The ratio of CO to (C0 2 +CO) is .8. What will be the composition of the 
gas by volume and weight, the pounds of carbon, air, and steam required per cubic 
foot of gas and the heat of combustion per cubic foot of gas? 

For an air steam ratio of 5, S will be £=.1667. Since CO -HCO+C0 2 ) =1 -x, 
= .8 or x = .2. 



The C0 2 by weight is 



The CO by weight is 



The H 2 by weight is 



The N 2 by weight is 



The C0 2 by volume is 



1667\ 



3. 66 X.2 / 1 .1667 
1.2 \5.759~ 1.5 



■ 



1+.1667+ 



•+ 



.1667 \ 
1.5 I 



1.2\5.759 

2.33 X . 8 /_1__, J^67\ 
1.2 \5-579 1.5 / 

>«-+&&?% 

.1667 X. 1677 
1.5 



1+.1667+ 



1_ J667\ 
5.759 1.5 / 

4^28 
5.759 



>+"*&&?&> 



= 12 A per cent. 



31.6 per cent. 



= 1.3 per cent. 



= 54.8 per cent. 



2 9.8 X.2 
1.2 



M_ + ^67\ 
\5.759 1.5 / 



29.8/ 1 
1.2 \5.7 



.1667X 56.32 29.8x.1667 
1.5 / 5.759 1.5 



= 7 per cent. 



The CO by volume is 



The H 2 by volume is 



The N 2 by volume is 



1 cu.ft. gas requires 



HEATING BY COMBUSTION 

29.8 X . 8/ 1 J667\ 
1.2 \5-759 1.5 / 



739 



29.8/ 1 ,1667\ 56.32 29.8 X. 16 67 
L 1.2 \5.759 1.5 / 5.759 1.5 J 



29.8 X. 16 67 
1.5 



29. 



■ ilM \ 56.32 29. 8 X. 1667 
1.5 



1.2\5.759 1.5 / 5.759 

56.32 
5.759 



S9.8/ 1 .1667\ 56.32 29. 8x.1667 
1.2\5.759 1.5 / + 5.759 + ~T5 

/ 1 ^667\J_ 

\5.759 + 3.5 /1.2 

29 - 8// 1 -1667\ 56.32 29 .8 X .1667 

1.5 



28.1 per cent. 



16.4 per cent. 



= 48.5 per cent. 



1.2 \5.759 1.5 / 5.759 

1 



1 cu.ft. gas requires 



1 cu.ft. gas requires 



1 T29.8/ 1 ,1667\ 56.32 29.8 X. 16 67] 
.1667 L 1.2 \5.579 1.5/ 5.759 + 1.5 



= .0116 lb. C. 



= .0083 lb. steam. 



■ 1667\ 56.32 



.0495 lb. air 



1.2\5.759 1.5 / 5.759 ' 1.5 

The heat of combustion of the gas per pound of carbon is, low value, 



14,544 



10,193 X. 2 +4351 ,„„„ /1545X.2-4298 1 
-+.166/ 



5.759 



/ 1545X.2 -4298 V 
\ 1-5 / 



/ 1 .1667 \ 
\ 5.759 1.5 / 



13471 B.T.U. 



and from the above, 1 cu.ft. of gas requires .0116 lb. of carbon. Hence the heat 
of combustion of the gas is 13,476 X. 0116 or 156 B.T.U. per cubic foot. 

Prob. 1. The following anlysis of coke producer gas was taken when the producer 
was receiving little steam. CO =28; H=4; C0 2 =CH 4 =4; N 2 =60. What was the 
ratio of steam to air? 

Prob. 2. If the air-steam ratio had been raised to 4, what would have been the 
composition of the gas for the same CO to C0 2 ratio? 

Prob. 3. For an air-steam ratio of 6 and a CO to (C0 2 + CO) ratio of .9, what 
will be the carbon, steam, and air required for 1000 cu.ft. of gas and what will be 
the heating value of the gas? 

Prob. 4. A coal used in a producer consists of 85 per cent fixed carbon, 5 per cent 
volatile, and 10 per cent ash. The CO to C0 2 ratio and the air-steam ratio are both 5. 
If the volatile has a heating value of 20,000 B.T.U. per pound and it is assumed that 
it comes off uniformly, what will be the heat value of the gas per pound and per cu.ft.? 



740 ENGINEERING THERMODYNAMICS 

Prob. 5. A coke producer blast is saturated at 140° F. What is the maximum 
CO to C0 2 ratio which could be had for a continuous run and the efficiency? 

Prob. 6. For a ratio of CO to C0 2 of 10, what would be the minimum steam ratio 
for a continuous process and the corresponding efficiency of reaction for a fixed 

carbon bed? 

Prob. 7. Coal No. 6 of Table CIV is used in a producer. The volatile is the same 
as gas No. 5 of Table CX and CO to C0 2 ratio is 6 with an air blast saturated at 
160° F. What will be the composition of the gas and the efficiency of the producer? 
500 B.T.U. per pound of coal is required to liberate the volatile. 

Prob. 8. Coal No. 86 of Table CIV is used in a producer. What will be the 
volume of gas per ton of coal and what will be the composition for the following 
assumptions: CO to C0 2 =5; S = .2; volatile of coal is the same as No. 12 of Table 

CX. 

Prob. 9. In Table CXVIII are given some producer tests. Compare the compo- 
sition of the gas as found with that as calculated and the actual efficiencies, Assume 
S to be .2. 

10. Combustion Effects. Final Temperature, Volume and Pressure for 
Explosive and Non-Explosive Combustion. Estimation of Air Weights and 
Heat Suppression Due to CO in Products from Volumetric Analysis. When 
used for industrial purposes fuels are burned in one of two radically different 
characteristic ways, designated as, (a) explosive, and, (b) non-explosive combus- 
tion, and while in every case there is a rise of temperature it may be accompanied 
by a volume or by a pressure change or both, rise of pressure being especially 
characteristic of explosive combustion. In any case the fundamental effect 
of combustion is a temperature rise which can be only estimated as to amount 
because of the great uncertainty of specific heats coupled with the dissociation 
phenomenon which retards further oxidation of fuel elements after the temper- 
ature has reached a certain value. Accordingly, all calculations of combustion 
effects must be regarded as more or less approximate and to the results, factors 
must be applied to bring them into accord with direct observation of effects, 
which factors measure by their magnitude the errors of the method and the 
uncertainty of fundamental physical constants. The application of such 
factors, which is necessary even with the most precise methods, opens the way 
for quick estimation of results by shorter methods, which are almost univer- 
sally used in engineering work concerned with explosions, especially those 
in the cylinders of internal combustion engines. 

Usually non-explosive combustion takes place at constant pressure and explo- 
sive at constant volume but these are not essential relations as either kind may 
proceed under any conditions of pressure or volume relation. The essential 
distinction between explosive and non-explosive combustion is to be found in 
the self-propagation or progress of the flame or reaction causing it, for the former, 
For combustion to be explosive there must be a mixture of fuel and oxygen which 
may be diluted with neutral gases like nitrogen, carbon dioxide or water vapor 
within certain limits. Moreover, the mixture must be of the most intimate 
and homogeneous sort though the fuel may be in the solid liquid or gaseous 



HEATING BY COMBUSTION 741 

state. Each particle of oxygen must be in contact with the fuel particle the 
combustion of which it is to support, so that if the fuel is solid it must be in 
powdered form suspended in the oxygen; if liquid, it must be sprayed in, in as 
fine a mist form as is possible. The ideal condition of the fuel for explosive com- 
bustion is, however, the gaseous or vapor form, mixed homogeneously with the 
supporting oxygen. Mixtures that comply with these specifications will explode, 
that is, combustion once started by an ignition of any part of the mass at a single 
point, will propagate itself throughout the entire mass if it is at rest. The flame 
will at any instant be on a surface ideally of no thickness, but actually, if there 
be some inactive gases present, of measurable thickness, which surface separates 
the burnt from the unburnt. This surface advances through the mixture 
at a speed entirely independent of any motion of the mixture mass, and the com- 
bustion proceeds at a rate determined by this speed of propagation peculiar 
to the mixture itself and not by any surrounding conditions. If the mixture 
be in a closed chamber the whole mass will burn in a short time ; if it be supplied 
to a combustion chamber continuously as is done sometimes in gas and oil 
furnaces and those consuming powdered coal, the combustion will proceed at 
the same rate against the current of supply, the velocity of which must be greater 
than the rate of propagation to prevent back flash, and be subsequantly reduced 
in the furnace below the rate of propagation to prevent blow off. 

If the mixture contain not enough oxygen to support combustion of all the 
fuel present, then the combustion may still be explosive within certain limits, 
to be discussed later, but the combustion cannot be complete when the explo- 
sion takes place in a closed chamber and can only be complete when supplied 
in a stream, as in Bunsen burners or from furnace nozzles or burners, provided 
a supporting atmosphere of extra air be supplied externally to the jet or flame. 
In such cases the rate of combustion will be less because some of the fuel must 
wait until it can find oxygen; the flame will accordingly be longer and fill greater 
volumes of the combustion chamber, the rate will be least and the flame volume 
greatest for no previous mixture, all air for combustion in this case being 
derived from the external supporting atmosphere. This latter condition 
is found in common illuminating gas flames and in the fires of coal with 
much volatile, which rises from the bed practically without air mixture, all 
air necessary for combustion being admitted above the fire, meeting the gas 
when it can and the last particles burning in an atmosphere much diluted with 
the products, of previously burnt fuel. Such conditions promote escape of 
unburnt gases, hydrocarbons, carbon monoxide or hydrogen, even when some 
oxygen is still unused and appears as free oxygen in the products of combustion. 

Similarly, for solid fuels the combustion proceeds as fast as the air can reach 
the fuel, the rate of combustion being fixed by the rate of air supply but not 
directly proportional thereto, because of the stream form of the air between 
the particles of fuel and the diluting, separating influence of previously burnt 
products tending to keep air and fuel apart. Thus, solid fuel combustion 
always must take place with excess air or be incomplete, for if just the right 
amount were supplied some oxygen would slip between the coal particles without 



742 ENGINEERING THERMODYNAMICS 

touching and the quantity of combustible gases formed above would always 
be greater than this slippage air could burn. More air supplied with a view 
to burning this gas is ineffective because of lack of means of pre-mixing, so that 
while the fuel might all be burnt partly in the solid and partly in gaseous form 
it can be done only with excess air, absence of free oxygen in the flue gas being 
according to universal experience accompanied by much unburnt fuel. 

While it is possible to calculate the weight of air necessary to complete the 
combustion of a pound of fuel from the reaction equation or similarly the volume 
of air per cubic foot of fuel, the above discussion indicates that such calcula- 
tions are not sufficient to permit a judgment of the effects of combustion to be 
expected in practical working with either explosive or non-explosive, contin- 
uous or intermittent fires. In all cases, however, the temperature rise should 
be given by the ratio of heat liberated by combustion to the product of the 
specific heat of the products, into the weight of products. The weight of the 
products will always be the weight of air per pound of fuel chemically necessary 
for combustion, plus the excess air per pound of fuel, plus one, and the heat 
liberated is the calorific power per pound of fuel if it is all burnt. When, how- 
ever, the air is insufficient to burn the fuel and in the case of solid fuel, some- 
times also when it is sufficient, the full calorific power cannot be developed and 
heat will be lost. 

Furthermore, the full heat of combustion will not be available for temperature 
rise if in the course of that rise the dissociation temperature for the products 
is reached, for after that time further union, heat liberation and temperature 
rise will cease. Abstraction of some heat will permit combustion to proceed 
but without any more rise of temperature. These considerations show that the 
calculation of temperature rise on combustion or the consequent increase of 
volume at constant pressure, or increase of pressure at constant volume, when 
based on purely physical constants, is hardly more than an academic exercise 
even when the constants are positively known, but decidedly so at the present 
time, when the dissociation constants and specific heats at high temperature are 
practically unknown in spite of much work to find them. 

Accordingly, to calculate probable effects of combustion experimentally 
determined factors must be relied upon. 
Let Q = B.T.U. per pound fuel; 

h and t\ = final and initial temperatures F. ; 
T2 and T\= final and initial temperatures absolute; 
V2 and Vi= final and initial volumes cu.ft. 
P2 and Pi = final and initial pressures lbs. sq.ft.; 
P2 and pi = final and initial pressures lbs. sq.in.; 

wp = lbs. products per lb. fuel = wt. air chemically necessary + 
wt. of excess air per lb. fuel+1; 
P = fraction of B.T.U. per lb. of fuel actually causing rise of 
temperature; 
Cvy Opt C v , C/** specific heats at and mean specific heats, at constant 
pressure and volume for the products of combustion. 



HEATING BY COMBUSTION 743 

Then 



T2—T\ = t2 — t\ = — ~-7, when the pressure is constant (a) 
= — 7T7, when the volume is constant (6) 



(811) 



For explosions at constant volume, 

P2_P2_T 2 _ / W \ 

Pi pi Ti 1+ Wc;rJ' (812) 

P2 - Pl =ft(^c7)> (813) 

Also for constant pressure combustion, 



' r '- v ''Wk) < 81 » 

These equations as set down are perfectly correct but of no practical use, without 
evaluating g, C p ' and C/. To illustrate the point: consider the combustion of 
carbon monoxide in oxygen, the temperature of which as measured by Mallard 
and Le Chatelier, was found to be 5800° F. The heat of combustion of CO 
burning to CO2 is 4369 B.T.U. per pound CO and the weight of products is 

44 

— = 1.57 lbs. Taking the specific heat of carbon dioxide at constant volume 

at .2025 the value at 32° and constant, and assuming g = l that is, all heat 
developed, 



This is pretty far away from the measured value of 5800, which latter is really 
only 40 per cent of that derived by the calculation based on constant specific 
heat. Introducing the variable specific heat the result is a little better but 
still fails to agree with the measurement by a pretty wide margin, as will appear 
from what follows. 

The most commonly accepted, but still uncertain form for the specific 
heat of gases at high temperatures in terms of the value at 32°, is given by 
a first or second degree equation, from which the mean value between 32° 



744 



ENGINEERING THERMODYNAMICS 



and any temperature can be derived, and this multiplied by the temperature 
range will give the corresponding quantity of heat. To solve this equation for 
the final temperature in terms of the quantity of heat causing it is troublesome 
by algebraic methods but easy by a chart such as Fig. 193. Here one set of 
the Mallard and Le Chatelier values for the mean specific heatof various 
gases given in Eq. (816) has been used to calculate the temperature rise above 
32° for various quantities of heat and plotted. For any heat increment per 
pound of gases there is a corresponding temperature increment that can be 
read off directly. Thus, for CO2, consider 1 lb. to receive 1000 B.T.U., 
starting at 32° F., the temperature rise would be 3290° F.3-2° F. = 3258°, 
whereas from 1000° F. as a starting point this same 1000 B.T.U.would yield 
a temperature of 3690° F. or a rise of 2690°. 





' .149 +.000048* 


for 


co 2 . 


.(a) 


Mean specific heat 1 


.433 +.000060* 


a 


H 2 . 


■(b) 


Mallard & Le Chatelier f = ■ 


.242 +.000024* 


a 


N 2 . 


■ (0 


Pressure constant. J 


.212 +.000021* 


u 


2 .. 


■(d) 




, 3. 3 +.000015* 


it 


H 2 . 


■(e) 



(816) 



Applying this to the combustion of CO to CO2, yielding —^=- = 2783 B.T.U. 

1.0/ 

per pound of CO2, the temperature rise above 32° is 6790 and the final tem- 
perature 6822° F., a little better than for constant specific heats but not good 
enough. 

Other values for the variable specific of gases are given by different experi- 
menters but are not quoted here as the results obtained by their use are no 
closer to the observed temperatures. 

In place of the mean specific heats of the products taken by the variability 
law for the several constituents, the specific heat of air may be substituted 
at its value for low temperatures C p =.170 and C p =.243 approximately. 
Then introducing a new constant C by the relation, (C -f- constant specific heat 
of air) = (@ -1- specific heat of products), the temperature rise will be given by 



T2 — T\ = *2 — *i = C ^Jl , for constant volume (a) 
.170w P 



-c- 



, Q 



.243w P 



, for constant pressure (6) 



(817) 



Estimation of temperature rise by this second method, Eq. (817), is possible 
whenever a direct observation on a similar case permits of the determination 
of the value of C for such cases. This is always possible in furnaces into which 
pyrometers may be inserted, but it is not possible for explosions in closed 
chambers because of the shortness of the time, but in such cases the pressures 
of explosion can be quite accurately determined, much closer than there is any 



\ 



HEATING BY COMBUSTION 



745 



)50^CO+-50St 
UI 2 Var.) 




500 



1000 



1500 2000 2500 

B.T.U. per Pound of Gas 



3000 



3500 



4000 



Fig. 193. — Relation Between Temperatures and Heat for Gases According to the Constant 

and Variable Specific Heat. 



746 ENGINEEEING THEKMODYxYAMICS 

necessity for and closer many times than the temperature. From such pressure 
observations the coefficient can be found from relations of the following kind. 

Let H = B.T.JJ. per cu.ft. gas at 32° and 14.7 lbs. sq.in. low value. 
" a = cu.ft. air per cu.ft. gas. 
" n = cu.ft. neutral added to air-gas mixture containing one cu.ft. gas. 

Then from Eq. (813) if the specific heat is assumed constant 

P P _*VPQ \-R W _ 77S(C p -c v ) w 

P2 ~ Pi "nteJ"Fi"cr — c v — Tx (818) 

But 

Tp- =B.T.U. per lb. fuel, divided by cubic feet of gases containing or formed 

from one pound of fuel, which is equal to B.T.U. per cubic foot mixture at 
pressure Pi and temperature T\. Hence 



Vi 



-fewi)**** « 

»-f.-S(Y-D 41+^X^x1?]. . . .(820) 

Introduce the constant C and the value of y for air according to the equa- 
tion of condition, Eq. (807), 

[y (for mixture) - 1]& = [y (for air) - 1]C = AC . . . (821) 
77 8X.4X49 2 Pl ( H \ _ 79 ~o r Vi ( H \ 

V initial temp. F. abs. / [ 14>7 lbs . per sq in abs j 

For explosions the pressure rise given oy Eqs. (820) and (822) have been 
determined hundreds of times for all the fuels Used in gas-engine cylinders, 
which include practically every one of the gases previously examined, as well 
as some vapors like kerosene, gasolene and alcohol, separately and mixed in 
various proportions. The pressure rise has been measured with initial pressures 
of one atmosphere in special experimental chambers and in the engine cylinders 
themselves by indicators, the mixture being compressed previous to explosion, 
and for all ranges of fuel and initial pressure, a most extraordinary constancy 
on the value of C has been observed, which may be summarized by Eq. (823). 

Heat suppression factor 1 _^ = { -45 minimum 1 (qo<i\ 

for pressure rise [ .55 maximum J ' 

To illustrate the procedure in determining the heat suppression factor the 
following calculation and experimental determinations for an oil gas, recently 



HEATING BY COMBUSTION 



747 



carried out with some care, will serve the purpose. In all cases the lower 
calorific power is taken. 



Gas analysis per cent by 
volume — 



CO 2 = 

C 2 H 4 = 20.93 

C 6 H 6 = 7.42 
H 2 = 6. 

C 2 H 4 = 62.15 
2 = 0.75 
CO= 0.3 
N 2 = 2.65 



Total, 100.00 



Of this C0 2 =0.00 
N 2 =2.65 



Total, Neutral, = 2.65 



Air required = 11.68 volumes per vol. gas, therefore, cu.ft. best mixture 
per cu.ft. gas =12.68, of which 

' Neutral in gas = .029 

Neutral in nitrogen in air = 9.24 
Neutral in mixture = 9.269 vols, in 12.68 vols, mixture = 73.2% 

by vol. Therefore the detonating or neutral free, gas mixture in the best 
mixture =26.8% by vol. The low heating value of the gas was 1060 B.T.U. 

per cu.ft., whence that of the best mixture is =89 B.T.U. per cu.ft. 

With these data the following mixture characteristics for various proportions 
of air and gas will follow for the explosion ranges, as given in Table XCII. 
These limiting proportions for explosive combustion, constituting the first 
and last columns of the table, were determined by direct experiment. 



Table XCII 
CHARACTERISTICS OF EXPLOSIVE MIXTURES OF PINTSCH OIL GAS AND AIR 



Volume of gas. . 
Volume of air. . 
Percentage of gas in 

mixture . 

Excess or neutral air 

Active air 

Excess or inactive gas 

Active gas 

Neutral in active air . 
Neutral in active gas. 

Total inactive 

°/o inactive or excess. . 
°/o detonating gas. . . . 
B.T.U. per cu.ft. of 

mixture (avail.) .... 
Calculated pres. rise. . 
Observed pres. rise. . . 
Ratio 

observed pres. p 
calculated pres. 



1 


1 


1 


1 


1 


1 


1 


1 


8 


9 


10 


11 


12 


13 


14 


15 


11.1 


10. 


9.08 


8.34 


7.7 


7.14 


6.67 


6.25 










.32 


1.32 


2.32 


3.32 


8 


9 


11 


11 


11.68 


11.68 


11.68 


11.68 


.315 


.229 


.144 


.057 










.685 


.771 


.856 


.943 


1 


1 


1 


1 


6.33 


7.12 


7.91 


8.69 


9.24 


9.24 


9.24 


9.24 


.0202 


.0227 


.0252 


.0278 


.0295 


.0295 


.0295 


.0295 


6.665 


7.372 


8.079 


8.777 


9.59 


10.59 


11.59 


12.59 


74.1 


73.72 


73.5 


73.0 


73.8 


75.6 


77.3 


78.7 


25.9 


26.28 


26.5 


27.00 


26.2 


24.4 


.22.7 


21.3 


80.6 


81.7 


82.5 


83 


81.5 


75.7 


71.6 


66.3 


163 


165 


166.6 


167.6 


164.6 


153 


144 6 


134 


73.5 


80 


81 


79.5 


77.5 


70.0 


64.5 


59 


.45 


.485 


.486 


.476 


.472 


.46 


.446 


.44 



16 

5.88 

4.32 

11.68 

1 

9.24 

.0295 
13.59 
79.9 
20.1 

62.5 
126 
50.5 

.40 



748 



ENGINEERING THERMODYNAMICS 



It is quite remarkable how small is the variation in the quantity of active 
or detonating mixture for all the ranges of explosive proportions and this ex- 
plains why the pressures vary to so small a degree. 

The relation of pressure rise to air-gas ratio in the mixture is clearly 
shown in Fig. 194 for this gas without the addition of neutral. On the same 
figure is added a set of curves on the results of a set of experiments con- 
ducted with carburetted water gas which had for its purpose the determi- 































































































80 






















T 


<o* 


eut 


^ 


,, 




































f 








% 


■ 


































£ 














































£/ 








































































60 






















































c 








1 






































w 






% 






































4^ 








\ 














































V 




























40 










c 


V 


^ A 


S> 






































4 






% 






































j 


7 






~x 


f 




































J& 




v 

O' 


& 




V 






































f» 


^ 




^ 


A 




























20 












& 

&/ 


^ c / 


k> 




































c 


7 < 
1 & 


V 




\: 


^ s 




































7 


g> 


Wa 


terC 


ras 








































S T eut 


ralA 


ddec 




"v 






















































































































< 


) 




2 




4 


\ 




6 




i 






10 




1 


I 




14 




.1 





• 



Parts Air per One Part Gas 
Fig. 194. — Air-gas Mixture Explosion Pressures. 



nation of the explosion pressures for various amounts of neutral added to 
air-gas mixtures of different proportions for the whole range of mixture which 
would explode. These curves show the same general form as that for the 
oil gas and air and indicate clearly that the maximum explosion pressure or 
pressure rise above atmosphere is always obtained for the combining propor- 
tions of air to gas and that the pressure rise decreases as the amount of 
neutral or inactive gases increases. 



HEATING BY COMBUSTION 749 

To facilitate calculations with characteristic analyses of common gases the 
data of Table CXXI at the end of the Chapter have been prepared and illustrates 
another remarkable thing regarding explosive mixtures of these gases with air, 
viz., the small variations in the B.T.U. per cu.ft. of standard air gas mixtures 
in spite of a tenfold or more variation in the B.T.U. per cu.ft. of gas, which 
explains why explosions of so called rich gases give rise to pressures but little 
greater than those for the so-called lean, poor or weak gases. The lowest 
value is about 60, that for blast furnace gas, the highest value is 107, that for 
kerosene, not twice as much, though the calorific of the fuel vapor per cu.ft. 
is for kerosene nearly ten times that for blast-furnace gas. 

In this table gasolene is assumed to be heptane, C7H16; kerosene to be 
dodecane, C12H26; alcohol to be pure methyl, C2H5OH and the hydrocarbon 
illuminants of other gases to be distributed as indicated in the table. It is 
very uncertain just what these hydrocarbons are, but it seems to make very little 
difference so far as the mixture heats are concerned, as the following calculations 
will show: 

For the paraffine series, C„H2 W +2- 



12n 6n , 

rer cent C = — — —x — rs = -s — r~r- W 
12n+2n+2 7n+l 

^ 2n+2 n+1 „. 

Per cent H2 = 77. — r-^ — m = s — ;— r • (0) 

12n+2n+2 7n+l v J 

Neglecting the heats of formation 
B.T.U. per lb.=per cent CXl4,544+per cent H 2 X 51,892 (low value) 

= 14 ,544X6n+5l',892(n+l) = 139,156n+51 ,892 

7n+l 7n+l 



(824) 



(825) 



Cubic feet vapor per pound (standard) = 9 _l o = -|4 _lo • • • ■ (826) 

139,156n+51,892 
B.T.U. per cubic foot } = 7n+l — _ =777>4w+ 289.9 . . .(827) 



vapor (standard) low J 358 

14n+2 

2>n-\- 1 
By the reaction formula, C»H2 n +2H ^ — 02 = nC02+(n+l)H20, it follows 



that 



Cubic feet air per cubic foot vapor = — — =7. 15nx+2.38 . . . (828) 
Cubic feet air -vapor mixture per cubic foot vapor = 7.15n+3.38 . . (829) 



750 ENGINEERING THERMODYNAMICS 

Therefore 

B.T.U. per cubic foot air-vapor mixture = _ * . . . (830) 

7 .i.on-j-o.oo 



= 101 app. for n= 1 
= 105 app. for n= 20 
= 108 app. for n = 100. 



This indicates that from the lightest, methane, to the heaviest of the series, 
the heat per cubic foot of combining or best mixture with air will be between 
102 and 110 approximately and the same will be true for any group of hydro- 
carbons such as make a real oil. 



For the ethylene series C«H2» 

B.T.U. per lb. =20053 

358 
Cubic feet vapor per lb. =-^— 



.'. B.T.U. per cu.ft. vapor = 785n app. (831) 



Cubic feet air per cubic foot vapor — l.lbn .... (832) 

Cubic feet mixture per cubic foot vapor = 71. 5n+l. . . (833) 
Therefore 

B.T.U. cubic feet of mixture = ' ^ ' (834) 

7.15n+l 

= 99 for n= 2 
= 106 for n = 20 
= 106 for n = 100 

This shows the same is true with even narrower limits for this series, hence for 
any hydrocarbon oil, B.T.U. per cubic foot best mixture with air measured at 
standard conditions, will be between 100 and 108, no matter what it is. 

The weight characteristics of the mixture are added to the volumetric in 
Table CXXI for convenience in the use of the relations indicated by the pre- 
ceding equations and summarized as: 

Temperature rise is proportional to B.T.U. per lb. of products or of mixture. (835) 

Pressure rise for explosions is proportional to B.T.U. per cu.ft of initial 
mixture (836) 

In furnace work the temperature rise determination, as well as the final 
temperature, is made along similar lines to that used for the pressure rise due 
to explosions in closed chambers, whether the fuel be solid coal or coke, liquid 
oil, or gas, unmixed, partially mixed with air or completely mixed in explosive 
proportions. High temperature attainment is always facilitated by initial 
heating of the blast or mixture supplied, as the combustion determines the tern- 



HEATING BY COMBUSTION 751 

perature rise and final temperature is the sum of initial temperature and tem- 
perature rise. The final temperature does not rise the same amount, however, 
as the initial temperature, except for low temperatures, indicating the operation 
of the dissociation limitation, but it does rise appreciably, showing that all heat 
suppression is not due to dissociation and may be charged to higher specific 
heats than are believed to hold. This, however, is at present a controversial 
point among physicists the full discussion of which is not worth while here 
in view of the prime purpose of getting numerical results that are reasonably 
close to the truth by whatever means are available. 

In furnaces, especially hand-fired boiler furnaces, the temperature is con- 
stantly fluctuating with the air supply, draft, condition of bed and all sorts of 
minor influences and the diagram, Fig. 195, by the U. S. Geological Survey, 
taken from one of their boiler tests illustrates this admirably. According to 
this diagram the furnace temperature rose to a maximum of about 3000° F. 
and fell to about 2000° F., as limits, due to operating conditions, the rise 
and fall being cyclic with firing. Opening the door, of course, chills the fire 
unless previously the gases had insufficient air. Depositing coal results in 
rapid distillation, as the coal was bituminous, in one case 36 per cent volatile 
and in the other 21 per cent, and the gases in burning raise the temperature 
in the combustion chamber beyond the bridge wall, while that of the bed drops 
due to less air passage and larger per cent of CO formed in it, as well as to the 
heat absorption of volatile distillation. The explanation of variations in simul- 
taneous temperatures in the bed, over it, and in the gas combustion space beyond 
would take too much space here, but are related to the combustion of gas com- 
pared with coke and the air supply above and below the fire. All temperatures 
in the furnace would be lower with anthracite coal especially in its smaller sizes. 

As carbon yields 14,544 B.T.U. per pound when completely burned to CO2 
and in so burning, requires 11.52 lbs. air per pound yielding 12.52 lbs. of products, 
the value of $ in equation (811), assuming the temperature rise to be at the 
high value of 2400° F., will be 



h-h 



Q 



.243w, 



2400 X. 243X12.52 _ ,.„ 

14,544 = ' 5 (837) 



It thus appears that for ordinarily high coal furnace temperatures the heat 
suppression coefficient is of the same order of magnitude as the corresponding 
factor for explosion pressures, but it must be noted that for lower temperatures 
its value will be higher as for example, when a little carbon, oil or gas is burned 
in a large excess of air, limiting the temperature rise. Thus, if there were 100 
per cent excess air in the furnace, the temperature rise would be only 1200° 
for the same heat suppression factor whereas experience shows that the 
temperature will be closer to 2000°, which must be explained somehow, and 
the most rational explanation is the dissociation one, according to which 
the temperature attainable cannot rise much after dissociation sets in. 



752 



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Fig. 195. — Temperatures in a Boiler Furnace. 



H ^ w 
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138 



*3 



HEATING BY COMBUSTION 753 

This discussion shows the essentially approximate nature of temperature 
rise calculations for furnaces and explosions and that the search for correct 
specific heats of mixtures of various constituents in the products at assumed 
temperatures from supposedly correct constants is at the present time a purely 
academic procedure, which in practical work is to be discouraged until such time 
as constants for all factors are determined, including dissociations, which when 
used will give results that at least nearly agree with direct measurements. 

According to the operation of the dissociation limit in furnaces, combustion 
cannot proceed after the attainment of a certain maximum temperature unless 
heat be abstracted, in which case it will proceed to keep up the temperature 
when it would otherwise fall, and there is evidence that this happens in both 
furnaces and gas-engine cylinders, though it is not quite absolute. It has been 
many times observed that the first part of the expansion line in gas engines 
does not follow the same expansion law as the last part, having a slope that 
indicates continued heat addition in spite of the certainty of simultaneous 
heat abstraction of the jacket. This, generally termed the after burning, is 
evidence of heat suppression, followed by liberation when temperature falls 
enough, but as some indicator cards 'do not show the action it is hardly 
conclusive. 

It might be expected in furnaces that if at the time dissociation is limiting 
combustion, the gases be suddenly cooled below their ignition temperature, 
so suddenly as to leave insufficient time for union which would take place by 
gradual heat abstraction, that they would show unburned fuel elements. This 
is always found to be the case in boiler fires and is one of the explanations of 
smoke and of the presence of CO and some hydrogen and hydrocarbons in the 
gases when smoke is much in evidence, even though there is free oxygen 
present at the same time, the sudden cooling being done by contact of flame 
with the cold boiler tubes and plates. In this connection the following analyiss 
of soft coal soots by Sexton is of interest, the free carbon having separated 
out from the hydrocarbon gases at the high temperatures, perhaps by a 
preliminary reduction to ethane and acetylene, and not having time to burn 
either from deficiency of oxygen or dissociation limitations while in the high 
temperature and being prevented later by quick chilling. 

; COMPOSITION OF BITUMINOUS COAL SOOT (Sexton) Per Cent by Weight 

I II III IV 

Carbon 39.0 86.94 68.5 75.3 

Hydrocarbons 14.3 3.3-5.2 4.4 3.9 

Sulphur 4.8 3.2 

Sulphuric acid 4.0 

Ash 36.67 8-9.7 22.7 16.3 

In no boiler furnace is combustion ever complete without excess oxygen 
indicated by free O2 in the analysis of flue gas, and seldom so, even with appre- 
ciable amounts of it, so that analysis of flue gases of boiler furnaces or even other 
furnace fires is a most important practical method of indicating the attainment 
of good or bad fire conditions, leading to correction of faults in adjustment of 



754 ENGINEERING THERMODYNAMICS 

air supply above and below the grates, and at the same time giving a measure 
of the amount of unburnt fuel escaping, the amount of excess air being used 
and the weight of products per pound of coal. Flue-gas analysis is, therefore, 
the indicator of furnace efficiency and it, together with the temperature of 
escaping gases, measures the heat discharged to the chimney and unavailable 
for evaporating water. 

The weight of combining air, excess air and total products, can be computed 
from the volumetric relations of the flue-gas constituents approximately for the 
fixed carbon of the fuel but not so easily for its volatile. Considering the fixed 
carbon alone, the fundamental relation for no excess air is given by Eq. (777), 
Section (7) which is, 

1 lb. C+ 5.759(1+2) lbs. of air give a gas consisting by volume of 

/ 1_ _\[^"_ wro 2 l ( l \ /29.8(C0+C0 2 )\ 

V56.32z+86.12y/ f 6 % 2 \l +s)of N 2 ° T \56.32s+86.12A56.32(l+aON 2 A 

The weight of air per pound of carbon is 5.759(1 +x) lbs. which can be evalu- 
ated numerically in terms of flue-gas volumetric relations when (l-f-z) can be 
found in terms of their CO and C0 2 contents. From these volumetric relations, 

% by volume N 2 = 56.32(l+x) _ 

% by volume CO+% by volume C0 2 29.8 1,wlltIj " ' K °™ } 

Hence, ' (l+^-^x^^^. . . (839) 

Denoting by N 2 , CO, and C0 2 the volumetric per cent of each, 

•'• PoundsairperpoundC= w(coTO) =3 - 1 (cof^)- • • (840) 

If there is excess air there should be no free CO in the flue gases, but as the gases 
flow in streams the oxygen may not have come in contact with the unburned 
CO, its presence indicating unused air though not necessarily in excess chemi- 
cally. It is, however, customary to consider all unused air as excess air 
whether chemically in excess or not. although uncombined air might better 
describe it. 

\ If A = weight of the excess or uncombined air in pounds per pound of 
carbon, then if N 2 ', CO', and C0 2 ' are volumetric per cents exclusive of the 
excess air of the flue gas, 

Pounds air per pound C = 3.1 (pfyxWw ) + A - • • • (8 41 ) 

But the presence of the 2 and N 2 of the extra air will change all the ratios. 
Each pound excess air per pound C will add 12.387 cu.ft. (std.) to the flue gas 



HEATING BY COMBUSTION 755 

of which 21 per cent or 2.701 cu.ft. is oxygen and 79 per cent or 9.686 cu.ft. is 
nitrogen. The total nitrogen will then be [56.32(1 +z)+9.686A] for A pounds 
excess air, and 2.701A cu.ft. oxygen will be with it. The total weight of air 
per pound C is = 5.759(1 +x)+ A, the evaluation of which requires an expres- 
sion for A and (l+#) in terms of volumetric relations, which is not possible 
exactly. By trial the following is found to express the relation fairly well: 

N 2 -f.055O 2 56.32(l+z)+9.686A + .055X2.7A 



or 



CO+CO2 29.8 

= 1.89(l+x)+(||^)A = 1.89(l+a;) + .33A 

= .33[5.7(l+z)+A], 
3N+.1650 2 



CO+CO2 



5.7(l+x)+A (842) 



Similar relations can be found for fuels consisting partly of hydrogen, but 
they are too complex to be of much help, so that the estimation of weight of air 
and excess air from the volumetric flue-gas analysis is possible by a simple 
formula, only for fuels that are all or mainly fixed carbon. The hydrogen relations 
can be worked out by expressions similar to those for producer gas and as 
flue-gas analysis is always made with cold gases none of the H2O formed will 
show in the analysis. There will be added 337 cu.ft. of nitrogen (std.) for each 
pound of hydrogen and it will be derived from 34.64 lbs. of air required. This 
will have the effect of increasing the nitrogen per cent and decreasing the CO2, 
CO and O2 per cents in the flue gases from high volatile coal, oil, or hydro- 
carbon gases, for the same disposition and perfection of carbon combustion. 

The heat loss due to unburned CO in the flue gas can be expressed in terms 
of the volumetric relations of CO to CO2. If the carbon burns partly to CO2 
and partly to CO and x represents the fraction burning to CO2, then (l — x) 
represents the fraction burning to CO and 10,193 (1 — x) is the heat lost due 
to CO per pound of carbon. 

B.T.U. per pound carbon, heat loss, due to unburnt CO = 10,193(1 —x) (843) 

But in the gases there will be 

29.8zcu.ft. of C0 2 per lb. C, 
29.8(1 -x) cu.ft. of CO per lb. C. 

Hence designating by CO and CO2 the respective per cents of each by volume 
CO 29.8(1-3) _ 29.8(1 -x) = 1 _ 



CO+C0 2 29.8x+29.8(l-z) 29.8 

Therefore 



B.T.U. per pound C, heat loss, due to unburned CO = 10,193 ( CQ 4_q ) (844) 



756 



ENGINEERING THERMODYNAMICS 



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O P«b 5 o0 .1° aSB^uaoaad 



HEATING BY COMBUSTION 757 

At this point it is of interest to examine the sort of constituents that are 
ordinarily found in boiler flue gases, and in Table CXXII at the end of the 
Chapter are some careful determinations as compiled by Uehling, that may be 
regarded as representative of the resultant of all the complex conditions and 
reactions in a boiler fire. His conclusions are most reliable and are quoted: 
"((There is a definite relation between the per cent of CO2 and O2 contained 
in the flue gas. The percentage of O2 falls almost exactly in the same ratio as 
CO2 goes up and vice versa. The analysis also shows conclusively that there 
is no relation between either O2 or CO2 and CO. High O2 or low CO2 is no 
evidence that CO will not be present or that the combustion is more complete 
with high CO2 or that low O2 means incomplete combustion. There is a ten- 
dency toward larger CO contents with larger CO2 contents." 

The total per cent of CO, CO2 and O2 together in the U. S. Geological Sur- 
vey tests varied from 16.7 to 20 per cent, and of 1130 analyses only one total 
fell below 17 per cent, four below 18 per cent, 20 between 18 and 20 per cent, 287 
between 19 and 20 per cent, 277 between 20 and 20.8 per cent. The best per cent 
of CO2 varies with the carbon hydrogen ratio of the fuel, being smaller for large 
carbon hydrogen ratios. As a matter of interest in comparison with gas pro- 
ducer work the ratios of CO to CO2 and of CO to (CO+CO2) is added to the 
tables and the curve Fig. 196, by Uehling is also added as it is of great assistance 
in getting a correct idea of the relations. 

Nothing could be more clearly shown than the utter impossibility of com- 
pletely burning without excess air, which carries off heat otherwise usefully 
available, than these curves and analyses, which represent more than usually 
good adjustment of fires, though expert adjustment with no end in view but 
perfect combustion, would improve results to some extent. In actual working, 
boiler capacity is just as important as efficient combustion and actual condi- 
tions represent attainable compromises. 

Prob. 1. The flue-gas analysis taken on a boiler trial gave 2 = 10 per cent; C0 2 = 10 
per cent; H 2 =80 per cent. If the coal used consisted of 80 per cent C and 20 per cent ash 
and the stack temperature was 400° F. above the room, what was the stack loss per 
pound of coal? 

Prob. 2. 1000 heat-units are added to a pound of gas at 32° F. at constant pressure. 
The gas consists of 30 per cent CO, 40 per cent N 2 , 5 per cent C0 2 , and 25 per cent H 2 . 
Calculate the final temperature on basis of constant specific heat and give what you 
consider a likely value for the actual temperature. 

Prob. 3. A pound of air is compressed from 1 to 5 atmospheres above atmosphere 
and then 1000 heat-units are added at constant volume. What will be the final tem- 
perature? 

Prob. 4. A gas requires 5 cu.ft. of air to completely burn it. Its heating value is 
900 B.T.U. per cubic foot under standard conditions. A mixture of one part of this 
gas with 5, 6 and 7 cu.ft. of air at a temperature of 100° F. and at a pressure of 25 lbs. per 
square inch gage is ignited in a closed vessel. What is the probable pressure after com- 
bustion? 

Prob. 5. A pound of coal gave the following analysis: C=70 per cent; H 2 =5 per 
cent; 2 =5 per cent; ash =20 per cent. If twice the air chemically needed were 



758 ENGINEERING THERMODYNAMICS 

supplied, what would be the maximum temperature as calculated on basis of constant 
specific heats? What would be the flue-gas analysis? 

Prob. 6. In a boiler test it was found that 20 per cent of the heat in the coal went 
up the flue as sensible heat and in unburned carbon. The coal contained 13,500 B.T.U. 
per pound. The flue gas analysis showed C0 2 = 12 per cent, CO =6 per cent, 2 =2 
per cent, N 2 =80 per cent. If the coal is assumed to be 80 per cent C and the boiler- 
room temperature is 80° F., what was the flue temperature? 

Prob. 7. The pressure in a gas-engine cylinder before ignition was 70 lbs. gage 
and after ignition 350 lbs. gage. The temperature before ignition was 300° F. For 
a mixture containing 110 B.T.U. per cubic foot under standard conditions, what was 
the ratio of observed to calculated pressure rise? 

Prob. 8. Take an analysis of coal gas from the Table CX and on the basis of there 
being .5 of the calculated pressure rise, find the pressure exerted on the side of a room 
50'XlO' if there should be an explosion of the gas in it with atmospheric pressure and 
temperature before the explosion and with best with mixture. 

11. Temperature of Ignition and Its Variation with Conditions. Limits 
of Proportion Air-Gas-Neutral, or Detonating Gas and Neutral, for Explosive 
Combustion of Mixtures. Limits of Adiabatic Compression for Self-Ignition 
of Mixtures. It is only with gaseous fuel previously mixed with measured air 
that best combustion results can be obtained, perfectly complete combustion 
without excess air, and this is easily managed in engine cylinders. Gas and oil 
and powdered coal fired boilers may approximate this, but with the burners in 
common use explosive mixtures cannot be used without endangering the 
settings should the flow be momentarily interrupted and then continue, and just 
such serious accidents have happened. With specially designed burners and 
furnaces, explosive combustion is perfectly feasible in boiler and other fires but 
as yet these are not widely used. Mixtures of fuel and air are explosive at the 
chemical proportions and for a considerable range beyond, usually more on 
the excess fuel side than that of excess air. Two of the most authoritative 
tables of limits, those of Eitner and Bunte, given below, are relatively 
but not absolutely correct. This is true because the limit of explosive pro- 
portions is said to be reached when a mixture will not explode and this is as 
much a question of how the ignition is attempted, as it is one of proportions. 
Flames will ignite mixtures that will not explode with electric sparks, some 
sparks will ignite mixtures that others will not, and mixtures will ignite when 
a flame is plunged into the mass when they cannot be ignited at a touch hole 
or at one corner of their chamber. 

When, however, the following experimental observations are considered, 
the complexity of the ignition becomes more apparent. With carbon it is 
known that the state determines the temperature at which combustion proper 
sets in, it being lower for porous varieties of charcoal, higher for denser vari- 
eties and very high for graphitic forms, but it is also known that all forms will 
react with oxygen, yielding carbon dioxide at some temperature. This leads 
to a suspicion that there is no such thing as the ignition temperature of carbon 
and probably also for other fuel elements and compounds. Consider the 



HEATING BY COMBUSTION 



759 



hydrocarbons which in flames burning without previous air mixture become 
decomposed; ethylene, for example, according to Sexton splitting up into 
acetylene and methane 3C 2 H4 = 2C 2 H2+2CH4 between 1500° F. and 1800° F. 
or ordinary furnace temperatures, and methanere acting at 2000° F., 2CH 4 = 
C 2 H 2 +3H 2 and finally acetylene C 2 H 2 = C2+H2 at 2430°F. Who can say 
in view of this possibility just what takes fire when ethylene is ignited, is it 
ethylene as a whole or is it methane or acetylene or hydrogen separated out 

































-p 


































1 
































































; 



































































































































































































1300 


















































































































































































































































































































































































































































































































































































































































































1200 










































































































































7 




























































L 






/ 






























I 






























• 




j 


( 




























/ 
































' 




/ 






























/ 


































1 


/ 






























/ 


































\ V 


f 




























I 




































*> 






























1 

































































1 




































1100 ' 






























































































/ 








































L 
































































1 


























/ 


/ 








































, 
























/ 










































\ 






















/ 












































' 






















/ 
































































7 














































\ 




















r 















































V 


















/ 














































1000 


s 
















/ 


















































\ 














/ 




















































\ 












V 
























































K 


^ 




S 












0= Hydrogen 

* = Carbon Monoxide 






























V 


i— 


















































































































































































































































































































































































900 i-l. 






L 










I 










i 


(T 


T" 






t 






. 

























Parts 2 per Part of \ c q 
Fig. 197. — Temperature of Ignition for H 2 , and CO, mixed with 2 . 



by the attempt to ignite? There is no answer to the question, but a little 
light can be thrown on it by some recent experiments prompted by previous 
work of Davy followed by Mallard and LeChatelier, who observed that for gaseous 
mixtures the ignition temperature true or apparent, was different for different 
proportions of air and fuel and likewise still different when neutrals were present. 
Results, however, were quite variable with the method of procedure and quite 



760 ENGINEERING THERMODYNAMICS 

inconsistent and it was not till Falk, by rapidly compressing a previously 
determined mixture, heating it by compression uniformly throughout to igni- 
tion, the temperature of which was calculated by the adiabatic law, that any 
progress could be said to have been made. His results are shown graphically 
in Fig. 197 for carbon monoxide-oxygen and hydrogen-oxygen mixtures, which 
are the principal fuel elements in power producer gas. 

From the curve it is clear that there is a regular variation of ignition temper- 
ature with proportion and a minimum point for each fuel which corresponds to 
the H2 + O2, proportions for hydrogen, indicating that it reacts in two stages 
first to H2O2, and later to 2H2O, whereas, for carbon monoxide the minimum 
point corresponds to the detonating mixture. 

When the combustible and its oxygen must be ignited in the presence of 
neutral matter, like excess oxygen or excess combustible which when not com- 
bining are in effect neutral, or real neutrals like nitrogen, the ignition tem- 
perature may be calculated from the two following formulas Eqs. (845) and 
(846), working out each independently of the other and accepting whichever 
is the lower value. 

Ignition temperature 

f [Ignition temperature for H2 with O2" alone] ) 



andN 2 2 = l |[ 54 X volume of inert gases "I '\ . (845) 

^ I L Vol. of H2 or O2 (whichever is smaller) J J 



Ignition temperature 

[Ignition temperature for CO with O2 alone] ] 

144 X volume of inert gases ! r . (846) 

Volume of CO J 



of CO with 

2 and N 2 



+ 



By the use of these equations the ignition temperature for five samples 
of hypothetical producer gase have been calculated as an example, the 
various steps being set down in the following Table XCIII. 

The results show that in all but the first case with H2 = 12.6 per cent, which 
increases to 26 per cent, it is the hydrogen that controls the ignition tem- 
perature and that the difference for wide range of proportions is really not 
great. 

The same apparatus was used for gasolene and ethyl alcohol and gave the 
following values which were fairly constant for explosive mixtures through a 
100 per cent range of proportions. 

(Ignition temperature of gasolene mixtures with air) = 986° F. (a) [ (qaj\ 
(Ignition temperature of alcohol mixtures with air) = 1292° F. (6) J ' 

These temperatures of ignition would never be reached by compression 
of cold charges adiabatically to the pressures used in engine cylinders, yet for 
even small compressions the charges ignite themselves or preignite, proving 
that the compression temperatures are very high in such cylinders and must 
become so by reason of heat added to the charge before or during compression. 



HEATING BY COMBUSTION 



761 



Table XCIII 
CALCULATED IGNITION TEMPERATURES FOR PRODUCER GAS 







Producer Gases, Composition by Volume. 


Volumes O2 
to Burn 100 
VolumesGas. 


Volumes Air 
to Burn 100 
Volumes Gas. 


Total 
Volume 
Mixture. 




C0 2 


CO 


H 2 


N 2 


1 


.0 


38.9 


12.6 






48.5 


25.75 


123.08 


223.08 


2 


3.7 


34.2 


16.1 






46.0 


25.15 


120.22 


220.22 


3 


7.2 


29.9 


19.4 






43.5 


24.65 


117.82 


217.82 


4 


10.7 


25.6 


22.7 






41.0 


24.15 


115.44 


215.44 


5 


13.6 


21.5 


25.8 






39.1 


23.65 


113.05 


213.05 


No. 


Volume of Inert Gas. 
When 


n' 


n 


Temperature for Zero 
Inert Gas from Fig. 197. 


Calculated Ignition Tem- 
perature for the Mixture. 




CO Burns. 


H2 Burns. 


For CO. 


For H 2 . 


Based on 
CO. 


Based on 
H 2 . 


1 


164.73 


204.18 


4.13 




16.2 




1112 


982 


1708 


1857 


2 


168.42 


196.07 


4.93 




12.18 




1118 


963 


1824 


1620 


3 


172.97 


188.72 


5.93 




9.73 




1128 


979 


1984 


1503 


4 


177.04 


181.39 


6.92 




7.99 




1148 


954 


2146 


1385 


15 


180.80 


174.35 


8.4 




6.76 




1175 


957 


2385 


1322 



As the temperature rise before compression is multiplied by compression, it 
may be assumed that this is the controlling factor and the following calcula- 
tion shows to what extent this may be expected to take place. 

Let T\, ti, and T2, to be absolute and scale temperatures before and after 
compression to ignition temperature from one atmosphere to P2 
atmospheres absolute, adiabatically. 
Then assuming the exponent, s = 1.4. 



Ti 



,29 



and Ti = 



(ft 



.29 



From this the initial temperature for mixtures that really preignite at a 
known compression in engine cylinders can be calculated and three such cases 
are given below. 

CALCULATION OF INITIAL TEMPERATURE FROM PREIGNITION PRESSURES 

IN GAS ENGINES 



Fuel. 


Compression 

Ratio in 

Cylinder when 

Preigniting, 

P2 

Pi 


Temperature of 
Ignition, Falk. 


Corresponding Initial Tem- 
perature. 




T 2 
Absolute F. 


Ti 

Absolute F. 


tx 
Scale F. 


Producer gas < 
Alcohol 


fCO 29.9 

H 2 19.4 

vC0 2 7.2 


9 

13 

7 


1080+460 = 1540 

1292+460 = 1752 
986+460 = 1446 


816 

833 

824 


356 

373 


Gasolene 


364 







762 



ENGINEERING THERMODYNAMICS 



This shows that in using these fuels properly mixed with air in cylinders 
the amount of compression allowable without self -ignition varies with the fuel 
mixture ignition temperature and with the initial temperature, which cannot be 
allowed for the three fuels above to rise over 350° F., and does reach pretty 
close to this value, probably 300°, as some heat will be derived from the walls 
during compression. This temperature, when charges are taken in at about 
60° F. on the average, is a measure of the heating effect of the hot walls during 
suction, coupled with the heat added by the residue hot gases from a previous 
explosion, and left in the clearance. 

In the following Table XCIV are given some of the usually practiced limits 
of compression in gas engines, which, of course, are subject to considerable 
variations with type of engine and its management, and which are controlled 
as much by cylinder heating as by the natural temperatures of ignition of the 
mixtures themselves. 

Table XCIV 
COMPRESSIONS COMMONLY USED IN GAS ENGINES 



Fuel. 



Gasolene. . . 
Gasolene. . . 
Kerosene . . . 
Kerosene. . . 
Natural gas . 
City gas ... . 
Producer gas 
Blast gas . . . 



Type of Engine. 

•* 

Automobile 

Stationary 

Hot bulb, 250-500 r.p.m 

Vaporized before entering cylinder 

Medium and large engines 

Small engines 

Medium engines 

Large engines 



Compression, 

Pounds per Square 

Inch Absolute. 



60-95 
70-100 
30-75 
45-85 
75-130 
60-110 
100-160 
120-190 



Average Compres- 
sion, Pounds per 
Square Inch Abs. 



75 

90 

60 

65 

115 

85 

135 

155 



The setting on fire of a combustible, or more scientifically, its ignition, is, 
therefore, not the simple operation that it might seem, especially when the fuel 
is in the form of explosive gaseous mixture. It has been the custom to say 
that the fuel, when raised to its ignition temperature in the presence of oxygen, 
will at once begin to burn and ignition temperature tables like the following 
Table XCV are common, though probably wrong, surely so for gases. 

Not only do the Falk data prove the controlling influence of proportions 
on ignition temperatures for gases but the following Table XCVI from Violette 
proves that for solid charcoal that trie ignition temperature is dependent on 
its origin, which may' be assumed to control its state. In this case the prime 
variable is the temperature at which the charcoal was formed by wood distilla- 
tion, which it has been pointed out controls the density and character of the 
residue and probably similar conditions maintain for other fuels. 

Temperatures of ignition for explosive gaseous mixtures are intimately 
related to the limiting proportions for explosive combustion, because inflamma- 
tion will propagate or pass of itself, from particle to particle in the mass, only 



HEATING BY COMBUSTION 



763 



when the heat liberated by combustion at one point is able to raise the next one to 
its temperature of ignition. It is impossible yet, to state algebraically just what 
this relation is, but in connection with it, an examination of some data on 
gas and air mixtures showing what proportions will and what will not explode 
is worth while. 

Table XCV 
IGNITION TEMPERATURES, °F 



Substance. 

Carbon, C 

Soft coal 

Anthracite 

Peat 

Lignite dust 

Hydrogen, H2 

Hydrogen, H 2 

Hydrogen, H 2 

Carbon monoxide, CO. . 
Carbon monoxide, CO. . 
Carbon monoxide, CO. . 
Methane, CH 4 



Ignition Temperature. 



752 

600 

750 

430 

300 

1077 

1124 

1031 

1253 

1347 

1211 

1212 



(Sexton) 



(Strohmeyer) 

(Olsen) 

(Meyer) 

(Le Chatelier) 

(Allen) 

(Meyer) 

(Le Chatelier) 

(Allen) 



Substance. 

Methane, CH 4 . . 
Methane, CH 4 . . 
Ethane, C 2 H 6 . . . 
Ethylene, C 2 H 4 . . 
Ethylene, C 2 H 4 . . 
Propylene, C 3 H 6 . 
Acetylene, C 2 H 2 . 
Acetylene, C 2 H 2 . 
Propane, C 3 H.\ . 
Alcohol, C 2 H 5 OH 
Coal gas 



Ignition Temperature. 



1201 
1213 
1141 
1124 
1124 

940 
1038 

896 
1017 
1292 
1100 



(Meyer) 
(LeChatelier) 
(Allen) 
(Allen) 
(Meyer) 
(Allen) 
(Allen) 
(Robinson) 



(Robinson) 



Table XCVI 

VARIATION OF IGNITION TEMPERATURE 
OF CHARCOAL— (Violette) 



Temperature of Distilla- 
tion at which Charcoal 
was made, ° F. 


Temperature of Ignition 

of the Charcoal in Air, 

F 


520 


650-680 


570 


680-715 


600 


680-700 


800 


750 


1800-3000 


1110-1500 


Above 3000 


2300 



In the previous section some curves were given to show the explosion pres- 
sures for various mixtures of air, gas and neutral or inert matter, which curves, 
stopped at points where the mixture was non-explosive, and the corresponding 
proportions can be read off directly. A series of values for the limiting pro- 
portions of air-gas mixtures is given in Table CXXIII, at the end of the Chapter, 
showing wide differences for different fuels with no apparent relation between 
them, nor are any controlling influences indicated, though some are really 
known to exist. 

Pressure and temperature variations act irregularly, but in general tend 
to widen the limits; rise of temperature and pressure, more particularly the 



764 



ENGINEEEING THEEMODYNAMICS 



latter, will render explosive, proportions otherwise not so, while neutral dilution 
always narrows the limits. A series of experiments by the writer indicated a 
tendency toward a limit of explosibility when the active or detonating mixture 
was equal to, or greater than, about 17 per cent of the total, or the neutral 
inactive gases not in excess of 83 per cent. This active mixture consists of 
the combustible constituents of the gas with just the right amount of oxygen 
to burn them, and is generally called the detonating gas or detonating mixture. 
All excess oxygen or excess fuel is classed as neutral and the following table 
shows the relations in various mixtures of (air-gas-neutral) of the detonating 
mixture to the whole mixture. 

This limitation of explosibility by limiting quantities of detonating mixture 
or total inactive constituents noted above for all ranges of one gas, was 
observed by Bunte to air-gas mixtures alone for all kinds of gases with a 
comparatively small range of variation, as in Table XCVII. 

Table XCVII 

PER CENT DETONATING MIXTURE IN TOTAL AIR-GAS MIXTURE AT 
EXPLOSIVE LIMITS OF PROPORTION (Bunte) 



Gas 

Carbon monoxide 

Hydrogen 

Water gas 

Acetylene 

Coal gas 

Ethylene 

Alcohol 

Marsh gas 

Ether 

Benzene 

Pentane 



When Air is in Excess. 



24.75 
14.17 
18.80 
11.72 
17.40 
16.40 
15.00 
18.30 
19.25 
19.87 
21.60 



When Gas is in Excess. 



15.78 
21.16 
20.94 
14.03 
31.15 
23.91 
24.17 
27.47 
22.61 
22.25 
22.47 



These figures taken in conjunction with those for large neutral additions 
suggest the possibility of a universal law for all explosive gaseous mixtures what- 
ever the fuel or the nature of constituents associated with it, that explosion is no 
longer possible when there are sufficient inactive particles of gas present to prevent 
a heat propagation that will allow one group of burning fuel and oxygen particles 
to set fire to the next separated from it by an insulating layer of dead gas or even 
allow the oxygen to reach the fuel, and that this per cent of inactive is nearly constant 
for all. 

One remarkable thing brought out by the table illustrating this general 
law is the narrowness of the range for rich hydrocarbons, especially on the 
excess air side. Practically no excess air is permissible, a fact that indicates 
the necessity in gasolene and kerosene engines for most accurate adjustment 
of proportions, without which there is almost a certainty that excess fuel will 
be used by an engine operator, all excess being direct waste. 



HEATING BY COMBUSTION 765 

Prob. 1. Acetylene gas is escaping into a room which is 15' X20 / Xl0 / . How 
many cubic feet of gas must escape into the room and mix with the air before there 
is a possibility of an explosion? Should there be' nothing to ignite the mixture, how 
many cubic feet of the gas must there be in the room before the upper limit of explosi- 
bility is reached? Should the gas have been coal gas, what would have been the 
quantities? 

Prob. 2. A sample of gas gave the following analysis by volume: CO =35 per cent; 
C0 2 =6 per cent; H 2 = 14 per cent; N 2 =35 per cent. If this gas be mixed with the 
combining proportion of air to what temperature must it be heated to cause it to 
ignite? 

Prob. 3. A gas engine is running on a natural gas which may be considered as all 
methane. The combining proportions of this gas with air are in the ratio of 9.5 to 1. 
Taking the limits of proportion for explosive mixture from the general law what would be 
the cubic feet of air per cubic foot of gas for the leanest and richest mixture which 
would still ignite? 

Prob. 4. Gasolene is stated to have an ignition temperature of 986° F. If the 
temperature of the mixture in the cylinder of an engine using this fuel is 400° F. at 
the beginning of compression, what will be the maximum compression which can be 
carried without preignition? The value of s may be taken as 1.4. 

Prob. 5. If instead of gasolene there be used in the engine of Prob. 4, the following 
gas, by how much could the compression be raised without preignition occurring? 
CO =37.6 per cent; H 2 = 11.3 per cent; C0 2 =3.7 per cent; N 2 =47.4 per cent. 

12. Rate of Combustion of Solid Fuels with Draft. Propagation Rates, 
Normal, and Detonating for Explosive Gaseous Mixtures. The last com- 
bustion characteristic to be examined is the rate of combustion or speed with 
which the combination, once started, proceeds. In boiler fires the coal rests 
on grates and the rate with which it may burn determines the amount of steam 
that the boiler can make per hour, assuming a constancy of heat losses to flues 
and setting. When gases or oils are burned under boilers or in other furnaces 
the rate of combustion determines the size of furnace necessary for the combus- 
tion of the requisite fuel for the duty expected of the heat, while in gas engine 
cylinders the time of combustion is almost vanishingly small if efficiency is to 
be secured. For the latter case the explosion should always be as near complete 
as possible before expansion begins, and should not start too soon toward the 
end of compression or the work of compression will be too great. To get an 
idea of the time available, assume an automobile engine running at 1200 R.P.M. 
or 20 revolutions per second, completing one revolution in .05 second. Then 
if ignition and combustion must take place during 18° of crank rotation near 

18 
the dead center or — = .05 of the revolution the time for the whole process to 

be completed is . 05 X. 05 = .0025 or about — — second, for a path not over 6 ins., 

400 

corresponding to a flame speed of .005 second per foot or 200 ft. per second. The 
very largest engine with a cylinder of 4 ft. diameter, turning at 60 R.P.M. or 1 
revolution per second, for the same crank angle requires .05 second, whence 



766 



ENGINEERING THERMODYNAMICS 



.05 
for 1 ft. of mixture the speed must be --— = .012 second approximately and the 

flame speed about 85 ft. per second. 

By elaborate apparatus Mallard and LeChatelier photographed the progress 
through a tube, of a flame passing through an explosive mixture of known char- 
acteristics. The photographic plate was moved across the tube during the flame 
passage giving a picture shown in Fig. 198 which is really a diagram .of speed of 
propagation. Horizontal distances represent time, while verticals represent length 
of tube traversed or linear travel of flame through the mixture. The slope 
of the line, or distance (vertical), divided by time (horizontal), is the speed 
of propagation and as the line varies in slope the speed of flame propagation 
is a variable quantity. It appears, however, that at the beginning of combustion 
in the open tube the speed is first uniform then undulating or wavelike and some- 
times at the end, very great, as indicated by the nearly vertical record line. 
This rapid mode of propagation is termed the explosive, or better, the detonating 
wave and is always reached in good mixtures if there is distance enough to be 
traversed, or if there is agitation of the mixture or pressure waves developed 
equivalent to agitation, and is quite common in engine cylinders. It was at 
first believed to follow the speed law of sound or other impact waves in gases, 
but Dixon has shown that this is not true, the speed being much greater than 
that of sound but it is constant. 

Therefore, in explosive mixtures the rate of combustion is at first uniform and 
slow, and rapidly accelerates to a high constant value. The following Table 
XCVIII, gives Berthelot's and Dixon's experimental determination in feet per 
second for the detonating waves, compared with sound waves. 

Table XCVIII 
VELOCITY OF EXPLOSIVE OR DETONATING WAVES 



Mixture. 


Velocity in Feet per Second. 


Authority. 




Detonating Wave. 


Sound Wave. 




2H 2 +0 2 

2CO+0 2 

CH 4 +20 2 

C 2 H 4 +30 2 


9220 
3583 
7504 
7248 
8143 
7202 
8961 
5693 
11588 
5591 


1696 
1076 
1142 
1050 
1060 
938 


Berthelot 
it 

1 1 

tt 


2C 2 H 2 +50 2 


1 1 


C 2 H 4 +20 2 

C 2 H 2 -{-0 2 

H 2 +C1 2 


1 1 
Dixon 






2H 2 +0 2 +6H 2 

2H 2 +0 2 +50 2 




i ( 




i t 







Various attempts have been made to derive a fundamental formula for the 
speed of the wave in terms of physical constants of the mixture, and while 



HEATING BY COMBUSTION 



767 



V_, 



Distance in Tube 



/ 



^■fe 



Distance in Tube 




Distance in Tube 

I 

1^ 



Distance in Tube 

IT ^ 



Distance in Tube 




Distance in Tube 




Distance in Tube 



k 



Distance in Tube 



~o 



tf 



ft 

I 

o 



o 
■ft 

X 

W 

fcJO 

2 
H 



ft 
o 



708 



ENGINEERING THERMODYNAMICS 



some are reasonably satisfactory no one of them seems to fit all cases or give 
results agreeing with measurements. If in cylinders, as often happens, the 
combustion seems to be too slow, then it is certain that the wave has not been 
set up and it is as a matter of fact undesirable to have it develop because of the 
accompanying shocks. 

For the slow, uniform propagation the best data on the conditions and 
rates are given by Mallard and LeChateiier, whose results show that the walls 
have an influence, the nature of which is a cooling one, abstracting heat at various 
rates compared to that of generation. When cooling is strong enough to take heat 
away as fast as it is generated, propagation of explosion is impossible and this 
always takes^place in narrow slits between plates, in very small tubes and in cold 
wire gauze screens. Rise of temperature always increases the rate, thus for H2 
and air with 30 per cent H2, 



r = 10.75 ft.sec. at 60° F. 
r = 14.26 ft.sec. at 212° F. 



The greatest single influence in fixing the rate of propagation is the nature and 
proportion of the mixture as shown by the following figures of Table XCIX. 

Table XCIX 

RATE OF PROPAGATION (UNIFORM) FEET PER SECOND FOR DIFFERENT 
PROPORTIONS IN MIXTURES AT ATMOSPHERIC PRESSURE 



H 2 per 100 Vols. 


r 


C 2 H 4 per 100 Vols. 


r 


11.65(C 2 H 4 )+z 


r 


6 





5.6 





x = . 5N 2 


1.38 


10 


1.97 


6.0 


.098 


" .0N 2 


.984 


20 


6.40 


10.0 


1.37 


" 1.4N 2 


.623 


30 


10.82 


12.0 


2.01 


" .5C0 2 


1.02 


40 


14.34 


14.0 


1.18 


" 1.0CO 2 


23.48 


50 


11.32 


16.0 


.33 






60 


7.54 


16.2 









70 


3.61 










80 














In all cases neutral dilution has a strong influence on reduction of rate of 
propagation, reducing it for all air-gas mixtures to zero, when the limit of 
proportion has been reached at somewhere about 83 per cent of total neutral. 
In general the uniform or slow normal rate of propagation r never exceeds 
14.1 ft. per second and occurs with 40 per cent of gas or 30 per cent excess H2; 
for C2H4 and air the maximum is 2 ft. per second and occurs in the mixture 
containing 12.2 per cent gas or an excess of 42 per cent; for CO and air the 
maximum 6.56 ft. per second, and for illuminating gas and air the maximum 
is 4.1 ft. per second, occurring when the mixture contains 15 per cent excess 
gas, all at atmospheric pressure. 



HEATING BY COMBUSTION 



769 



These figures show that in engine cylinders, pressure and temperature rise 
by compression, are necessary to make the mixtures burn in the time available, 
as the rate of combustion at atmospheric pressure is not high enough. Of 
course, if the detonating wave is set up, there will be more than enough time. 

Rate of combustion of coal on grates is measured in pounds per square 
foot per hour, and if the coal burned as fast as the air flowed through the 
bed, the rate should increase directly with the square root of the draft, but 
be different for every different condition of porosity of bed, thickness of bed, 
or in general, resistance to flow. As a matter of fact coals with much volatile 
are burned as much above the grate, in the gasified form of hydrocarbon and 
carbon monoxide as on it. The gas thus made may easily equal in weight that 
of the fixed carbon left on the grate. In this case the rate of combustion will 
not vary with the square root of the draft, and especially if the coal cakes some 
the rate will be constantly varying from minute to minute. The following 
figures, Table C, are commonly quoted, but judged from present-day practice 
are considerably wrong, as later figures will show. 

Table C 
RATES OF COMBUSTION OF COAL 



Type of Furnace. 



Stationary boilers 
Marine boilers. . . 

Locomotive 

Lead smelting. . . 
Copper smelting. . 
Puddling smelting 
Steel smelting. . . . 



Pounds Coal per Hour per Square Feet Grate. 



Rankine. 



4- 16 
16- 24 
40-120 



Griiner. 



8- 20 

81-102 
12- 16 
15- 30 
20- 30 
41- 81 



On the assumption that the coal will burn as fast as the air passes through 
the bed, and that air will pass with a velocity proportional to the square root 
of the draft measured between ash pit and furnace, then 

(Lbs. coal sq.ft. grate per hr.) =CV (draft in inches of water). . . (848) 

The constant C may be determined from experiment approximately, as 
it is not a true constant for a number of reasons. Consider a given coal on a 
given grate, then C will vary with thickness of fire, age of the fire, amount and 
condition of ash and clinker and amount of cake on the surface if the coal 
is of the caking variety. A change of size of coal with no other change will 
affect the air flow and the speed of carbon union with the air that does pass, 
and a change in proportion of surface and bottom air will also modify C, as 
will a change in grate. A grate that breaks up a bed like the step stoker will 



770 



ENGINEERING THERMODYNAMICS 




co § 3 - eo 

( a ^A\ JO S3tj?m) ^Td qsy pub aouuunj uaaAvpq pojmtoH ywa JO aw(\j 



HEATING BY COMBUSTION 



771 



pass much more air than one that does not like a chain grate, and a fireman 
that hand slices often, lets more air pass than one that does not. A coal that 
has much volatile may burn as much above as on the grate, and its rate of 
combustion be very different at the same draft from another with less volatile, 
everything else being the same. 

In spite of all these influences the rate does follow fairly well the relation 
of Eq. (848), though the constant must be selected with care. One general 
statement of the relations for a variety of coals is given by the Stirling Boiler 
Company, compiled from tests on their boiler and is reproduced in Fig. 199 
with some additions. Assuming the middle rate reported for each coal to be 
correct, the square root curve is drawn through this point for higher and lower 























































































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"Rate of Combustion libs, per Sq. Ft. Per Hr. 



50 



Fig. 200. — Rate of Combustion of Anthracite Coal. 

drafts and from it the following constants, Table CI, are found for hand- 
fired furnaces: 

Table CI 

CONSTANTS OF PROPORTION ["C" in Eq. (834)] FOR 
RATE OF COAL COMBUSTION] 



Values of C in Rate of 
Combustion, Eq. (848). 


■ ■ 1 

Coal name. 


17.3 
24.5 
31.5 
81.6 
50.2 
59.3 


Anthracite rice 
No. 1, Anthracite buckwheat 
Pea anthracite 
Rnn-of-mine bituminous 
Run-of-mine semi-bituminous 
Bituminous slack 



772 



ENGINEERING THERMODYNAMICS 



To show how far actual cases depart from these averages, the results of 
some special determinations are separately plotted in Fig. 200 for anthracites, 
and in Fig. 201 for bituminous coals, from Tables CXXIV and CXXV at the end 
of the Chapter. The highest rates and drafts are used on locomotives, next 

































































































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Rate of Combustion Lbs. per Sq. Ft. Per Hr. 

Fig, 201. — Rate of Combustion of Bituminous Coal. 



120 



140 



on war vessels, especially torpedo boats, and the lowest on stationary boilers 
with natural draft, and all cases are illustrated by test data. 

Prob. 1. A boiler rated at 180 H.P. has 60 sq.ft. of grate area. If 70 per cent of 
the heat of the coal is absorbed by the water, what must be the rate of combustion 
for No. 1 buckwheat containing 11,000 B.T.U. per pound. What draught will be 
required for the rate? When the capacity was raised to 270 HP. the efficiency fell 
to 60 per cent. What draught would then be required to burn the necessary coal? 

Prob. 2. A gasolene engine has a cylinder 5x5 ins. and runs at 1000 R.P.M. 
Ignition occurs 12 degrees ahead of dead center. What must be the velocity of flame 
travel to give complete inflammation at dead center with cylindrical clearance 20% of 
the stroke in length? 

Prob. 3. A stationary boiler had a grate area of 60 sq.ft. and was rated at 180 H.P. 
A locomotive boiler having a grate area of the same amount was rated at 500 H.P. 
Assuming that both boilers were fired with run-of-mine bituminous and that the 
respective efficiencies were 65 and 50 per cent, what draughts would be needed in each? 



HEATING BT COMBUSTION 773' 

Prob. 4. The coal No. 16 in the general coal table at the end of this chapter is being 
fired at the rate of 50 lbs. per square foot of grate surface. What must be the rate 
of firing for coal No. 121 of the same table to produce the same heat? 

Prob. 5. 100,000 ft. of natural gas are burned per hour under a boiler. Assume a 
natural gas from the general table at the end of the chapter and an anthracite coal 
from the coal table and find what rate of combustion must be maintained with the 
coal on a grate area of 100 sq.ft. to develop the same heat? 

Prob. 6. Coal No. 100 of the general table is being burned at the rate of 75 lbs. 
per hour per square foot of grate surface on a grate containing 120 sq.ft. What draught 
will be needed? What grate area will be needed to produce the same heat with anthra- 
cite pea coal of 12,000 B.T.U. per pound burned at the same rate? 

Prob. 7. Ten barrels of fuel oil are burned per hour under a boiler. At what rate 
must bituminous coal be burned under another boiler having 60 sq.ft. of heating surface 
to produce the same heat? Take any fuel oil from oil table and bituminous coal from 
coal table. 

Prob. 8. No. 2 buckwheat anthracite, having a heating value of 11,600 B.T.U. , 
is being burned with a draught of .7 in. on a grate with 80 sq.ft. of surface. On another 
grate run-of-mine bituminous is being burned with .5 in. draught. If the calorific 
power of the latter is 14,000 and the rate of combustion follows the square root law, 
what is the necessary area of the second grate for equal heat generation? 

Prob. 9. W^hat would be the relative rates of combustion for equal generation for 
coals Nos. 10, 25, 61, 113, 162, 185, 201 of the general table? 

13. Steam Boiler Evaporative Capacity and Horse-power. Horse-power 
Units, Equivalent Rates of Evaporation and of Heat Absorption. Factors of 
Evaporation. Relation between Absorption Rates and Rates of Heat Genera- 
tion. Influence of Heating and Grate Surface, Calorific Power of Fuels, Rates 
of Combustion and Furnace Losses. It seems a little curious to have to confess 
that the steam boiler, which has been built in one form or another for nearly 
two hundred years, is the least understood to-day of all the important structural 
elements of plants dealing with heat for power or other purposes, yet such is 
the fact. Burning coal under and passing the furnace gases over surfaces 
holding boiling water looks like the simplest sort of process, and if it made no 
difference how much steam could be made from a square foot of surface or 
from a pound of coal it would be simple. 

The difficulties in dealing with boiler processes are all concerned with the 
establishment of some fundamentally sound relations for the capacity of 
heating surface to absorb heat in terms of the amount of heat developed by 
the burning of coal, oil or gas in the furnace, when the dimensions are known. 
After all these years of experience, and in spite of much theorizing and thou- 
sands of recorded tests it is not possible to-day to calculate from any fundamental 
relations either the amount of steam that will be developed per hour with a 
given fuel, or the weight of steam that will be produced per pound of fuel. Of 
course, from empiric relations this can be done, because nothing is simpler 
than a comparison of the boiler in question and its fuel, with exactly or nearly 
similar ones for which tests have been made, yet even here it is possible to 
make blunders by forgetting differences in operating conditions such as avail- 



774 ENGINEERING THERMODYNAMICS 

able draft or methods of firing. It must be said, therefore, that there is no 
generally accepted fundamental theory of either steam boiler capacity or 
efficiency, but there are some relations between certain important factors, that 
are established, so a discussion of steam boilers must be almost entirely con- 
fined to these relations which cannot yet be grouped together to constitute a 
general theory of the subject. To illustrate, it is possible to explain why a 
given boiler gives more steam with one coal than with another, or more steam 
at .4 in. draft than with .3 in. draft with the same coal; why, when twice the coal 
is burned, less than twice the steam is made and even to account for and measure 
the losses that are responsible for the difference; also in some cases, why with 
the same coal and draft one boiler makes more steam than another, either 
in pounds per hour or in pounds of steam per pound of coal, but it is not possible 
to compare two different kinds of boilers as to type and size with different 
methods of firing different coals with different drafts, in such a way as will 
explain or permit of prediction of difference in results due to all influences acting 
together. To state it a little differently there is no absolute measure of boiler 
performance as to capacity or efficiency as a basis of comparison to measure 
the goodness of a boiler as a boiler; comparisons must, therefore, be between 
one and another boiler, or one and another service condition; one boiler may 
be said to be better than another, or one condition more favorable and another 
worse, for the results desired, but hardly more than this is possible. 

In this section will be discussed the various factors that seem to be deter- 
mining influences in fixing the capacity or as it is generally measured in terms 
of horse- power, the horse-power of boilers and what determines it. Boiler 
horse-power like gas-producer horse-power is scientifically a bad term, but 
custom has sanctioned it and it will always be used, probably with more and 
more precision as to definition in the future than has been the case in the past. 
The term appears in the literature of the subject first during the period when 
steam pressures were low, about 70 lbs. gage and when steam engines were not as 
economical as they are now nor as different in type and in steam consumption. 
At that time it was easier to discuss the average steam consumption of engines 
than now, and the number that departed from the average was not great; this 
average being about 30 lbs. per hour per horse-power. Accordingly a boiler 
was said to have 100 horse-power capacity when it could make steadily 3000 
lbs. of steam per hour and thus was boiler horse-power defined. In time all 
sorts of variations appeared, better and worse engines were built, each best 
adapted to some sort of service, higher steam pressures and boiler-feed tem- 
peratures were also used, so that no longer did the making of 30 lbs. of steam 
per hour, take from the fire the same amount of heat as when boiler pressures 
were uniformly lower and feed temperatures more constant. Also engines were 
built that operated with a lower water rate, less than half the old rate, while 
others, chiefly small or cheap ones and direct-acting pumps, often used three or 
four times the old average water rate. In short, variations in both engine and 
boiler conditions made it absurd to talk of average water' rates of steam engines, 
and introduced correspondingly large differences in the amount of heat necessary 



HEATING BY COMBUSTION 775 

for the production of a pound of steam, so that an adjustment of boiler horse- 
power definition became necessary. This was done in America by the American 
Society of Mechanical Engineers, adopting a double definition which was 

(a) The evaporation of 34.5 lbs. of water per hour from and at 212° F. ; 
(6) The absorption by the "water between feed condition and that of the 
steam leaving the boiler, of 33,305 B.T.U. per hour per pound. 

This last heat definition was most fortunate as it is an absolute unit, and whether 
it has any relation to engine requirements or not, is a matter of no importance 
whatever. It was believed to be the equivalent of the weight definition and 

OO OAK 

would be, if the latent heat of evaporation at 212° were =965.36, and 

34.5 

also equivalent to the evaporation of 30 lbs. from 100° F. feed water to steam 
at 70 lbs. gage. The language of the committee of the Society adopting the 
standard in 1885 is worth quoting: " Your committee, after due consideration, 
has determined to accept the Centennial Standard, and to recommend that 
in all standard trials (boiler tests) the commercial horse-power be taken as an 
evaporation of 30 lbs. of water per hour from a feed-water temperature of 100° 
F. into steam (dry saturated) at 70 lbs. gage pressure (above standard atmos- 
phere) which shall be considered to be equal to 34 J units of evaporation; that 
is, to 34J lbs. of water evaporated from a feed-water temperature of 212° F. 
into steam (dry saturated) at the same temperature. This standard is equal 
to 33,305 thermal units per hour." A later committee, 1899, retained the 34J 
lbs. from and at, but changed the heat equivalent to 33,317 B.T.U. per hour, 
based on a latent heat of 965.7 so that 965.7X34.5 = 33,317 B.T.U. 

Now that the latent heat at 212° F. is by recent research known to be higher 
than 965.7, and the use of superheat is quite general though not contemplated by 
that committee, it is necessary to once more reconcile the double definition, one 
part with the other or to adopt a new standard, though the old ones must be 
understood by anyone trying to interpret old published test data, based on them. 
Considering the history of the term, it is believed that the retention of the 
34J lbs. evaporation from and at 212° is desirable, and according to the generally 
accepted Marks and Davis Steam Tables the latent heat corresponding is 970.4, 
therefore, the heat equivalent of a boiler horse-power is 970.4X34.5 = 33,478.8 
B.T.U. per hour. This unit is adopted in this work and is applicable as well 
to superheated, as to wet or dry saturated steam. Therefore, 

- , „ , ( Absorption by water and steam in the boiler . 

1 boiler horse-power = - r 00 Ano ■ ^ TT , y. . (849) 

^ of 33,478.8 B.T.U. per hour 



With the heat basis as a standard, the weight of water evaporated per hour 
per boiler horse-power will, of course, vary regularly with the initial water 
temperature and final steam condition, and to facilitate practical work a factor 



776 ENGINEERING THERMODYNAMICS 

of evaporation is calculated for reduction of weights according to the relation 
of Eq. (851) derived by the following relations illustrating its use. 



Lbs. water evaporated\ 
per hour per B.H.P. / 



X 



/B.T.U. per lb. steam above feed\ _" 
\ water temperature / ' 



or 



Lbs. water evaporatedX _ 34.5X970.4 

k per hour per B.H.P. / "B.T.U. per lb. of steam above feed temp. ^ 

34.5 

(&) 



B.T.U. per lb. of steam above feed temp. 
970.4 

34 1 5 

Factor of evaporation 



Hence 



,-r, , » , . \ B.T.U. per lb. of steam above feed temp. 
(Factor of evaporation) = - crm£ ■ 

Heat of liquid at feed 
temp, above 32° F. 



Total heat per lb. of 
steam above 32° F. 



(c) 

(a) 
(6) 



(850) 



(851) 



970.4 



The values of the factor of evaporation and equivalent pounds of water 
per hour per boiler horse-power are to be found from the steam tables or 
directly from the curves, Fig. 202, which also give the heat per pound for dry 
saturated, wet or superheated steam above any feed-water temperature by the 
following simple operations. Each of the upper curves gives directly the total 
heat per pound of steam above 32° and the distance between them and the 
lower curve intercept, that for any feed -water temperature, by a vertical 
distance. If, therefore, AB be the total heat for the steam above 32° at 
100 lbs. per sq.in. absolute and 20° superheat and DE the heat of liquid at 
200° F. feed temperature above 32°, then AC the vertical distance between 
these two points is the heat per pound of steam above the feed temperature 
200° F. for 100 lbs. steam with 20° superheat. This can be marked on a slip of 
paper and read off on the extra scale to the right in terms of, heat in B.T.U., or 
factor of evaporation, or actual weight of water that must be evaporated per 
hour to give a boiler horse-power. 

A sort of commercial rating of boilers has grown up as a convenience to 
purchase and sale, based solely on the surface of tubes and plates exposed to 
the heat, or so-called heating surface, and the figure used is either 10 or 12 sq.ft. 
for common forms, but varies from 8 to 16 for others. This is the result also 
of old average evaporations per square foot of surface for stationary boilers and 
suggests the dependence of capacity on the heat-absorbing power of the surface 
which in such ratings is assumed to be constant. To show what is the evapora- 
tion in pounds per hour per square foot of surface when 10 or 12 sq.ft. are allowed 



HEATING BY COMBUSTION 



77 



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250 300 

'0.15 0.250.5 t 2 3 5 10 15 20 30 40 50 75" 100 150 200 
Upper Line temperature for Saturation. 

Lower Line= Absolute Pressures 



Fig. 202. — Heat per Pound of Steam above Feed Temperature. Evaporation per Hour per 
Boiler Horse-power. Factor of Evaporation. 



778 ENGINEERING THERMODYNAMICS 

per horse-power, with the corresponding rate of heat absorption in B.T.U. per 
hour per square foot, some extra scales are added to the right of Fig. 202 show- 
ing all relations at a glance. 

The figures on evaporation per square foot of heating surface per hour, 
or better still, the heat absorbed per square foot of heating surface per hour, 
is a proper basis for comparing two sets of data for different boilers at equiva- 
lent capacities. Thus different boilers may be said to be operating at the 
same capacity when their heating surface rates of heat absorption are the same 
though one may be developing 50 and the other 1000 boiler horse-power 
actually. The rate of heat absorption by boiler heating surface is a quantity 
that must be dependent on two factors or terms, one an absolute figure and 
the other expressing the rate of change with generation. It is clear that as 
heat is generated faster it must certainly be absorbed faster unless the capacity 
of the water side of the heating surface for heat is lessened at the same time and 
this is not the case with ordinary boilers, but not beyond the bounds of possibility. 

The experimental data on the transfer of heat showed conclusively that the 
capacity of water in tubes or tanks to absorb heat very much exceeded the 
capacity of gases to give it to the surface under a wide range of conditions, so 
it is not unreasonable to suppose that as gases bring more and more heat to water 
surfaces, it will be taken up by them at a rate which is some function of the 
rate of supply. If, therefore, by faster fuel combustion there is a regularly 
increasing quantity of heat available for the water in the form of hot gases, 
then more will be absorbed per hour, more steam made per hour, and more boiler 
horse-power developed per square foot of heating surface, and for the whole 
boiler, and these things must be functions of the rates of combustion and heat 
generation. In two different boilers the rate of increase of absorption may or 
may not bear the same relation to the rate of increase in generation, but it is 
quite possible for them to be similar in this respect,but at a given rate of genera- 
tion in each it is more likely that the actual rates of absorption should differ 
by reason of the differences in structure. This latter difference should require 
a different surface per horse-power for the same amount of heat generated, 
even though a doubling of the rate of heat development produced the same 
fractional increase in horse-power of both. 

Heat generation is, of course, essential to absorption, but it is necessary 
to distinguish between apparent and real generation in seeking relations between 
absorption and generation in accordance with the following definitions. 

Apparent heat generation in\ _ /Lbs. fuel supplied\ /B.T.U. per lb. of\ / 8 r 9 x 
boiler fires, B.T.U. per hr. / \ per hr. / \ fuel as fired / v 

Real heat generation in boil-\ _ /Lbs. fuel supplied\ /B.T.U. per lb. of\ 
er fires B.T.U. per hr. / \ per hr. / \ fuel as fired / 



?)]■ • (853) 



/Fraction lost N 
\ in furnace 



HEATING BY COMBUSTION 779 

The fraction of heat lost in furnaces is that part of the calorific power repre- 
sented by unburnt fuel in ashes, in cinders, in soot discharged through the 
flues, by unburned gases, carbon monoxide, hydrogen and hydrocarbons, by 
evaporation of moisture in the fuel and that conducted away from the hot gases 
by the furnace walls and settings between the point of firing and the place 
where absorption begins. This fraction is always appreciable and often large 
though very difficult to exactly measure, 20 per cent being not uncommon, in 
which case only 80 per cent of the calorific possibilities of the coal are available 
at all for absorption. These losses are mentioned here because of their 
importance but will be more fully discussed in the next section with others 
which together make the whole boiler efficiency less than 100 per cent. 

With some fraction of the fuel heat available for absorption in the form of 
hot gases, it may be absorbed in one of two characteristic ways, first as radiant 
heat from the glowing coal, brickwork, or incandescent floating carbon particles 
in hydrocarbon flames, and second by actual contact of gases with surface. It is 
convenient and usual to characterize these as fire-box and tube absorption, 
respectively, although some tubes may get radiant heat as well as conducted heat, 
and all internal fire-boxes get some gas contact heat. Now, the rate at which 
fire-boxes, actual or equivalent, absorb heat, being governed by the laws of 
radiation, is proportional to the fourth power of the temperature difference, 
according to the Stefan and Bolzmann law, so it would seem to be quite inde- 
pendent of the rate of generation or rate of fuel combustion except as this may 
affect the temperature and extent of the radiant matter. This independence 
of radiant heat absorption with respect to r.ate of combustion or the constancy 
of radiant heat for all good fires seems to be really established. On the other 
hand the absorption by tubes or plates, of heat from hot gases passing through 
or around them, is governed by the laws of heat transfer from hot gases to 
water whatever they may be. It is certain that higher mean temperatures 
of the gases should increase the rate of absorption per hour, and it is also certain 
that any influences that make the dead gas film thinner, or promote contact 
between fresh hot gases and the tube itself, will increase the rate of absorption. 
One such influence on reduction of dead the film is the velocity of the gases, which 
is roughly proportional to the amount of air supply to the furnace, or the rate 
of combustion. Whatever these laws are, they should be the same in kind for 
all boilers, but comparing any two, certain factors affecting the rate of absorptio?i 
are fixed by the construction and may be very different in the two cases. Fo-i 
example, the disposition of the heating surface may be such as to offer 
a long, narrow passage-way for the gases between the surfaces, or a short, 
wide one, or a narrow short one, or a wide long one. If the gases passed the 
water surface in infinitely thin streams, it seems likely that they would give 
up their heat almost instantly, so the length of such a passage would not make 
much difference. On the other hand if the passage were wide, the gases at 
the center of the stream might have to wait a long time to come into surface con- 
tact and might never do so, in which case they must give up their heat through 
surrounding gas layers which would take a long time and require long tubes. 



780 



ENGINEERING THERMODYNAMICS 



It appears, therefore, as the disposition of the heating surface, the propor- 
tions of the gas passages between the absorbing surface, the relations of fire- 
box to flue surface and such matters as are fixed by construction, may vary 
through almost infinite ranges, that, however well known might be the funda- 
mental laws of radiant heat flow and of transfer by conduction and convection, 
the rate of heat absorption in boilers would resist any definite generalization 
though offering a most attractive field for theorizing. This makes it seem all 
the more remarkable that experimental results should show such a consistency 
as is demonstrated below, even though the constants in the relation differ 
because of the structural influences noted above. 

In selecting boiler data for discussion and analysis, one cannot be too careful 
in avoiding a most natural but improper and unscientific tendency, notable 
in boiler literature, which is, to pick out tests that seem to prove a precon- 
ceived theory, selecting the evidence to prove the case. The tests used here 
are believed to be free from improper manipulation, and to be fairly represen- 
tative, though to be quite sure it would be necessary to analyze every authentic 
test in existence, which is beyond the scope of such a work as this. 

The locomotive boiler is typical of internal fire-box construction and of 
operation with greatest " forcing," that is, highest rates of combustion and 
highest rates of evaporation per square foot of heating surface, while the ordi- 
nary stationary water-tube boiler is typical of much slower rates of operating 
conditions and of practically all tube construction. Accordingly, the first 
two sets of data selected for study are the report of Goss (Bulletin 402, U.S. 
G.S.) on 18 tests of an American locomotive boiler at the Purdue testing plant, 
with experienced and skillful experimenters, using two different, but both 
good quality coals, under four different boiler pressures and with different 
rates of combustion for each, all high. The boiler was of the extended wagon- 
top type, fire-box 72y§- ins. long, 34J ins. wide, 79 ins. deep, with fire-tubes 
2 ins. in diameter and 16 ft. 5 ins. long, making the total heating surface of 
the boiler proper 1023 sq.ft. It was also equipped with a Cole return-tube 
superheater, 1J ins. outside diameter, 32 loops each 17.27 ft. long, adding 193 
sq.ft. of superheating surface to 1023 sq.ft. of boiler surface, making the total 
heat absorbing surface 1216 sq.ft. The nature of the two different coals used 
is indicated by the proximate analysis and calorific power, one with twice the 
volatile of the other. 

COALS USED IN GOSS LOCOMOTIVE BOILER TEST 





Moisture, 
Per Cent. 


Volatile, 
Per Cent. 


Fixed C, 
Per Cent. 


Ash, 
Per Cent. 


B.T.U. per Pound. 


Coal 


Coal Dry. 


Combustible. 


A 


1.89 
3.10 


31.94 
15.23 


57.71 

72.75 


8.46 
8.92 


14047 

14347 


15372 


B 


15802 







HEATING BY COMBUSTION 



781 



The first characteristic to be examined is the heat-absorbing capacity of 
the heating surface with respect to the heat developed in the furnace, which is 
plotted in Fig. 203, to coordinates representing horizontally, the rate heat 
developed and available for absorption per square foot of heating surface, and 
vertically the rate of heat absorption by heating surface both in B.T.U. per hour. 

The experimental points all lie on straight lines whether the water surface 
be considered alone, the superheater surface alone, or the entire heating sur- 
face as one, within what may be regarded as the limit of experimental error. 
Taking the line representing the whole surface as characteristic of the whole 
heat-absorbing process or the boiler itself as a heat absorber, it is represented 
algebraically by Eq. (854) below. 



B.T.U. absorbed 
per hour per sq.ft. 
H.S. 

Boiler horse-power 
per sq.ft. H.S. 



= 1450+. 66 



= .0433 +.0000197 



B.T.U. developed in fire and 
available for absorption 
per hour per sq.ft. H.S. 

B.T.U. developed in 
fire and available for 
absorption per hour 
per sq.ft. H.S. 



(a) 



(6) 



► (854) 



Similar results are found for a very different type of boiler, the Heine water 
tube, by the tests of the U. S. Geological Survey, though it was operated at very 
much lower rates, and on a great variety of coals ranging from low grade 
lignite to good bituminous. Over two hundred tests are represented in the data 
of Fig. 204, which is an almost perfect straight line within the limits of exper- 
imental procedure, and it is worth noting that the different coals are well dis- 
tributed over the curve, showing how independent is the heating surface per- 
formance, of the sort of fuel used. The boiler had 2031 sq.ft. of heating and 
40.55 sq.ft. of grate surface. 

This line is represented by Eq. (855). 



B.T.U. absorbed 
per hour per sq.ft. 
H.S. 



Boiler horse-power 
per sq.ft. H.S. 



f B.T.U. developed in fire and 
615+.511 | available for absorption per 
[ hour per sq.ft. H.S. 

B.T.U, developed in 
fire and available for 
absorption per hour 
per sq.ft. H.S. 



= .0185+.0000152 



(a) 



(6) 



■ (855) 



It thus appears, that from two totally different boilers, driven at very differ- 
ent rates, 4000 B.T.U. per sq.ft. heating surface maximum, absorbed in the Heine 
water-tube and nearly 15,000 maximum in the locomotive fire-tube boiler or 
about 4 to 1, and with all sorts of coal in the former, some about the same as 
in the latter, that the heat absorbed varies by a straight-line law with the amount 



782 



ENGINEERING THERMODYNAMICS 



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Fig. 203. — Relation between Rate of Heat Absorption oo B.H.P. and Rate of Heat Generation 
Available for Absorption in a Locomotive Boiler. 



HEATING BY COMBUSTION 



783 



developed, for both. Furthermore, the slope is nearly the same for both, though 
the constant for the water tube is about half that of the locomotive fire tube, 
which means that the rate of increase of absorption with increase of develop- 
ment or evolution of heat is nearly the same for both. 

Extending the investigation a little further, more confirmatory results are 
obtained and for this three boilers are selected, the first representing the largest 
and the second the smallest for which authentic tests are available, while the 
third gives data for oil fuel to further check the influence of kind of fuel. 

The largest boiler is that of the Detroit Edison Co., tests for which are 
reported by Jacobus, having a heating surface of 23,650 sq.ft. and of the curved 
tube Stirling type, fired with coal mechanically, on both Roney and Taylor 



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Fig. 204. — Relation between Rate of Heat Absorption or B.H.P and Rate of Heat Generation 
Available for Absorption for Heine Water Tube Boiler. 

stokers. To the same coordinates as were used for the other cases the test 
data are plotted in Fig. 205 and no better straight-line relation is possible. This 
line is represented by the linear relation Eq. (856). 

B.T.U. developed in fire 



B.T.U. absorbed 
1 per hour persq.ft. 
I H.S. 



Boiler horse-power 
I per sq.ft. H.S. 



235.3+.8235 



= .0703 +.0000245 



and available for absorp- 
tion per hour per sq.ft. 
H.S. 

B.T.U. developed in 
fire and available for 
absorption per hour 
per sq.ft. H.S. 



(a) 



(b) 



(856) 



7U 



ENGINEERING THERMODYNAMICS 




4000 5000 0000 7000 8000 

Bate of Heat Generation Available for Absorption per Hour .per 
Square Foot of Heating Surface =x 

Fig. 205.— Relation between Rate of Heat Absorption or B.H.P. and Rate of Heat Generation 
Available for Absorption for 23650 sq.ft. Stirling Boiler. Detroit Edison Co. 



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Rate of Heat Generation Available for Absorption per Hour 
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Fig. 206. — Relation between Rate of Heat Absorption or B.H.P. and Rate of Heat Generation 
Available for Absorption for White Steam Car Boiler. 



HEATING BY COMBUSTION 



785 



The smallest boiler is represented by one used on the White steam auto- 
mobile and for which tests are reported by Carpenter. This is of the " flash " 
type, gasolene fuel and carrying 45.8 sq.ft. of heating surface, the steam being 
of high pressure 200-500 lbs. gage and high superheat 400° F. These results 
are plotted in Fig. 206 and the resulting straight line is represented by Eq. (857). 



B.T.U absorbed 
per hour per sq.ft. 
H.S. 

Boiler horse-power 
per sq.ft. H.S. 



B.T.U. developed in fire and avail- 1 
= .78 \ able for absorption per hour per [ (a) 
sq.ft. H.S. J^ 

[B.T.U. developed in fire and ] 
026 \ available for absorption per > (b) 
hour per sq.ft. H.S. J 



(857) 



To show how independent of the fuel are these relations, the tests of a Hohen- 
stein water-tube boiler fired with liquid fuel under the direction of a special 



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Fig. 207. — Relation between Rate of Heat Absorption or B.H.P. and Rate of Heat Generation 
Available for Absorption for Hohenstein Oil Fired Boiler. 



board representing the U. S. Navy, are plotted in Fig. 207. This boiler had 
2130 sq.ft. of heating surface and was worked under quite a wide range of 
conditions, nevertheless a straight-line relation maintains as given by Eq. (858). 



786 



ENGINEERING THERMODYNAMICS 



B.T.U. absorbed 
per hour per sq.ft. 
H.S. 



Boiler horse-power 
per sq.ft. H.S. 



= 1367.8+.613 



.0408 +.0000183 



B.T.U. developed in 
fire and available for 
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per sq.ft. H.S. 

B.T.U. developed in 
fire and available for 
absorption per hour 
per sq.ft. H.S. 



(a) 



(&) 



(858) 



There appears from the above examination of good data, to be no doubt 
of the existance of a linear relation between heat absorbed and heat available 
for absorption in boilers, regardless of boiler structure, of kind of fuel or rate 
of forcing. The only difference is to be found in the two constants and for 
this there is adequate explanation. When rate of heat development or evolution 
becomes zero there appears to be some absorption in all but the White auto- 
mobile boiler, because the line does not pass through zero, but cuts the coor- 
dinate of absorption rate at a finite value. This value then must be a measure 
of the constant rate of absorption due to radiant heat, whether any hot gases 
carry heat to the absorbing surface or not. It is natural that this should differ 
for different boilers as it is a structural feature, depending on the exposure of 
heating surface to heat rays from both incandescent fuel and brickwork. The 
rate at which the rate of absorption increases with the rate of evolution of heat is 
a result of the proportions of the gas passages through the heating surface tubes, 
or around them, again a structural factor for which existing data are insuffi- 
cient to permit a numerical valuation in terms of dimensions alone. 

It is often impossible to secure knowledge of furnace losses giving the amount 
of heat not available for absorption but still supplied in the fuel, so it is worth 
while to examine the relations between heat absorption and heat supplied in 
fuel or apparently available. This heat supplied in the fuel is the product of 
the weight fired per hour, and the calorific power of the fuel as fired. The fur- 
nace loss, so called, is the difference between this product and that available 
for absorption in the form of hot gases approaching the heating surface. Alge- 
braically these relations are given by Eqs. (852) and (853). 

Accordingly if the furnace loss as a fraction of the heat supplied be represented 
by L, then, 



Fraction of combustion heat really generated = 1— L, 



. (859) 



Whence 



B.T.U. developed in 
in fire and avail- 
able for absorption 
per hr. per sq.ft. 
H.S. 



f Lbs. of fuel sup- 1 [ 

(l-L)X plied per hour | X | 

I per sq.ft. H.S. J I 



B.T.U. per 
lb. of fuel 
as fired 



(860) 



HEATING BY COMBUSTION 



787 



Substituting this in the general linear equation between rate of absorption 
and rate of true generation, Eq. (861), gives the following relation between rate 
of absorption and rate of apparent generation, Eq. (862). 



B.T.U. absorbed per 
hour per sq.ft. H.S. 



f B.T.U. developed in fire and avail- 
a-\-b \ able for absorption per hour per 
I sq.ft. H.S. 



= a+6(l-L) 



Lbs. of fuel sup- 
plied per hour 
per sq.ft. H.S. 



X 



B.T.U. 

per lb. 

of fuel 
as fired 



(861) 



(862) 



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35 7 1.05 1.4 1.75 2.10 

Pounds of Dry Coal Fired per Hour per Square Foot of Heating Surface 
5513 11025 16538 22050 27503 33075 

Heat Apparently Developed in B.T.U. per Hour per Square Foot of Heating Surface ~x 

208. — Relation between Rate of Heat Absorption or B.H.P., and Rate of Combustion 
or Apparent Heat Generation for a Locomotive Boiler. 



If now L is a constant, then b being also a constant, 6(1— L) is likewise a 
constant, so that, this relation would also be linear and experimental determi- 
nations should be on a straight line. Should it happen in any case that a curve 
results, then this is in itself a proof of a variable fractional furnace heat loss. As a 



788 



ENGINEERING THERMODYNAMICS 



general rule a straight line does result, so that constant fractional furnace heat 
losses are rather the rule than the exception. Figs. 208 and 209 demonstrate 
this clearly. It should be noted that as the ratio of heating surface to grate 
surface is fixed in any given boiler, the weight of coal per hour per square foot 
of H.S. is directly proportional to the rate of combustion. For a fuel of constant 
calorific power the last term of Eq. (862) will be constant, and a linear relation 
is established between rate of absorption or boiler horse-power and rate of com- 
bustion, a most valuable relation and true if, 

(a) The furnace losses are a constant fraction of the heat in the fuel as 

supplied; 
(6) The calorific power of the fuel is constant. 

Referring to Fig. 208 representing the data of the Goss tests of a locomotive 
boiler, vertical distances representing rate of heat absorption and boiler horse- 
power to a double scale, horizontals, rate of combustion and equivalent rate 
of heat apparently developed, it is clear that a straight line is a fair represen- 
tation of the results. The difference in the slopes for the lines representing 
the two different fuels shows clearly the effect of different furnace losses in one 
case compared with the other, but in each the fractional loss is constant as 
proved by the linear relation. | 

A similar conclusion is indicated by the four lines of Fig. 209, each repre- 
senting a series of tests on a U. S. Navy water-tube boiler intended respectively 
for U.S.S. Denver, Cincinnati, Nebraska and Virginia taken from the records 
of the Navy Department. These boilers had the following characteristics. 
That of the Denver was a Hohenstein, 2174 sq.ft. H.S. for six tests and 2130 
sq.ft. for the rest, and 50.14 sq.ft. G.S. The coal used was Pocahontas run- 
of-mine. The Cincinnati and Nebraska had Babcock and Wilcox marine- 
form boilers of 2640 and 4682 sq.ft. H.S., 63.25 and 111.72 sq.ft. G.S., 
respectively, and were operated under both closed ash-pit and closed 
fire-room systems of draft with picked Pocahontas coal. A Niclausse boiler 
was tested for the Virginia, having 1852.56 sq.ft. of H.S. and 45.225 sq.ft. 
of G.S. 

These straight-line relations give the following values for the constants 
of Eq. (862). 

VALUES OF CONSTANTS FOR LOCOMOTIVE AND MARINE BOILER CAPACITY. 



Boiler. 


Value of Constants, in Eq. (862). 




a. 


b(l-L). 


B. & W., U.S.S., Cincinnati 

B, & W., U.S.S. Nebraska 

Niclausse, U.S.S. Virginia 

Hohenstein, U.S.S. Denver 

Locomotive, coal A 


1350 
50 
1760 
1090 
1010 
3950 


.606 

.722 

.566 

.56 

.504 


Locomotive, coal B 


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HEATING BY COMBUSTION 



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o Hohenstein Boiler For U.S.S. Denver 
X New Babcock & Wilcox Boiler For U.S.S. Cincinnati 
• Babcock & Wilcox Boiler For U.S.S. Neoraska 
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.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 1.1 1.2 1.3 1.4 1.5 

Hate of Combustion in lbs. per Hour per Square foot of Heating Surface. 
1524 3048 4572 6096 7620 9144 10868 12192 13716 15240 16761 18288 19812 21338 
Heat Apparently Generated in B.T.U.per Hour per Square Foot of Heating Surface 



Fig. 209.— Relation between Rate of Heat Absorption or B.H.P., and Rate of Combustion 
or Rate of Apparent Heat Generation for Four U. S. Navy Water Tube Boilers. 



790 



ENGINEERING THERMODYNAMICS 



It has been pointed out that when the furnace losses are not a constant 
fraction of the heat supplied in the form of fuel, then the curve between rate 
of absorption or boiler horse-power and rate of combustion or of apparent 
heat generation is really a curved, not a straight line, and in Fig. 210 one such 
case is given. This case represents a large number of tests on the Hohenstein 
boiler by the U. S. Navy testing board, with oil fuel for which a linear relation 
was found to hold bet we an absorption and true generation rates allowing for 
furnace losses (Fig. 207) . 



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.B.T.U. Apparently Generatedjper-Hbur per;Sq,Et. of Heating Surfaces # 
(lbs. of oil bunted per-Sq.Efc.'Heafing S«ia©face per Ho.uxx.B..X:U.. perilb,) 

Fig. 210. — Relation between Rate of Heat Absorption or B.H.P. and Rate of Combustion 
or Approximate Heat Generation for a Hohenstein Oil Fired Boiler. 

A second-degree equation fits fairly well the average curve through the 
experimental results and is given by Eq. (863): 



B.T.U. absorbed per 
hour per sq.ft. H.S. 



294+. 662 



[J Lbs. oil per hour 
{ per sq.ft. H.S. 



X 



B.T.U. per 

lb. oil 



.00455 



J Lbs. oil per hour 1 
1 per sq.ft. H.S. J 



X 



B.T.U. per } 1 
lb. oil ] J 



(863) 



It is the last term that represents the increasing furnace loss and to show 
this more clearly the relations of both Fig. 210 and Fig. 207 are replotted 
together in Fig. 211. Vertical distances as before, represent rate of absorption 
or boiler horse-power, but two horizontal scales are used, the lower representing 
apparent heat generation for the lower curve, and the upper scale real generation 
and available for absorption, for the upper curve. The horizontal distance 
between the two curves when referred to the lower scale is the furnace loss in 
B.T.U. per hour per sq.ft. of H.S. at the time when the real and apparent 



HEATING BY COMBUSTION 



791 



rates of generation have the values represented by the points of intersection. 
Thus, when the real rate of generation of heat available for absorption is 15,000 
B.T.U. per hour, point D, the furnace loss is represented by AB, and the 
apparent generation rate is 17,650 B.T.U. per hour, point C. 



.50779 



,-~i>000 

170001— 



B.T.U. Available for Absorption per Sq.Ft. of Heating Surface per Hour=a? 

10000 15000 20000 25000 



30000 



8 ,41805 « 15000 

f I 

a .3S831 « 13000 



a 



co 



• .32857 &j H00O 
to 5 

s- C. 
P-.2G8S3 g 9000 
ft ^ 

£ 1 

g 20909 -3 7000 



u 



o. 14935 • 5000 
H 

.08961 3000 























D 














,0 
















































fi 


<>y 






^ 




































J 


^ 








^ 


S^ 


































n 


■^ 

y? 






vV 


^ 


r^ 




































# 


£? 






^ 


£>- 




















































































k 




— 


Pfi 


5 




















































































? 


^ 








<— 




->j 
































■< 


$ 


r > 


y/\ 


V 








V 


~T 


Fu 

Ab 


■na< 
son 


tioi 


OSS 

iRa 


teL 






















V 




^ 


V 












































<r 


'^ 




















































































































|c 


: 

























'SODO 10000 15000 20000 25000 

B.T.U.. Apparently Generated per Hour per Sq.Ft. of Heating Surfaced 
(lbs. of oil burned per Sq.Ft. Heating Surface penllourx B.T.U. per lb.) 



30000 



Fig. 211. — Relation between Heat Absorbed, Apparently Generated, Available and Heat 

Lost in Furnace. 

The general fundamental equations (864 a,b,c,) may be changed as to form 
into others following containing different but related variables, and in some one 
of these different forms they apply to existing test data. 

B.T.U. developed in fur- 
nace and available for ab- 
sorption per hour per sq.ft. 
of heating surface 

B.T.U. developed 
in furnace and 
b 



B.T.U. absorbed per hour 
per square foot of heat- 
• ing surface 



= a+b 



(a) 



Lbs. evaporation from 
and at 212° F. per 
square foot of H.S. per 
hour 



970.4+9: 



f Boiler horse-power per 
i square foot of heating 
I surface 



33479 ' 33479 



available for ab- 
sorption per hr. 
per sq.ft. heating 
surface 

B.T.U. developed 
in furnace and 
available for ab- 
sorption per hr. 
per sq.ft. of heat- 
ing surface 



l (6) 



(c) 



(864) 



792 



ENGINEERING THERMODYNAMICS 



Multiplying throughout by the heating surface in square feet gives three 
more formulas Eq. (865): 



B.T.U. ab- 
sorbed per 
hour 

Lbs. evapo- 
rated per hr. 
from and at 
212° F. 

Boiler horse- 
power 



Sq.ft. of | 
heating \ +6 
surface J 



Sq.ft. of 
heating 
surface 



970.4 



33479 



Putting 

B.T.U. generated 
per hour 



Sq.ft. of ) 
heating j- 
surface J 



Lbs. fuel 
per hour 



B.T.U. developed 
in furnace and 
available for ab- 
sorption per hr. 



970.4 



33479 



B.T.U. developed 
in furnace and 
available for ab- 
sorption per hr. 

B.T.U. developed 
in furnace and 
available for ab- 
sorption per hr. 



(a) 



(b) 



(c) 



(865) 



X 



I" /B.T.U. per\ _ /B.T.U. furnace V 
L\ lb. fuel / \lossperlb. fuel /_ 



three more formulas result, involving the fuel consumption rate, Eq. (866) : 



B.T.U. absorbed If Sq.ft. of heating 
per hour J } surface 

Lbs. of fuel 1 | /B.T.U. per\ _ /B.T.U. furnace\ 
per hr. J } \ lb.of fuel / \ loss per lb. fuel / 

Lbs. evaporated per hr. 1 _ a J Sq.ft. of heat- 
from and at 212° F. J ~ 970.4 { ing surface 



+b 



1 970.4 

Boiler horse- 
power 

4 



Lbs. of fuel 
per hr. 

a 



B.T.U. per 
lb. of fuel 



)- 



33,479 
Lbs. of fuel 
33,479 I per hr. 



B.T.U. furnace\ ' 
loss per lb. / 

Sq. ft. of heat- 1 
ing surface J 

YB.T.U. per\_ /B.T.U. furnace\ 
lb. of fuel / \ loss per lb. fuel / 



(a). 



(6) 



(c) 



(86G N 



Into these equations the grate surface may enter through the rate of fuel 
consumption by Eq. (867): 

(Lbs. fuel per hr.) = (lbs. fuel per sq.ft. of grate surface per hr.) X (sq.ft. grate 
surface) 
= (rate of combustion) X (sq.ft. of grate surface.). . . (867) 



HEATING BY COMBUSTION 



793 



Whence 



B.T.U. absorbed per 
hour 



+& 



Sq.ft. of 
grate 
surface 



f Rate of ) 
j com- 
l bustion J 



Sq.ft. of heat- 
ing surface 



B.T.U per 
lb. of fuel 



B.T.U. fur- 
nace loss 
per lb. of 
fuel 



Lbs. evaporated per hr. 1 a 

from and at 212° F. J ~ 970.4 
Rate of 



970.4 



(Sq.ft. of 
\ grate 
[ surface 



com- 
bustion 



Sq.ft. of heat- 
ing surface 

/B.T 



Boiler horse-power [ = 



33,479 



[Sq.ft. of 
grate 
surface 



33,479 
Rate of 
com- 
bustion 



B.T.U. per 

lb. of fuel 



Sq.ft. of heat- 
ing surface 

B.T.U. per 
lb. of fuel 



nace loss 
per lb. fuel. 



B.T.U. fur- N 

nace loss 

k per lb. fuel/ 



(a) 



(6) 



(c) 



(868) 



As the ratio of heating surface to grate surface is frequently used as a 
variable it may be introduced conveniently by Eq. (869). 



Ratio of heating surface to grate surface = R = 



Whence 



Sq.ft. heat. sur. A) 
Sq.ft. grate sur. A t 



(869) 



(B.T.U. absorbed per hour) 
, [Rate of 
a-f-B \ com- 



i bustion 



'B.T.UA 
per lb. I 
of fuel/ 



'B.T.U. furnace^ 

loss per lb. of 

k fuel / 



(Lbs. evaporation per hr. from and at 212° F.) 



970.4 



, [Rate oV 
a+ -n \ com- 

l bustion , 



(Boiler horse-power) 

7 fRate of 



An 



33,479 



a+ 



R 



com- 
bustion 



'B.T.U. 

per lb. 
> of fuel 



B.T.U. N 

per lb. 
of fuel. 



'B.T.U. furnace\ 

loss per lb. of J 

v fuel / 



'B.T.U. furnace^ 

loss per lb. of 

. fuel j 



(a) 



(b) 



(c) 



(870) 



One of the best analytical investigations of the laws of boiler-surface heat 
absorption is that of Professor John Perry, who arrives at conclusions very 



794 



ENGINEERING THERMODYNAMICS 



similar to the above. According to his mathematical analysis, based on the 
kinetic theory of gases and using the fundamental idea suggested by Osborn 
Reynolds, the flue part of a boiler, that is, all surface getting its heat from hot gas 
contact, will absorb always the same fraction of the heat that gets to it. The frac- 
tion depends on the relation of surface to cross-section of gas passage, that is, 
its mean hydraulic depth and on nothing else, except a constant of propor- 
tionality, the nature of which is somewhat hazy. 



Heat absorbed by flue 
Heat supplied to flue 



/Constant depending on dimensions of gas 
\ passage between heating surface 



= C. (871) 



From an analysis of the data of an old French locomotive test made on a 
boiler made up of sections, the evaporation in each of which could be measured, 
Perry concludes that the absorption or the evaporation in the fire-box alone 
is given by 



B. T. U. absorbed 



per hour by fire-box \ =Ax ^ rate surface) +#X (pounds coal per hour). (872) 



Assuming that the flues can absorb a constant fraction of what is left 
after the fire-box has taken out some of the heat of combustion from the hot 
gases, then neglecting furnace losses, 



[B.T.U. ab- 
sorbed per 
hour by 
flues 



= Cxi 



=cx 



'B.T.U. developed in s 
furnace and avail- 
able for absorption 
per hour 



B.T.U. developed in s 
furnace and avail- 
able for absorption 
per hour 



/B.T.U. absorbed\ 
— by fire-box per I (a) 
\ hour / 



. I Grate \ _ ^/Lbs. coal\ 
\ surface/ \ per hour/ 



(6) 



(873) 



Total B.T.U. absorbed 
per hour by boiler 



B.T.U. absorbed per hour 
by (fire-box + flues). 



A (Grate surface) -\-B (Lbs. coal per hour) 

[ /B.T.U. developed in furnace\ I A (Grate surface) \ 
+Cj and available for absorption J I +2?(Lbs. coal ) !■ ( a ) 
[ \ per hour / \ per hour) / J 

/B.T.U. developed in fur- \ f A (1 - C) (Grate surface) } 
■ Cl nace and available for )+ j ,r>(i /C)Lbs. coal \ [ (b) 
\ absorption per hour / ( \ per hour 



(874) 



HEADING BY COMBUSTION 795 

This Eq. (874), reached by Perry partly by mathematical analyses starting 
with the kinetic theory of gases and partly from fire-box evaporation data 
for one boiler, is substantially the same as Eq. (865a), which is purely empiric, 
and based on many tests of different boilers, conducted with far more than 
usual accuracy. To show this similarity it is only necessary to assume that 



fB.T.U. developed in furnace 
MX | and available for absorption 
1 per hour 



Lbs. coal per hour. 



Also that 

p . _ Heating surface 

Ratio of heating surface to grate surface 

which on substitution in (Eq. (8746) gives 

,_ _ TT . / B.T.U. developed in furnace and \ 

(B.T.U. absorbed per hour)=G ., , , , f 

\ available for absorption per hour / 

A(l-C) / S( l- ft - of \ /B.T.U. developed in fur-\ 

H b — I heating \-\-B(l — C)M\ nace and available for J 

\ surface/ \ absorption per hour / 

A(l-C) / Sq ' ft - of \ /B.T.U. developed in fur-\ 

= — ^-p — -I heating \ + \B{l-C)M+C}l nace and available for J 

\ surface/ \ absorption per hour * / 



(875) 



As the ratio of heating surface to grate surface R is a constant for one boiler, 
the two coefficients above become the constants (a) and (b) of Eq. (865a). To 
state it otherwise, his conclusion is identical with the new facts in the case if 
the ratio R is constant and if the furnace losses are a constant fraction of the 
apparent generation. If the approximate identity above noted is real then 
his most interesting conclusion, the most promising of all boiler transfer 
theories, is supported and flues always will take out a constant fraction of the 
heat carried to them by the hot gases, the fraction being larger, as they are longer 
and the cross-section of gas passage smaller compared to the heating surface 
at its perimeter or edge. 

Prob. 1. A boiler receives 10,500 lbs. of water per hour at a temperature of 50° F. 
and turns it into steam at 150 lbs. per square inch gage, with 100° of superheat. What 
boiler horse-power is being developed and what is the factor of evaporation for this 
case? 

Prob. 2. A boiler has a heating surface of 1050 sq.ft. On the basis of 10 sq.ft. per 
horse-power, how many pounds of steam will it make per hour from feed-water at 
150° F. and steam 95 per cent dry at 125 lbs. per square inch gage? 

Prob. 3. A boiler with 1000 sq.ft. of heating surface makes 4000 lbs. of steam per 
hour from feed-water at 200° F. and steam at 200 lbs. gage with 3 per cent moisture. 
What is the number of square feet of heating surface per horse-power? 



7fi6 ENGINEERING THERMODYNAMICS 

Prob. 4. What would be the factors of evaporation for the following cases: 



Feed Water 


Steam Pressure 


Moisture in 


Temperature. 


Gage. 


Per Cent or 
Superheat in ° F. 


50 


110 


5% 


150 


50 


100° 


212 








250 


200 


150° 


300 


400 


3% 


125 


100 


50° 


212 





10% 


100 


150 


5% 


70 


125 


200° 


212 





500° 



Prob. 5. 3000 lbs. of coal having a heating value of 12,500 B.T.U. per lb. are fired per 
hour. If the cinder loss is 5 per cent, CO loss 3 per cent, moisture loss \ per cent, and 
radiation and other furnace losses are 5 per cent, what is the rate of real generation per 
hour in B.T.U.? 

Prob. 6. If 15 per cent of the heat really generated is lost to the stack and the 
efficiency of the heating surface is 80 per cent, what will be the horse-power of boiler, 
efficiency of boiler as a whole, and pounds of steam made per hour from feed water 
at 100° F., and steam at 100 lbs. gage dry, and saturated? 

Prob. 7. For a boiler, the constants of Eq. (856a) were, a— 1000 and, b= .55. 
The ratio of heating surface to grate surface was 60 and grate was 60 sq.ft. in area. 
The coal fired per hour was 1500 lbs. and 90 per cent of the heat in the coal was actu- 
ally generated. If the B.T.U. per pound of coal were 13,450, what was the horse- 
power of boiler? 

Prob. 8. A boiler has 3500 sq.ft. of heating surface and a grate surface of 50 ft. 
The constants in the straight-line equation are a = 1000 and b = .5. Show how the 
horsepower will vary with the rate of combustion for coal containing 12,000 B.T.U. 
per pound and a constant furnace loss of 15 per cent. 

Prob. 9. A boiler follows the same law as that of the Heine boiler, but with a constant 
(a) equal to 60 per cent of that of the Heine. If the furnace efficiency is 90 per cent and 
constant what must be the rate of combustion per boiler horse-power for |, f, 1, and 
1| times the rated capacity? The boiler has 15,000 sq.ft. of heating surface, a ratio 
of heating surface to grate surface of 45, and is rated on 10 sq.ft. of heating surface. 
The coal used is No. 20 of the general table. 

Prob. 10. The rate of combustion in a boiler varies from 20 to 80 lbs. per hour per 
square foot of grate surface. The grate area is 150 sq.ft. and ratio of heating surface to 
grate surface is 60. For a constant furnace loss of 15 per cent and a coal having a 
heating value of 14,000 heat-units, what will be the variation in horse-power, the 
relation being the same as for the Goss tests with Coal A? 

14. Steam-boiler Efficiency, Furnace and Heating-surface Efficiency. 
Heat Balances and Variation in Heat Distribution. Evaporation and Losses 
per Pound of Fuel. Defining the efficiency of a boiler, inclusive of grate, furnace, 
setting and heating surface, as the ratio of the heat that is actually absorbed 
and retained by the steam per pound of coal, to the calorific power of the coal 



HEATING BY COMBUSTION 797 

as fired, it may be said that there is no fundamental theory or absolute standard, 
as was the case for evaporative capacity, but like it there are known and estab- 
lished relations between the factors that together make it less than 100 per cent. 
Study of boiler efficiency must be based, therefore, not on what might be expected 
of the heat absorption surface but negatively, as it were, on the nature and extent 
of the losses or the several parts of the original heats that do not get into the water 
or steam, whether operating, to reduce the amount available for absorption 
before the gases reach the surface or, on the other hand, to leave a residue 
of sensible heat in the gases after they have swept the surface. 

Boiler heat losses can be divided, grouped and classified in many different 
ways of course, but the following is a very useful one. 

Boiler-heat Loss Classification 

1. Furnace and Setting Loss. This includes all those amounts of heat 
per pound of coal that tend to reduce the sensible heat of the gases available 
for heating surface absorption except radiant heat taken up directly by the 
heating surface, such as: 

(a) That necessary to evaporate the moisture in the coal; 

(6) That due to burning hydrogen to vapor instead of to water, the 

difference between high and low calorific power of hydrogen burnt 

per pound of coal; 

(c) That radiated (1) from furnace and setting before absorption by 

water or steam; 

(d) That due to unburned gases CO, H2 and hydrocarbons in the flue 

gases; 

(e) That due to unburned fixed carbon, in ash dropping through grate, 

or in soot and cinders in the flues, or discharged from the stack. 

2. Flue Loss. This includes the amount of sensible heat still carried by 
the gases that have swept over the heating surface, and is the product 
of the total weight of gases per pound of coal, including water vapor and 
excess air, into the mean specific heat and excess of temperature over that 
of the air supply. Obviously, this is more or less governed by the steam pres- 
sure and temperature as the leaving gases can never be made cooler than the 
last boiler surface they touch and if this is superheater surface the temperature 
may be high, but if it is feed-water supply surface it may be low. Otherwise 
this loss is governed by the control of air supply, so that too much excess is not 
to be used, but also and more fundamentally by the absorbing laws of heating 
surface, as it is the heat residue after absorption. 

On the assumption that all the rest of the heat gets into the steam and water, 
all the heat is thus accounted for. That which gets into the water and steam 
may be regarded as the useful effect, but this may not be warranted if some steam 
and water leakages exist or radiation occurs from the steam or water surfaces 
themselves. In these cases there may be a difference of opinion as to whether 
the losses should be credited or debited because from the standpoint of boiler 



798 



ENGINEEEING THEKMODYNAMICS 



goodness as a heat absorber all heat carried off by leakages is credit heat, for it 
has been taken from gases, but on the other hand from the steam user's stand- 
point this heat is of no value as it cannot run an engine. The steam and water 
surface radiation is, of course, also a steam user's loss, but the boiler abs rbed 
it from the gases and considering the boiler as an absorber rather than a con- 
server of heat, it is a credit. No confusion is likely to result in practical work 
if these items are clearly understood, circumstances in each case will indicate 
clearly where the items belong. 

The preceding classification indicates that there may be more than one standard 
of efficiency of boilers or rather that boiler efficiency is divisible into parts, which 
is a valuable way of keeping the processes clearly divided and of establishing 
a basis of analysis. 

Boiler-efficiency Definitions 

1. Furnace Efficiency E f , as a term is properly applied to the ratio of the 
sensible heat of the gases before heating surface absorbtion per pound of fuel 
to the calorific power of the fuel, or 



E t - 



(B.T.U. per lb. fuel) — (Furnace and setting loss per lb. fuel) 



1- 



B.T.U. per lb. fuel 
Furnace and setting losses per lb. fuel 
B.T.U. per lb. fuel 



(a) 



(&) 



(876) 



2. Heating-surface Efficiency E s as a term is properly applied to the ratio of 
the heat absorbed by the boiler water and steam to the sensible heat brought 
to the heating surface by the hot gases, radiated to it per pound of fuel. As 
the heat usefully absorbed is the difference between the sensible heat of gases 
as developed in the furnace and the flue loss this may be set down in two ways : 

(B.T.U. absorbed by water or steam per lb. fuel) 



E 



(B.T.U. per lb. fuel) — (furnace and setting losses per lb. fuel)' 

(B.T.U. per lb. fuel) — (furnace and setting losses per lb. fuel)— (flue losses per lb. fuel) 
(B.T.U. per lb. fuel) — (furnace and setting losses per lb. fuel) 

Flue loss per lb. fuel 



(a) 



(b) 



(c) 



877) 



(B.T.U. per lb. fuel) — (furnace and setting losses per lb. fuel) 

3. Boiler Efficiency Et, is most commonly applied to the ratio of heat absorbed 
by water and steam per pound of fuel to its calorific power per pound or it is the 
product of heating surface and furnace efficiencies. 

B.T.U. absorbed by water per lb. fuel 



Et 



-»-[ 



B.T.U. per lb. fuel . . . . . 

furn ace and setting losses per lb. fuel "! 
B.T.U. per lb. fuel J 

Flue loss per lb. fuel"] 
~ _ B.T.U. per lb. fuel J 



= E f XE s . 



(a) 

(&) 

(c)J 



(878) 



HEATING BY COMBUSTION 



799 



The experimental determination of all these losses and efficiencies is con- 
tinually going on, as it has for many years, and from the data, boiler designers 
and makers with the assistance of inventors — who, however, generally ignore 
the test data available — all have continuously sought to improve results and 
are still trying. As a consequence one might expect to find modern boilers 
turned out by the best engineers ever so much more efficient than those built 
at the country cross-roads plate shop, but this is not the case, nor can it be said 
that one type is any more efficient than another, if small portable and other 
special forms be excluded, nor are recent designs more efficient than old ones. 
This is a most striking situation and is very nicely illustrated by the summaries 
made by Donkin after studying some four hundred tests of all kinds of boilers 
located everywhere and which are quoted from his book in Table CII. 

Table CII 
BOILER EFFICIENCY SUMMARIES (Donkin) 



Type of Boiler. 



Water tube, 1^-inch tubes 

Locomotive 

Lancashire 

Two-story r 

Two-story 

Dry back 

Return-smoke tube 

Cornish 

Cornish 

Wet back 

Elephant 

Water tube, 4-inch tubes . 

Lancashire 

Cornish 

Lancashire 

Dry back 

Lancashire 3-flue 

Elephant 

Lancashire 

Vertical 



No. of Tests. 



6 
37 
10 

9 
29 
24 
11 
25 

9 

6 

7 
49 
40 

3 
107 

6 



Mean of Best I Lowest of One 

Two Efficiencies Test Efficiency, 

Per Cent. Per Cent. 



81.4 
83.3 
74.4 
76.1 
79.8 
75.7 
81.2 
81.7 
81.0 
69.6 
70.8 
77.5 
73.0 
65.9 
79.5 
73.4 
66.7 
65.5 
74.3 
76.5 



66.6 
53.7 
65.6 
57.6 
55.9 
64.7 
56.6 
53.0 
55.0 
62.0 
58.9 
50.0 
51.9 
60.0 
42.1 
54.8 
52.0 
54.9 
45.9 
44.2 



Mean of All 

Tests, Efficiency 

Per Cent. 



77.4 
72.5 
72.0 
70.3 
69.2 
69.2 
68.7 
68.0 
67.0 
66.0 
65.3 
64.9 
64.2 
62.7 
62.4 
61.0 
59.4 
58.5 
57.3 
56.2 



These data certainly demonstrate that boiler efficiency, however many things 
may determine it, cannot be associated with boiler type, nor, it might be added, 
with size either. Admitting that in some cases the efficiencies may be high, 
that is, somewhere near 80 per cent, and that in other cases thay may be low, 
50 per cent or even less, the causes of such a possible range from good to bad 
when the average good performance of all types of boilers may be and is about 
the same, are worth investigating even if prediction of efficiency for specific 
conditions is impossible. The charge of bad management will not account 
for all the facts even though a poor fireman may make the performance of the 
best boiler as poor as the worst one over designed, and there can be absolutely 



800 



ENGINEERING THERMODYNAMICS 



no question as to very considerable differences in efficiency with the most skill- 
ful management, that can be traced partly to form and proportions of the boiler 
structure and partly to operating conditions more or less beyond the control 
of the operator. In a great many cases gains or losses in efficiency are found 
chargeable to almost absurdly irrelevant things and the literature of the sub- 
ject, especially that circulated by makers of boiler attachments such as grates,* 
firedoors, stokers, furnaces, baffles, draft apparatus, air preheaters, tube retard- 
ers and many other minor items, is full of such false conclusions, due no doubt 
to unskillful or over skillful testing that failed to discover which one of a possible 
dozen or so variables that may effect boiler efficiency was responsible for an 
improvement obtained, when the particular device was in use and confidently 
credited without qualification to it. Such unwarranted conclusions would be 
impossible if boiler tests were made with sufficient thoroughness to absolutely 
fix all conditions and permit of tracing the heat losses, then, and then only, 
could a change in efficiency be connected with the change in that one condition 
either of service or structure that really caused it. 

A really complete statement of results of a boiler performance accounting 
for supply and distribution of heat, termed the heat balance of the boiler, is 
almost impossible, yet may be pretty closely approximated by careful experi- 
mental work. The degree to which the approximation fails to represent the 
true distribution is indicated by an item containing unaccounted for losses 
often grouped with leakage of steam and water, and radiation. To illustrate, 
three different heat balances are reported in Table CIII below for, first — the 
U. S. Geological Survey Heine water tube boiler of 210 HP. and 2031 sq.ft. 
of heating surface as reported by Breckenridge; second — the fire tube loco- 
motive boiler of 400 H.P. and 1216 sq.ft. of heating surface as reported by 
Goss, and, third — a Stirling water tube boiler. 

Table CIII 
THREE EXAMPLES OF HEAT BALANCE FOR BOILERS 



Distribution of Heat of Coal, Per Cent. 


510 H.P. 

Stirling. 


210 H.P. 

Heine. 


400 H.P. 
Locomotive. 


1 Absorbed by water and steam 


74.87 

.24 

10.73 

8.24 


60.30 

.26 

15.19 

3.97 


57 00 


2. Loss due to vaporization of coal moisture 


5 00 


3. Loss due to sensible heat of stack gases 

4 Radiation leakage and unaccounted for 


14.00 

7 00 






Total per cent of fuel heat developed 


94.08 

.42 

5.50 


79.72 

2.17 

18.11 


83 00 


5. Loss due to unburned CO in stack gases 


1 00 


6. Loss due to other forms of incomplete combustion, H2, 
C m H ra in gases and fuel in ash cinder, in flues setting 
and discharged from stack 


16.00 


Total per cent of fuel heat not developed 


5.92 


20.28 


17 00 








100.00 


100.00 


100.00 



HEATING BY COMBUSTION 801 

About some of these items there may be a difference of possible interpre- 
tation, especially as to whether a given loss represents heat developed or 
undeveloped and more particularly as to whether heat developed, as indicated, 
is really available for boiler heating-surface absorption. For example, when 
the fuel contains hydrogen, the steam from which does not get a chance to 
cool below 212° F., the low value of it only is developed whereas the high value 
is reported by the calorimeter as the calorific power of the fuel. Again, 
the heat absorbed by vaporizing moisture in coal is developed heat but it is 
unavailable for heating surface absorption because it is taken up by coal water. 

To avoid conflicts in commercial transactions in which the efficiencies 
and losses in boilers are sometimes the subject of legal guarantees, standards 
must be available as subjects of agreement whether correct scientifically or 
not, and such standards are established by the American Society of Mechanical 
Engineers, which should be consulted by everyone interested. Assuming 
that the loss due to vaporization of coal moisture is unavailable for absorption, 
and that the items of radiation, leakage and " unaccounted for," in which 
nearly all errors are concentrated, represent available heat, the furnace and 
heating-surface efficiencies can be estimated and it must be understood that 
it is never possible to do more than estimate these. Charging all losses repre- 
senting heat not available for absorption against furnace efficiency, these 
efficiencies decrease. 

Stirling 

Efficiency of furnace per cent 94 

" " heating surf ace per cent . . 81 

" boiler complete per cent . 75 

These figures are given not as typical of these types of boilers or of any par- 
ticular service conditions, though they are as fair as any, but rather to show 
how division of performance may be made. In actual service and even during 
the conduct of tests, conditions are continually changing, so that a whole series 
of tests under conditions that seem identical will produce different results 
and show that the conditions were not identical, and the causes of variation 
are due almost entirely to the impossibility of fire control or the maintenance 
of steady conditions therein. The ash is constantly accumulating and 
more in one spot than in another because air passes most freely where 
bed resistance is least, and burns the coal faster at that point, the thickness 
of fire is varying, coal cannot be supplied with absolute uniformity and the 
air supply is not only varying all the time but the ratio of top to bottom 
air and side leakage vary more; all these things change furnace and fire condi- 
tions so as to disturb the heat balance. 

Even if the furnace efficiency were kept constant through a more perfect 
fire and air control than are possible except in tests, the conditions for heat 
absorption by the heating surface may vary considerably. Probably the 
greatest of the disturbing influences that tend to baffle attempts at analysis 
and often ignored even by first-class investigators, is the delayed and long-con- 



!eine 


Locomotive 


79 


78 


76 


73 


60 


57 



802 



ENGINEERING THERMODYNAMICS 



tinned combustion of the gases beyond the fire-bed or the relation of flame to com- 
pletely burned hot gases. 

All theories of heating-surface absorption that have ever been proposed 
are based on the assumption that the gases passing over the heating surface 
are immediately cooled thereby and continuously and regularly more cooled 
according to some, assumed law. Pyrometric investigations in flues show 
that so long as there is a clear, visible flame there is practically no cooling and 
even where the flame becomes of the flickering, irregular sort, the cooling of 
the gases is very much slower than beyond the limit of flame, or, in other words, 
generation of heat continuing during part of, sometimes all of, the absorption 
period, is a fact that makes it practically impossible to theorize on absorption 
laws. 

It must not be understood, however, that this theorizing is of no value 
because it is, and attention has already been called to Perry's idea of absorp- 
tion as most useful, and reasonably in agreement with observation on rate 
of heat absorption related to rate of heat generation. The efficiency of heating 
surface being the ratio of the amount absorbed to that generated and available 
for absorption, can be derived from the relations between rates of generation 
and absorption whether these relations are derived from a theory or represent 
average experimental data. According to the experimental data analyzed 
in the last section the relation between absorption and generation rates could 
be expressed in either of the following ways: 



B.T.U. absorbed per hr. per 
sq.ft. of heating surface 



B.T.U. absorbed per hr. per 
sq.ft. of grate surface 



= a-\-b ■ 



= aR+b 



B.T.U. developed in fire and 
available for absorption 
per hr. per sq.ft. of heat- 
ing surface 



B.T.U. developed in fire 
and available for absorp- 
tion per hr. per sq.ft. of 
grate surface 



from which two possible expressions for efficiency of the heating surface follow 
both of the same general form Eq. (865) since efficiency is 



Es ■ 



B.T.U absorbed 



B.T.U. developed in fire and available for absorption 

a 



(B.T.U. developed in fire and available for absorption per hr. per sq.ft. H.S.) 

aR 



= b + 



(B.T.U. developed in fire and available for absorption per hr. per sq.ft. G.S.) 



(b) 



. (879) 



These are equations of an equilateral hyperbola asymptotic to 1006 per 
cent efficiency as indicated by AB in Fig. 212. They seem to be at fault because 
they show that efficiency msiy exceed 100 per cent with small amounts of 



HEATING BY COMBUSTION 803 

generation, which means, however, only that the relation is not true for very 
small generations. For the working ranges of generation rates the heating surface 
efficiency line is substantially straight and falling with increase of generation. 
According to Perry's division of absorption between fire-box and flue or their 
equivalents for the particular structure in question, the efficiency of the flue 
part of the heating surface is independent of the rate of heat supply to it, so 
that all variations in the efficiency of the whole heating surface must be 
chargeable to the part receiving radiant heat. Just how true this is, or just 
how the whole efficiency of heating surface should, and really does vary with 
rate of generation cannot be settled with any available data, though there is 
enough to prove almost anything with a little skillful stretching. 

The curve AB of Fig. 212 does, however, within the limits of the experi- 
mental points from which it was plotted, those for the Hohenstein oil-fuel 
boiler, represent the facts of these experiments, and comparison with Eq. (879) 
for heating surface efficiency in terms of losses indicates a most interesting 
meaning for the vertical distance from the curve AB to the 100 per cent line. 
This distance must stand for the flue loss per pound of fuel divided by the 
generation per pound fuel or the calorific power less the furnace and setting 
loss and shows that for these experiments, the ratio is increasing with increase 
of total generation or rate of combustion. 

If on the same diagram, the line CD represents the overall boiler efficiency 
then the vertical distance from it to the 100 per cent line must represent the sum 
of the losses in the flue as sensible heat, together with the furnace and setting 
losses. Similarly, the vertical distance from CD to AB must represent the 
furnace and setting losses. If the two curves are parallel these latter losses 
are constant in amount; if they converge toward the right they must decrease 
with increase in heat generation, and if they diverge the losses are increasing 
with increase of production. In this particular case the losses first increase 
and then decrease and are separately plotted below, line EF. 

In all cases the flue losses increase in quantity with increase in heat genera- 
tion according to these data and it is worth while to investigate what meaning 
may be attached to this. 

Setting down the expression for flue loss which is represented by the distance 
from the heating-surface efficiency line AB to the 100 per cent line, in symbols 

Flue loss per lb. fuel in sensible heat 
B.T.U. per lb. fuel — furnace and setting losses per lb. fuel 

B.T.U. generated per square foot grate' 

For the above to be true either the numerator must increase or denominator 
decrease, that is, with increase of combustion rate the sensible heat per pound of 
fuel must increase or the furnace and setting losses per pound of fuel must increase, 
or both. This shows clearly how variations in the furnace conditions affect 
the heating-surface efficiency in spite of any supposed constancy of the efficiency 



804 



ENGINEERING THERMODYNAMICS 



5000 



B.T.U. Apparently Generated per hour per Square Foot of Heating Surface = Z 
(Pounds of oil burned perSq. Ft. Heating Surface perHr.x B.T.U. per lb.) 
10000 15000 20000 25000 3000O 



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Furnace Efficiency (per cent) = 100 £f 

Fig. 212. — Curves showing the Relations which Exist between Quantities Entering into 

Boiler Efficiency. 



HEATING BY COMBUSTION 805 

of the flue part. As the experiments from which the conclusion is derived 
include all possible ranges in rate of combustion and kind of coal and several 
very different types of boiler it must be accepted in spite of the apparent failure 
of the law of relation for low rates of generation where the curve passes the 
100 per cent line. 

The flue loss in sensible heat is the product of specific heat of gases, their 
weight per pound of fuel and the excess of temperature over the air supply and 
if the flue part of the heating surface takes out always the same part C then 

Heat absorbed by flues = C X (heat supplied to flue) .... (881 ) 

Flue loss - (1 - C) X heat supplied to flue, . . (882) 



or 



or 



CV'-V^stack - ^air) — (1 — C)C plc(tf UTriac(i — £ air ) (883) 



[(stack temp.) — (air temp.)] = constant Xf (furnace temp.) — (air temp.)] (884) 

provided the weight of gases per pound of coal, w, is constant. 'Experimental 
observation proves that the stack temperature always rises with increase in rate 
of combustion unless excess air be supplied at the same time, and as a rule the 
excess air and weight of gases per pound of fuel decreases with increased rate 
of combustion or increased draft if the fire is thick enough and free from holes, 
otherwise, and this is most common with moving-bar stokers and with anthracite 
small sizes, whether hand or mechanically fired, the weights of gases per pound 
of coal decreases fast and the stack temperature increases not as fast as it should 
or not at all. 

In the U. S. Geological Survey series on the Heine boiler the combustion 
chamber temperature increased somewhat irregularly with rate of combustion 
and rate of heat evolution as shown in Fig. 213, for a considerable variety of 
coals and at the same time the flue temperature bore the relation to combustion 
chamber temperature indicated in Fig. 214. Comparing individual readings, 
the ratio of excess temperature of combustion chamber over the boiler tem- 
perature, to the excess of flue temperature, the ratio was found to range from 
4 to 14, the best average being 7.3. This could hardly be called a constant 
ratio and the reason may be due to the fact that the flame extended between 
the tubes more in some cases than in others, or it may be that the expectation 
of constancy is based on wrong hypotheses, no one can say which, and yet this 
is the best data on the subject in existence. 

Examples of specific relations like the above could be cited almost without 
limit for other variables, and at the end, the problem would be little if any nearer 
to solution, that is, the problem of finding and expressing these losses or the 
efficiency of a boiler in terms of such variables of construction, or operation, 
as evidently have some influence in making it what it is. 



806 



ENGINEERING THERMODYNAMICS 




O m °12 U 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 
Pounds of Dry Coal Burned per Sq.Ft. of Grate per Hour 



~-T 2600 

a 

a 

§ 2500 

H 

3 2400 

3 2300 

O 

a 2200 
o 

| 2100 
| 2000 



































































































































































































































































































17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34, 35 
10,000's of B.T..U... supposedly evolyed per Sa.Et. of Grate per Haur 



Fig. 213. — Relation of Combustion Chamber Temperature and Rate of Combustion or Rate 

of Heat Evolution. 



E=i 610 



u 
"8 600 

<D 
P. 

s 

© 
H 590 



570 



. „^- 

zzzEzEEzzzzzzzzzz 

• H 1 1 I I 1 1 1 1 14441^ 



560 

1900 2000 



2100 2200 2300 2400 2500 2600 2700 2800 
Combustion-Chamber Temperature (°F.) 



Fig. 214. — Relation between Flue Gas Temperature and Combustion Chamber Temperature. 



HEATING BY COMBUSTION 



807 



However difficult division or analyses of losses may be, there is little doubt 
as to the sort of effect an increase or decrease of any one factor may have and 
it is, furthermore, usually possible to explain an overall efficiency curve when 
obtained, which will be clear from a few examples. 

These overall efficiency curves are not always given in terms of per cent 
efficiency, nor is capacity, in terms of B.T.U. absorbed per hour, but for 
efficiency the evaporation actual or equivalent per pound of fuel is more com- 
mon, and for capacity boiler horse-power or per cent boiler rating or rate of com- 
bustion. One such curve of considerable significance is that of Fig. 215, repre- 
senting the results from the Interborough Railway boilers equipped with two 
Roney stoker furnaces, one 70 per cent of the grate area of the other and burning 
good semi-bituminous coal, first on one grate alone and then on both together. 
Increased capacity causes a decrease in efficiency regularly along a line 

















































































































































































































































































































































































































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400 500 600 700 800 900 JO00 

B.H.P. 

Fig. 215.— Relation between Horse-power and Draft or Efficiency. 



1100 



almost straight with one grate, but when for a given capacity the second grate 
is used, the efficiency becomes higher after which it again decreases along a 
nearly parallel line, that is, in accord with a similar law. The only way this 
could happen is through a change in furnace conditions, probably traceable 
to excess air variations, as the draft suddenly decreases for a given amount of 
generation, when passing from the single to the double grate -as is indicated by 
the draft curves. If there are holes or air leaks, as is the case with such stokers, 
this would cause a decrease of excess air and an increase of efficiency exactly as 
occurred when the draft changed from a high to low value for the same or even 
greater total rate of heat evolution. In the Goss locomotive tests the evaporation 
is reported per pound of dry coal as a function of equivalent evaporation per 
square foot of heating surface, and shows a similar nearly straight line relation 
as in Fig. 208, and different for the two coals used. The reason for this 
appears at once from Fig. 216, which shows the whole heat balance plotted 



808 



ENGINEERING THERMODYNAMICS 




4 5 6 7 8 o 9 10 11 12 13 14 15 16 

Evaporation From & at 212°F, per Square Foot of Heating Surface per Hour. 

Fig. 216. — Heat Balance for Locomotive Boiler Working Under Various Rates of Evaporation. 



HEATING BY COMBUSTION 



809 



ro* 



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60# 





o 


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o 

e 



15 20 

Lbs. Air per Lb. of Coal 



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Fig. 217. — Influence of Various Factors on Boiler Efficiency 



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Sq. Ft. of Grate Surface 



12 3 4 5 6 7 

Evaporation per Sq. Ft. of Heating 
Surface per Hr. 




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Fig. 218. — Influence of Various Factors on Boiler Efficiency. 



HEATING BY COMBUSTION 811 

to the same base and is mainly due to a difference in cinder loss which in 
both cases increases with boiler capacity and draft that produces it. The 
essential difference between this fire and that last discussed is in its very 
superior thickness, lack of agitation by stoker bars, and lack of air holes at sides 
or end. 

Another good illustration of the controlling nature of some one condition 
is given by Myers, reporting results of boiler performance with tan bark as a 
fuel in leather-making works, the chief characteristic of the fuel being the 
large amount of moisture in proportion to the calorific power, which makes it 
unburnable in an ordinary furnace. The fuel showed about 9500 B.T.U. per 
pound when dry and ten samples contained 65^ per cent moisture average, 
leaving only 2665 B.T.U. per pound of fuel as fired, and which gave a boiler 
efficiency over all of 71 per cent when referred to dry tan bark and 54 per cent 
when referred to the bark as fired. Use of this wet fuel absolutely prevents 
the attainment of high efficiencies even in the best apparatus because of the 
moisture heat loss which is the controlling condition. 

Another illustration of this fact is furnished by a test of the well-known 
B. and W. water tube boiler by the Pittsburgh Testing Laboratory, using 
blast-furnace gas fuel and which yielded a little over 54 per cent efficiency 
because of insufficient air and premixture with the gas, the flue gas analysis 
showing as high as 20 per cent escaping unburnt. With exactly the same boiler, 
except as to furnace and combustion chamber, an efficiency in excess of 80 per 
cent is not difficult with an oil fuel if care be exercised to control air supply in 
proportion to oil and means be provided to mix the oil and air; failure to do 
this may yield just as bad results as noted for the blast-furnace gas. These 
little incidents could be multiplied without end but will suffice to show that 
boiler efficiency is controlled not so much by the boiler, as by the furnace, not so 
much by boiler designer as by fireman, but the analysis shows that there is a 
problem of design, more concerned, however, with high capacity of surface, 
and therefore, cost and size of boiler, than with efficiency improvements. 

A number of boiler tests are given in Figs. 217 and 218 with some one item 
of importance, selected to show the effect of various conditions of service and 
fuels in the same and different boilers, all of which are self explanatory in view 
of this discussion. 

Prob. 1. A furnace is fired with coal containing 12,500 B.T.U. per pound and 
needing 14 lbs. of air per pound of coal. In the ash there is 5 per cent of the weight 
of the coal in unburned carbon; the coal has 5 per cent moisture with it; there is 
3 per cent of CO in the flue gas; the stack temperature is 500° above that of the room, 
which is 80° F., and twice the combining proportion of air is supplied. For the above 
data and allowing for 5 per cent radiation, of which 4 per cent is from the setting and 
1 per cent from steam surface, what is the boiler efficiency? 

Prob. 2. A boiler makes 6 lbs. of steam per hour at 100 lbs. gage pressure from 
feed-water at 60° F. per pound of coal fired. The coal is No. 66 of the general table 
and the ash loss is 8 per cent, radiation 5 per cent, and the CO loss 2 per cent. What 
is the heating surface efficiency, the furnace efficiency, the boiler efficiency. 



812 ENGINEERING THERMODYNAMICS 

Prob. 3. The heating surface efficiency of a boiler is 80 per cent when making 7 lbs. 
of steam per hour, 97 per cent dry from feed at 100° F. per lb. coal. The coal used 
contains 12,500 B.T.U. per pound. What is the furnace loss and boiler efficiency? 

Prob. 4. The efficiency of a certain boiler is 72 per cent. The steam pressure is 
125 lbs. gage, feed-water 180° F. and there is 125° of superheat. The coal used is 
No. 87 of the table and 100 per cent excess air is supplied. Flue temperature is 100° 
above steam temperature and the room is 70° F. The evaporation per pound of coal 
is 6.5. What is the heating surface and furnace efficiency? 

Prob. 5. If in the Hohenstein boiler test, for which curves are plotted in Fig. 212, 
the oil used had a heating value of 18,000 B.T.U. per pound and the heating surface 
was 2130, what was the efficiency of the boiler when 1.5 tons of oil were burned per 
hour, and what the furnace efficiency and the heating-surface efficiency? 



HEATING BY COMBUSTION 813 



GENERAL PROBLEMS ON CHAPTER V 

Prob. 1. Natural gas consisting practically of nothing but methane is used in a gas 
engine. What would be the greatest amount of air which could be mixed with the gas 
and still have the engine run? The least? For the best mixture? To what pressure 
would, the gas have to be compressed to cause self ignition, the compression being 
adiabatic and the temperature at the beginning being 300° F.? 

Prob. 2. In a boiler test it was found that the room temperature was 70° F. and 
the stack temperature 450° F. The coal was of the following composition: H = .10, 
O = .04, C=.70, ash = .16. The flue-gas analysis was: C0 2 =8 per cent; CO =4 per 
cent; 2 =8 per cent; N 2 =80 percent. What was the B.T.U. loss per pound of coal 
due to sensible heat, to unburned carbon in flue gases and to moisture in coal? 

Prob. 3. A boiler efficiency of over 80 per cent is unusual. What would be the 
highest evaporation from and at 212° F. that might reasonably be expected from a 
pound of any one oil? 

Prob. 4. The average flue gas analysis during a boiler test showed C0 2 =.12; 
O 2 = .04; CO = .01; N 2 = .80. The coal was 80 per cent carbon and 20 per cent ash. 
W T hat was the amount of air used per pound of coal and what was the stack loss if 
the flue temperature was 450° F. and boiler room 90° F.? 

Prob. 5. For a bed temperature of 1700° F., what would be the CO to C0 2 ratio 
for air gas and water gas according to Boudouard's and Bunte's experiments? How 
much steam should be decomposed? 

Prob. 6. A boiler has an efficiency of 70 per cent and is steaming at the rate of 150 
boiler H.P. with feed-water at 60° F., steam pressure 125 lbs. gage, and 100° of super- 
heat. The grate area is 215 sq.ft. For pea coal of 12,000 B.T.U. per pound, what 
must be the draft for this condition? Should the rating be forced up to 200 H.P., how 
would the draft have to be changed? 

Prob. 7. A certain coal is said to evaporate 7 lbs. of water from a feed temperature 
of 60° and at a pressure of 150 lbs. gage to 96 per cent steam. What would be 
the equivalent evaporation from and at 212° F. for this coal? 

Prob. 8. Coal costing $1.00 per ton made 5 cu.ft. of coal gas, similar to No. 8, Table 
CX, per pound of coal, and the gas sold for $1.50 per million B.T.U. The other expenses 
of manufacture were such as to just balance the profit on the gas. At what price 
must the coke be sold per ton to net 10 per cent of the receipts for the gas? 

Prob. 9. In a certain boiler it is found that 68 per cent of the heat of the fuel gets 
into the steam. How many pounds of steam could be made per pound of coal No. 62, if 
the heating value were that as found from the ultimate analysis? How would the results 
differ if the heating value was that as determined from Eq. (757), from Eq. (759), 
from Eq. (773)? 

Prob. 10. Which of the following boilers is developing the greatest boiler horse- 
power if the actual pounds of water evaporated per hour is the same for each? 



814 ENGINEERING THERMODYNAMICS 



Feed Temp, 


Press Ga. 


Quality 


or^Superheat 


40 


100 


95 




60 


150 




100° 


60 


100 




100° 


225 


100 




100° 


225 


100 


95 




40 


200 


95 




40 


200 




100° 



Prob. 11. Taking an analysis of one of each of the following gases from the tables 
and assuming the actual pressure rise to be .5 of the calculated, find which would give 
the greatest pressure when burned with the correct amount of air at constant volume, 
starting with a pressure of 20 lbs. per square inch absolute and a temperature of 
200° F. Natural gas, coal gas, air gas, carburetted water gas, producer gas. 

Prob. 12. In a given boiler the heat absorbed per hour bears a straight line relation 
to that available, as follows: 

(Absorbed) =1000 +.6 (available). 

When 1000 lbs. of bituminous coal yielding 14,000 B.T.U. per pound are being 
burned per hour with an unburned fuel loss of 8 per cent, what will be the boiler horse- 
power? What should the grate surface be for a draft of .5 in. of water? 

Prob. 13. Coal No. 20 of Table CIV is used in a gas producer having a blast 
saturated with steam at 130° F. What will be the weight of coal required per million 
B.T.U. in the gas if the CO to C0 2 ratio is 6, and the volatile is the same as gas No. 5, 
Table CX, and requires 500 B.T.U. per pound of coal for its liberation? What will 
be the producer efficiency and the heating power of the gas? 

Prob. 14. A boiler is being fired with coal No. 14 of the general coal table. The 
heating value of the coal is based upon formula (763). How many pounds of steam 
at 100 lbs. per square inch gage with 100° superheat may be made per pound of coal 
from feed-water at 60° F., with a boiler efficiency of 72 per cent? 

Prob. 15. A gas engine running on producer gas showed a thermal efficiency of 
20 per cent. The producer used no steam and the ratio of CO to C0 2 was 5.5. What 
was the economy in cubic feet of gas per hour per horse-power and in pounds of coke 
per hour per horse-power if the fuel were all fixed carbon? 

Prob. 16. Air and steam are supplied to a gas producer, the gas from which is used 
in a water-cooled gas engine, direct-connected to an electric generator, the current 
from which supplies motors and lamps. How is the heat originally in the coal dis- 
tributed? Estimate quantities. 

Prob. 17. When gasolene explodes what is the per cent of gasolene present in the 
air mixture? What for alcohol? 

Prob. 18. An automobile engine is delivering 30 net horse-power and has a thermal 
efficiency of 15 per cent. What will be its radius of operation at this load on a 15- 
gallon tank of gasolene, for a one-ton car with an average resistance of 20 per cent of 
its weight? 

Prob. 19. A gas engine is running on carburetted water gas. The pressure before 
ignition is 80 lbs. gage and after ignition is 400 lbs. gage. The temperature before 
ignition is 800° F. Assuming the gas to be one in the table find the ratio of observed 
to calculated pressure rise. 



HEATING BY COMBUSTION 815 

Prob. 20. A water gas was found to have this analysis: H 2 =50 per cent: CO =40 
per cent; CH 4 = .5 per cent; C0 2 =5 per cent; N 2 =4.5 per cent. How does this value 
of H 2 compare with that as found from the general equation in terms of the carbon 
monoxide, dioxide ratio? Explain the difference found. 

Prob. 21. How much run-of-mine bituminous coal would be burned on a 100-ft. 
grate with 1-in., 1^-in., and 3-in. draft? What boiler horse-power would be 
developed in a Heine type with a ratio of heating to grate surface of 50? 

Prob. 22. What would be the factor of evaporation for the following cases? What 
weight of water must be evaporated per hour per boiler horse-power? 

Quality or Superheat 



;ed Temp. 


Pressure 


Que 


40 


100 


.90 


150 


100 


1.00 


200 


100 




50 


200 


.95 


50 


200 





150° 

100° 

Prob. 23. In beehive coke ovens all the gas made is wasted. Considering that the 
gas made is No. 4 of Table CX, and that 4 cu.ft. are made per pound of coal, calculate 
the heat-units which are being thrown away per ton of coal? How many horse-power 
hours could be developed in a gas engine of 20 per cent thermal efficiency, by the gas? 

Prob. 24. What are the thermal efficiencies of the two following systems: A steam 
engine is delivering one horse-power hour for 1.6 lbs. of coal while a gas-engine producer 
set delivers the same power on 1.1 lbs. of coal. The coal has the following composition: 
C=.7; O = .025; N=.02; H=.12; S=.05; ash = .13. 

Prob. 25. A boiler plant consisting of 12 boilers is supplied by one feed pump. 
The combined capacity of the boilers is 2000 H.P., the feed temperature is. 180° F., 
pressure 120 lbs. gage, quality 98 per cent. What must be the displacement of the 
pump, allowing 25 per cent extra for slippage and leakage? 

Prob. 26. It is found that for a given boiler test the loss due to moisture in the 
coal is 2 per cent, that due to CO, etc., in flue gas is 5 per cent, cinder and soot loss 
3 per cent, and radiation from fire 2 per cent. What is the furnace efficiency? If 
the flue loss is 10 per cent, what is the heating surface efficiency and boiler efficiency? 

Prob. 27. Coke is used in a producer and may be regarded as 88 per cent C and 
12 per cent ash. The steam air ratio in the blast is .3 and the CO to C0 2 ratio in the 
gas is 5.5. What will be the analysis of the gas by weight and volume and its heating 
value? What will be the efficiency of the producer? 

Prob. 28. It is desired to sell coal No. 23 of the coal table, gas No. 5, of the natural 
gas table, and oil No. 12 of the oil table, at one dollar per million B.T.U. What 
must be the price per ton of coal, 1000 cu.ft. of gas and per barrel (50 gals.) of oil? 

Prob. 29. A plant requires 1,000,000 B.T.U. per hour in the form of air gas. A 
set of samples taken from the gas line show an average value for the ratio of CO to 
C0 2 of 4. How many pounds of coal will be needed per hour by the producer? How 
many cubic feet of air per hour must be supplied as blast? 

Prob. 30. Oil is burned under a battery of boilers. The plant is equipped with 
the usual heat-saving appliances such as economizer, feed-water heater, condenser, 
and superheater. Trace the heat in the oil to its ultimate disposition Estimate quan- 
tities. 

Prob. 31. In the case of a leak of the following gases into a room which would 
earliest create a condition favorable for an explosion — Acetylene, benzene vapor, Blaugas, 



816 ENGINEERING THERMODYNAMICS 

ether, gasolene, methane? Which would pass the danger zone first? Assume mixing 
means to be present. 

<■ Prob. 32. For car lighting Pintsch gas is carried in tanks under a pressure of 10 
atmospheres. How does a cubic foot of it at this pressure compare in heating value 
to an equal volume of kerosene under normal pressure? 

Prob. 33. A producer is blasted with an air blast saturated at 150° F., and the CO 
C0 2 ratio is 5. What would be the composition of the gas by weight and volume and 
its calorific power? What the thermal efficiency from coke and gas yield per ton of 
coke? 

Prob. 34. A locomotive has a fire-box of 65 sq.ft. area and when running at full 
speed has an available draft of 6 ins. of water. The coal used is run-of-mine bitu- 
minous of 13,500 B.T.U. per pound. With an efficiency of 60 per cent what boiler 
horse-power could be developed? For a feed-water temperature of 40° F., a boiler 
pressure of 190 lbs. gage and a quality of 98 per cent how much water would be 
required for a two-hour run? How much coal would be needed for the same time? 
Check the horse-power for a 50 to 1 ratio of H.S. to G.S. by the absorption law of the 
Goss test. 

Prob. 35. A barrel of crude oil (50 gals.) is worth $1.10 at a certain point, while 
bituminous coal is worth $6.00 per ton at the same point. If the former gives a 
boiler efficiency of 85 per cent and the latter 75 per cent, see if it is cheaper to burn 
oil than coal. 

Prob. 36. A ton of coal of which 85 per cent was carbon is fed per hour to a producer. 
If there were 5 times as much CO in the gas as C0 2 and no H 2 , how many pounds and 
cubic feet of gas were made per hour? How much sensible heat must be carried away 
in the scrubber water? 

Prob. 37. Coal No. 15 of Table CIV is used in a producer. The heat required to 
drive off the volatile is 600 B.T.U. per pound of coal and the volatile may be taken 
as having the composition of gas No. 5 of Table CX. For a CO to C0 2 ratio of 5 and 
a value S = .3, what will be the composition and heating value of the gas? 

Prob. 38. Air is compressed to as high a pressure as 2000 lbs. per square inch. 
Should it be possible to compress natural gas No. 2 of the gas table to the same pressure 
without liquefaction what would be the B.T.U. per 1000 cubic feet? 

Prob. 39. What will be the cost per 1000 lbs. of dry steam made at 100 lbs. gage 
pressure from water at 60° F., when the boiler efficiency is 70 per cent, with (a) coal 
No. 29 of the coal table at $4.00 per ton, (6) oil No. 15 of the oil table at 75 cents per 
barrel, and (c) gas No. 3 of the natural gas table at 20 cents 1000 cubic feet? 

Prob. 40. What will be the efficiency of fixed carbon gasification in the case of the 
following producer gases: Air gas when CO to C0 2 =5, and CO to C0 2 =6, water gas 
when CO to C0 2 is 8. 

Prob. 41. Classify coal Nos. 11, 21, 30, 36, 47, 62, 81, 91, 110, and 121, according 
to the classification of Frazer, Muck, and Campbell. 

Prob. 42. A gas engine is running on producer gas No. 7 of the table. Assuming 
that the best mixture of gas and air is used, what will be the pressure after explosion 
for a compression pressure of 125 lbs. gage? What would be the effect of supplying 
25 per cent excess air, 25 per cent deficiency of air? 

Prob. 43. A large gas engine is running on producer gas No. 5 of the producer gas 
table. Taking the proper value for the usual compression of such an engine and 
assuming the compression to be adiabatic with y =1.4, how hot would the gases have 
to be at the beginning of compression to cause self-ignition? 



HEATING BY COMBUSTION 817 

Prob. 44. A coke producer is blasted with air saturated at 120° F. What is the 
maximum amount of hydrogen which could exist in the gas formed? Would this be 
when the CO or C0 2 was high? For the case of CO to C0 2 = 7 and the same value 
of S, what would be the per cent of hydrogen by volume and the heating value of 
the gas? 

Prob. 45. For the combining proportions of air and oil gas No. 1, from the table, 
what temperature would be required to cause ignition? 

Prob. 46. The ratio of heat absorbed by a boiler to that available is represented 
by equation of the first degree and for this case (a) =700 and (6) =.62. What will 
be the boiler horse-power when 90 per cent of the heat of coal No. 126 of the general 
table is available for absorption and 1500 lbs. are burned per hour? What the thermal 
efficiency of heating surface and boiler? 

Prob. 47. If the furnace losses are 10 per cent and the flue losses 15 per cent what 
will be the boiler efficiency and horse-power for a boiler under which coal No. 22 of 
the general table is burned at the rate of 400 lbs. every 10 minutes? What will be 
the amount of feed- water required for this boiler if its temperature is 200° F. and the 
steam pressure and temperature are 150 lbs. gage and 410° F. respectively ? 

Prob. 48. Coal No. 7, of the general coal table, is used in a producer. The volatile 
is the same as gas No. 5 of Table CX and 450 B.T.U. per pound of coal are needed 
to liberate it. The CO to C0 2 ratio is 6 and the blast is saturated at 140° F. The 
gas is used in an engine which requires 100 ft. of it per horse-power per hour. The 
same coal is burned in a steam plant and a horse-power hour secured for 1.4 lbs. 
Which plant shows the best efficiency? 



818 



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TABLES 



831 



CO CO 

"tfl CO t-H Tt< 

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Wcoeocort<TtiTtico 



CO 00 CM © 
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^O^WIMNiOH 

looiNiCHaiH© 

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l— t T-H 00 Ttl 

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lOONM 

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COiCMNOJOOCO 
NiONrHNCDiOIN 

cOcOcOcOcOcOcOco 



t^ CO 

CO CT> 

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OS OS CO iO 

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CO rH 1> 00 

co o o *o 

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CM OS 
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co co 

00 00 



l>O000rt<OSOsr^© 
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14679 
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TABLES 



835 



Table CVI 
PARAFFINES (C n H 2 „+ 2 ) FROM PENNSYLVANIA PETROLEUM 



Name. 



Formula. 



Boiling-point. 


°C. 


o K 


-25 


-13 





32 


38 


100.4 


30 


86 


69 


156.2 


61 


141.8 


97.5 


207.5 


91 


195.8 


125 


257 


118 


224.4 


136 


276.8 


173 


343.4 


182 


359.6 


198 


388.4 


216 


420.8 


238 


460.4 


258 


496.4 


280 


536. 


205 


401. 


234 


453. 


370 


698 



Specific Gravity 
at 32° F. 



Molec- 
ular 
Weight 
Approx. 



Composition 
by Weight. 



%C. %H 



Gas 



Liquid 



Solid 



Methane 

Ethane 

Propane 

Butane 

b Pentane normal . 



Pentane iso 

Hexane normal . 

Hexane iso 

Heptane normal 
Heptane iso. . . . 
Octane normal. . 

Octane iso 

Nonane 

Decane 



Endecane . 
Dodecane. 



Tridecane. . . 
Tetradecane. 
Pentadecane , 
Hexadecane . 
Octodecane . 
Eicosane. . . . 
Tricosane . . . 



Paraffine (myricle) 
Paraffine (ceryl) . . 



CH 4 
C2H6 
C3H8 
C4H10 
C&Hi2 

C5H12 
CeHu 
CeHi4 
C7H16 
C 7 H 16 
CsHis 
CsHis 
C9H20 
C10H22 

C11H24 

C12H26 

C13H28 
C14H30 
C15H32 
C16H34 
C18H38 
C20H42 
C23H48 
C25H62 
C27H56 
C30H62 



.446 

.536 

.60 

.627 at 57 

.628 

. 658 at 68 

.664 

.683 at 68 

.699 

.702 at 68 

.703 

.718 at 68 

.741 

.73 at 68 

.757 

.774 at -15 

.765 

.773 at -10 

.776 

.792 

.775 at 39 

.775 at 64 

.778 at 99 
.779 at 118 



16 
30 

44 

58 

72 

72 
86 
86 
100 
100 
114 
114 
128 
142 

156 

170 

184 
198 
212 
226 
254 
282 
324 
352 
380 
422 



75 

80.12 
81.84 
82.76 
83.33 

83.33 
83.76 
83.76 
84.00 
84.00 
84.21 
84.21 
84.38 
84.51 

84.62 

84.71 

84.78 
84.85 
84.92 
84.96 
85.02 
85.10 
85.18 
85.23 
85.26 
85.31 



25 

19.98 

18.16 

17.24 

16.67 

16.67 
16.24 
16.24 
16.00 
16.00' 
15.79 
15.79 
15.62 
15.49 

15.38 

15.29 

15.22 
15.15 
15.08 
15.04 
14.98 
14.90 
14.82 
14.77 
14.74 
14.69 



ETHYLENES (C„H 2ra ) AND NAPHTHALENES (C w H 2ra _ 6 +H 6 ) FROM RUSSIAN 

PETROLEUM 



Ethylenes 
Ethylene. . . 
Propylene. . 
Butylene. . . 
Amylene. . . 
Hexylene. . . 
Heptylene . 
Octylene. 



Naphthalenes . 



Nonylene . 
Diamylene 



Oct. Naphthalene 



Dodeca Naphthalene 



Triamylene . . 
Tetraamylene 



C2H4 


gas 




C 3 H 6 


gas 




C 4 H 8 


1 


33.8 


C5H10 


36 


96.8 


CeHi 2 


70 


158 


C7H14 


84 


183.2 


CsHi6 


119 


246.2 


CgHio+H6 


136 


276.5 


C9H18 







C10H20 


161 


321.8 


C11H22 


180 


356 


C12H24 






CwHig+He 


196 


384.8 


C14H28 


240 


464. 


C15H30 


248 


478.4 


C20H40 


over 


over 




390 


734 



.635 



.76 
.714 
.733 
.771 



.777 



803 



28 


85.7 


42 


85.7 


56 


85.7 


70 


85.7 


84 


85.7 


98 


85.7 


112 


85.7 


106+6 


85.7 


126 


85.7 


140 


85.7 


154 


85.7 


168 


85.7 


162+6 


85.7 


196 


85.7 


210 


85.7 


280 


85.7 



14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 



836 



ENGINEERING THERMODYNAMICS 



Table CVII 

CALORIFIC POWER OF MINERAL OILS BY CALORIMETER AND 
CALCULATION BY DENSITY FORMULA OF SHERMAN AND KROPFF 



No. 



Class of Oil. 



Sp.gr. 

at 
15° C. 



Degree B6. 



B.T.U. per Pound. 



Calo- 
rimeter. 



Calcul. 
!.&K.Form. 



1 

2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 



Gasolene 

Gasolene 

Gasolene 

Gasolene 

Kerosene 

California, refined 
West Virginia, crude . 
Kerosene 

Ohio, crude 

Pennsylvania, crude. . 
California, refined 

Kansas, refined 

West Virginia, crude . 
California, refined 
West Virginia, crude . 

Pennsylvania, crude. . 

Ohio 

Indian Territory .... 

Indian Territory .... 
California, refined . . . 
Kansas, crude 

Kansas, crude 

Illinois, crude 

California, refined. . . 
Pennsylvania, fuel oil 

Pennsylvania, fuel oil 
Indian Territory .... 

Kansas, crude 

Pennsylvania, fuel oil 
Kansas, crude 



71 
7175 

,72 

.7709 

.7830 

.7850 

.7945 

.795 

.7964 

.8048 

.8059 

.8080 

.8103 

.8237 

.8248 

.8261 

.8321 

.8324 

.8418 

.8421 

.8436 

.8466 

.8500 

.8510 

.8514 

.8534 

.8580 

.8597 

.8616 

.8640 

.8648 

.8660 

.8670 

.8690 

.8708 

.8712 

.8745 

.8773 

.8800 

.8807 

.8810 



67.2 

65.1 

64.4 

51.6 

48.8 

48.35 

46.2 

46.1 

45.8 

44.0 

43.7 

43.2 

42.8 

40.0 

39.7 

39.5 

38.2 

38.2 

36.3 

36.25 

36.0 

35.4 

34.7 

34.5 

34.45 

34.05 

33.20 

32.8 

32.5 

32.05 

31.9 

31.65 

31.5 

31.1 

30.8 

30.7 

30.1 

29.6 

29.0 

29.0 

28.9 



21120 
20389 
20527 
20038 
20018 
20014 
20030 
20135 
20236 
20068 
20057 
19802 
19963 
19766 
19827 
20021 
19757 
19782 
19710 
19795 
19924 
19685 
19715 
19724 
19701 
19784 
19389 
19379 
19741 
19555 
19656 
19555 
19530 
19534 
19654 
19614 
19354 
19428 
19447 
19435 
19435 



20938 
20854 
20726 
20314 
20206 
20194 
20098 
20094 
20082 
20010 
19998 
19979 
19962 
19850 
19838 
19830 
19778 
19778 
19702 
19698 
19690 
19666 
19638 
19630 
19630 
19610 
19578 
19562 
19550 
19530 
19526 
19516 
19510 
19494 
19482 
19478 
19454 
19434 
19410 
19410 
19406 



TABLES 



837 



Table CVII — Continued 

CALORIFIC POWER OF HYDROCARBON OILS BY CALORIMETER AND 
CALCULATION BY DENSITY FORMULA OF SHERMAN AND KROPFF 



No. 



42 
43 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
60 
61 
62 
63 
64 



Class of Oil. 



Kansas, crude. . 

Indian Territory 

Indian Territory 
Texas, crude. . . 

Kansas, crude. . 

Kansas, crude. . 
Texas, crude . . . 
Texas, crude. . . 
Texas, crude . . . 
California, crude 

Fuel oil 

California, crude 
California, crude 
Texas, crude. . . 
California, crude 



Sp.gr. 

at 
15 °C. 



.8820 
.8828 
.8833 
.8860 
.8862 
.8900 
.8914 
.8970 
.9007 
.9050 
.9065 
.9066 
.9087 
.9114 
.9137 
.9153 
.9155 
.9158 
.9170 
.9179 
.9182 
.9336 
.9644 



Degrees Be\ 



28.75 

28.7 

28.5 

28.0 

28.0 

27.3 

27.1 

26.1 

25.4 

24.7 

24.45 

24.4 

24.1 

23.6 

23.2 

22.95 

22.9 

22.9 

22.7 

22.5 

22.5 

20.0 

15.2 



B.T.U. per Pound. 



Calo- 
rimeter. 



19643 
19249 
19474 
19454 
19372 
19418 
19242 
19355 
19359 
19228 
19352 
19089 
19282 
19303 
19028 
19246 
19008 
18572 
19103 
18779 
18985 
19080 
18589 



Calcul. 
S.&K. Form 



19400 
19396 
19390 
19370 
19370 
19342 
19332 
19294 
19267 
19238 
19228 
19226 
19213 
19194 
19178 
19168 
19166 
19166 
19157 
19150 
19149 
19048 
18858 



+ 



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TABLES 



841 



Table CIX 
COMPOSITION OF NATURAL GASES 



Source. 



Authority. 



Volumetric Analysis. 



2 . CH 4 . C 2 H 6 . H 2 . CO. C2H4. i Ni. C0 2 



West Virginia 

Kansas 

Caucasus 

Caucasus 

Kokomo, Ind 

Kokomo, Ind 

St. Mary's, Ohio . . 

Marion, Ind. 

Marion, Ind 

Findlay, Ohio 

Findlay, Ohio 

English 

Russian 

Caucasus 

Anderson, Ind. . . . 
Anderson, Ind. . . . 

Ohio 

Fostoria, Ohio. . . . 

Muncie, Ind 

Muncie, Ind 

Findlay, Ohio 

Caucasus 

Caucasus 

Leechburg, Pa. . . . 
Penna. & W. Va. . . 
West Virginia 

Butler County, Pa 
Butler County, Pa 

U. S 

Pittsburgh, Pa. . . . 

Penna 

Pittsburgh, Pa 

U. S 

U. S 

u. s 

u. s , 



Report Gas Eng. 

Com. N. E. L.A... 
Report Gas Eng. 

Com. N. E. L. A... 

Bunsen 

Bunsen 

Levin 

Eng. &M. J 

Levin 

Lucke 

Eng. & M. J 

Levin 

Eng. & M. J 

Levin 

Lewes 

Lewes 

Bunsen 

Eng. &M. J 

Levin 

Lewes 

Eng. & M. J 

Levin 

Eng. & M. J 

GiU 

Lucke 

Bunsen 

Bunsen. 

Hoyle 

Allen & Burrell 

Report Gas Eng. 

Com. N. E. L. A... 

Hoyle 

Hoyle 

Ford 

Levin 

Juptner 

Hoyle 

Ford 

Ford 

Ford 

Ford 



99.5 



25 



.15 



1.1 



81.5 

80.11 

75.44 

72.18 

72.18 

67.0 

67.0 

65.75 

60.7 

57.85 

49.58 



35 



1.7 
1.42 

2.74 



1.64 
1.84 



1.86 
2.01 
1.89 
1.89 
2.5 
2.35 
2.3 
2.18 
.94 

4.79 



.2 
13.5 

6.1 
20.6 
20. 
22. 
22. 
26.12 
29.03 

9.64 
35.92 



.73 
.73 

.55 
.4 
.45 
.5 
.5 
.93 
3.50 
.26 



1. 
1. 

.6 

.6 

.8 
.58 
1.0 
.4 



4.39 



5.72 
18.12 

3.0 
1.0 
5.0 



12.3 



1.2 



29 
29 



.3 
.3 

.25 



2.18 

2.18 

.26 

.26 

.75 
.20 

.25 
.3 
2.6 



35 



.6 



55 



66 
34 

8 

.8 
6 
6 
6 



23.41 



842 



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844 



ENGINEERING THERMODYNAMICS 



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TABLES 



845 



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846 



ENGINEERING THERMODYNAMICS 



Table CXI 

COMPOSITION OF UNITED STATES COKE 

(Mainly from U. S. Geological Survey Reports) 



Origin. 



Moist- 



Vol- 
atile. 



Fixed 
Carbon. 



Ash. 



From Connelsville bituminous coal, 72 hours roasting. 
From Connelsville bituminous coal, 48 hours roasting. 
Foundry Ganley Mountain, U.S. Geological Survey. . . 

Foundry Milwaukee Solvay, U.S.G.S 

From Connelsville, U.S.G.S 

From Alabama. coal, U.S.G.S. No. 1 

From Arkansas coal, U.S.G.S. No. 6 

From Illinois coal, U.S.G.S. No. 2 

From Illinois coal, U.S.G.S. No. 3 

From Indiana coal, U.S.G.S. No. 1 

From Indian Territory, U.S.G.S. No. 2 

From Iowa, U.S.G.S. No. 1 

From Iowa, U.S.G.S. No. 3 

From Kentucky, U.S.G.S. No. 1 

From Kentucky, U.S.G.S. No. 4 

From Missouri, U.S.G.S. No. 2 

From West Virginia, U.S.G.S. No. 1 

From West Virginia, U.S.G.S. No. 2 • 

From West Virginia, U.S.G.S. No. 3 

From West Virginia, U.S.G.S. No. 4 

From West Virginia, U.S.G.S. No. 5 

From West Virginia, U.S.G.S. No. 6 

From West Virginia, U.S.G.S. No. 10 

From West Virginia, U.S.G.S. No. 12 

Connelsville average of 3, J. B. Proctor 

Chattanooga, Tenn., average of 4, J. B. Proctor. . . 

Birmingham, Ala., average of 4, J. B. Proctor 

Pocahontas, Va., average of 3, J. B. Proctor 

New River, W. Va., average of 8, J. B. Proctor . . . 
Big Stone Gap, Ky., average of 7, J. B. Proctor. . . . 

Alabama, run-of-mine, foundry, Moldenke 

Alabama washed slack, foundry, Moldenke 

Colorado washed slack, foundry, Moldenke 

Illinois washed slack, foundry, Moldenke 

Pennsylvania washed slack, foundry, Moldenke. . . . 
Pennsylvania washed slack, foundry, Moldenke. . . 

Tennessee, foundry, Moldenke 

Tennessee, foundry, Moldenke 

Virginia, foundry, Moldenke 

Virginia, foundry, Moldenke 

West Virginia, foundry, Moldenke 

West Virginia, foundry, Moldneke 

Proposed standard foundry coke specification 



.23 
.19 

.75 

.27 

.18 

.33 

30 

57 

96 

16 

60 

11 

80 

.51 

.52 

2.18 

.40 

.59 

.38 

.20 

.42 

1.00 

.60 

1.00 



1.32 
.51 
.35 

.48 

.32 

.72 

2.85 

2.83 

.44 

1.24 

1.85 

1.79 

1.95 

.84 

.73 

1.82 

1.95 

1.31 

.87 

1.15 

.43 

1.85 

.55 

.75 



1.34 


1.03 


.75 


.75 


.44 


1.31 


2.78 


.74 


.23 


.29 


.91 


2.26 


.22 


.11 


1.67 


1.6 


.16 


.80 


1.52 


1.67 


.67 


.46 


.60 


2.35 


.5 


.75 



88.18 

89.6 

86.38 

89.63 

88.75 

82.63 

78.84 

75.42 

87.08 

84.81 

80.25 

77.01 

78.64 

93.25 

86.40 

81.34 

87.47 

86.70 

84.48 

85.42 

84.34 

89.60' 

90.34 

90.37 

88.96 

80.51 

87.29 

92.53 

92.38 

93.23 

83.35 

86.00 

82.18 

83.35 

92.53 

80.84 

92.44 

76.87 

93.24 

88.52 

95.47 

84.09 

89.75 



10.27 
9.7 
12.52 
9.62 
10.75 
16.32 
17.01 
20.18 
11.52 
13.19 
15.30 
19.09 
17.61 
5.40 
12.35 
14.66 
.18 
11.40 
14.27 
13.23 
14.81 
7.55 
8.51 
7.88 
9.74 
16.34 
10.54 
5.74 
7.21 
5.69 
14.28 
11.50 
16.07 
13.13 
6.95 
15.99 
7.23 
19.86 
5.80 
"8.29 
4.00 
12.96 
9.0 



TABLES 



847 



Table CXII 



FRACTIONATION TESTS OF KEROSENES AND PETROLEUMS 





Class and Density of Original. 


Volumetric 
Per Cent 
Distilled. 


Temperature of 
Distillation. 


Density of 

Distillate, 

60° F. 


Density, 
Baum6. 


No. 


Deg. F. at 
Beginning. 


Deg. F. at 
End. 






23 


257 


302 


.748 


57.21 






11 


302 


347 


.767 


52.5 






8 


347 


392 


.783 


49.0 




American kerosene 


9 


392 


437 


.794 


46.5 


1 


Robinson 


10 


437 


482 


.807 


43.5 


Sp.gr. .797 


16 


482 


527 


.821 


40.8 




Be. 45.67 


7 


527 


572 


.831 


38.8 






3 


572 


680 


.836 


37.5 






Left as res. 














13 


680 




.843 


36.5 






9 


239 


284 


.786 


48.2 






18 


284 


329 


.799 


45.4 






20 


329 


374 


.816 


41.6 




Russian kerosene 


13 


374 


419 


.829 


38.9 


2 


Robinson 


18 


419 


464 


.831 


38.5 


Sp.gr. .825 


12 


464 


509 


.845 


36.8 




Be. 39.9 


6 


509 


554 


.857 


33.5 






1 


554 


680 


.864 


32.2 






Left as res. 














3 


680 




.877 


29.8 






25 


293 


338 






3 


American kerosene 

Robinson 

Sp.gr. 


23 

28 
13 


338 
383 

428 


383 
428 
473 








7 


473 


518 










3 


518 


572 






4 


Alsatian petroleum 
Engler & Schestopal 
Sp.gr. .801 
Be. 44.8 


.08 
30.35 
44.7 
20.2 

3.8 


302 
392 
482 
572 


302 
392 

482 
572 
608 








"Kaiser " oil 


29.7 


302 


392 






5 


Engler & Schestopal 
Sp.gr. .795 


32.3 
26.3 


392 

482 


482 
572 








Be. 46.1 

i 


11.7 


572 


608 






6 


Pennsylvania kerosene 
Maschinenfabrik, Augsburg 


15.8 

22 

19.25 


302 
392 


302 
392 

482 








Sp.gr. .800 
Be. 45 


16.8 
26.15 


482 
572 


572 
608 







848 



ENGINEERING THERMODYNAMICS 



Table CXII — Continued 
FRACTIONATION TESTS OF KEROSENES AND PETROLEUMS 





Class and Density of Original. 


Volumetric 
Per Cent 
Distilled. 


Temperature of 
Distillation. 


Density of 

Distillate, 

60° F. 


Density, 
Bauml. 


No. 


Deg. F. at 
Beginning. 


Deg. F. at 
End. | 


7 


Bavarian, crude 
Maschinenfabrik, Augsburg 
Sp.gr. .827 
Be. 39.5 


14.7 

12.1 

10.7 

9.1 

7.8 


302 
392 

482 
572 


302 
392 

482 
572 
608 










28.2 




302 








Roumanian, lamp 


15.2 


302 


392 






8_ 


Maschinenfabrik, Augsburg 


12.8 


392 


482 






Sp.gr. .815 


24.3 


482 


572 








Be. 41.9 


7.7 
5.4 


572 

608 


608 














302 








Galician, solar 




302 


392 






9 


Maschinenfabrik, Augsburg 


2.7 


392 


482 






Sp.gr. .874 


65.1 


482 


572 








Be. 30.2 


14 


572 
608 


608 










20.1 




302 








Hungarian, blue 


15 


302 


392 






10 


Maschinenfabrik, Augsburg 


12.6 


392 


482 






Sp.gr. .836 


12.2 


482 


572 








Be. 37.5 


4.3 
2.4 


572 
608 


608 














302 








German, red 


3.7 


302 


392 






11 


Maschinenfabrik, Augsburg 


38.9 


392 


482 






Sp.gr. .870 


39.3 


482 


572 








Be. 30.9 


12.3 


572 

608 


608 














302 








German, yellow parif. oil 


2.4 


302 


392 






12 


Maschinenfabrik, Augsburg 


55.0 


392 


482 






Sp.gr. .860 


37.0 


482 


572 








Be. 32.8 


2.8 


572 
608 


608 










13.8 




302 








German, solar 


57.4 


302 


392 






13 


Maschinenfabrik, Augsburg 


25.5 


* 392 


482 






Sp.gr. .860 


1.0 


482 


572 








Be. 32.8 


.1 


572 


608 













608 









TABLES 



849 



Table CXII — Continued 
FRACTIONATION TESTS OF KEROSENES AND PETROLEUMS 





Class and Density of Original. 


Volumetric 
Per Cent 
Distilled. 


Temperature of 
Distillation. 


Density of 

Distillate, 

60° F. 


Density, 
Baumi. 


No. 


Deg. F. at 
Beginning. 


Deg. F. at 
End. 


14 


German, benzol 
Maschinenfabrik, Augsburg 
Sp.gr. .873 Be. 30.5 


68 
28.7 


212 
302 


212 
302 






15 


Beaumont, Texas 
Richardson & Wallace 
Sp.gr. .912 
Be. 23 . 5 


2.5 
40.0 
20.0 
25.0 


230 
302 

572 
752 


302 
572 

752 


.8749 

.9089 

.9182 


30.1 
24.2 
23.6 


16 


Ohio 

Mabey & Noble 
Sp.gr. .829 
Be. 38.9 


23.0 
21.0 
21.0 
27.0 


185 
302 
572 
752 


302 
572 

752 


.7297 
.8014 
.8404 
.8643 


62.3 
45.1 

36.8 
32.2 


17 


Pennsylvania 
Sp.gr. .914 
Be. 23.2 


21.0 
41.0 
14.0 
23.0 


176 
302 
572 

752 


302 
572 

752 


.7188 
.7984 
.8334 
Paraffine 


-65.2 

45.8 
38.3 


18 


Virginia, petroleum, heavy 
B. Redwood 
Sp.gr. at 32° F. .873, Be. 30.5 


1.0 

1.3 

12.0 


212 

284 


212 

284 
356 






19 


Virginia, petroleum, light 
B. Redwood 
Sp.gr. 32° F. .8412 
Be. 36.6 


1.3 
4.3 
11.0 
17.7 
25.2 
28.5 


212 

248 
284 
320 
356 


212 

248 
284 
320 
356 
392 






20 


Pennsylvania, light 
B. Redwood 
Sp.gr. at 32° F. .816 
Be. 41.6 


4.3 
10.7 
16.0 
23.7 
28.7 
31.0 


212 

248 
284 
320 
356 


212 
248 
284 
320 
356 
392 






21 


Penn., heavy, B. Redwood 
Sp.gr. at 32° F. .886. Be. 


12.0 


500 


500 
536 






22 


Java, petroleum 
B. Redwood 
Sp.gr. at 32° F. .923 
Be. 21.8 


1.0 
1.0 

"7.7 
15.0 
22.3 
24.3 


212 
248 
320 
356 
392 
428 


212 

248 
320 
356 
392 
428 
464 







850 



ENGINEERING THERMODYNAMICS 



Table CXII — Concluded 
FRACTIONATION TESTS OF KEROSENES AND PETROLEUMS 









Temperature of 












Distillation. 








Class and Density of Original. 


Volumetric 
Per Cent 




Density of 
Distillate, 


Density, 


No. 










Distilled. 


Deg. ■ F. at 
Beginning. 


Deg. F. at 
End. 


60° F. 








.8 




212 








Java, petroleum 


3 


212 


248 






23 


B. Redwood 


9.3 


248 


284 






Sp.gr. at 32° F. .827 


16.3 


284 


320 








Be. 39.2 


22.0 

27.8 


320 
356 


356 
392 








Java, petroleum 






392 






24 


B. Redwood 


2.3 


392 


428 






Sp.gr. at 32° F. .972 


4 


428 


464 








Be. 14 


9 


464 


500 










2.1 




212 










4.6 


2-12 


248 








East Galicia, petroleum 


8.7 


248 


284 






25 


B. Redwood 


13.7 


284 


320 






Sp.gr. at 32° F. .870 


14.3 


320 


356 








Be. 30.9 


21.7 
25.3 
32.3 


356 
392 

428 


392 

428 
464 










4 


212 


248 










9.8 


248 


284 








West Galicia, petroleum 


14.3 


284 


320 






26 


B. Redwood 


23.3 


320 


356 






Sp.gr. at 32° F. .885 


27 


356 


392 








Be\ 28.1 


30.7 
36.7 


302 

428 


428 
464 







TABLES 



851 



Table CXIII 
FRACTIONATION TESTS OF GASOLENES 





Class and Density of Original. 


Volumetric 
Per Cent 
Distilled. 


Temp, of Distillation. 


Density of 

Distillate, 

60° F. 


Density, 
Baumi. 


No. 


Deg. F. at 
Beginning. 


Deg. F. at 
End. 


1 


Gasolene [Blount] 
Sp.gr. .739 
Be\ 59.5 


39 

49 
7.5 
3.5 


158 
212 
248 
271 


212 

248 
271 


.722 
.748 
.757 
.767 


63.9 
57.2 
55.0 
52.6 


2 


Gasolene [Blount] 
Sp.gr. .736 
Be. 60.2 


48 
37 
11.5 
2.5 


158 
212 
248 
271 


212 
248 
271 


.727 
.747 
.762 
.767 


62.5 
57.5 
53.9 
52.6 


3 


Gasolene [Blount] 
Sp.gr. .717 
Be. 65.3 


65.5 

26.5 

4.5 

2.5 


149 
212 
248 
271 


212 
248 
271 


.708 
.742 
.754 
.769 


67.9 
58.8 
55.8 
52.2 


4 


Gasolene [Blount] 
Sp.gr. .716 
Be. 65.5 


69.0 
22.0 

4.5 

3 


149 
212 
248 
271 


212 
248 
271 


.707 
.743 
.751 
.770 


68 
58.5 
56.5 
51.9 


5 


Gasolene [Blount] 
Sp.gr. .716 
Be. 65.5 


65.0 

26.0 

5.0 

2.5 


145 
212 
248 
271 


212 
248 
271 


.704 
.742 
.753 

.772 


68.9 
58.9 
56 
51.5 


6 


Gasolene [Blount] 
Sp.gr. .717 
Be\ 65.3 ' 


70.0 

24.0 

3.0 

1.5 


149 
212 
248 
271 


212 
248 
271 


.71 
.744 
.753 
.769 


67.2 
58.2 
55.9 
52 


7 


Gasolene [Blount] 
Sp.gr. .719 
Be. 64.7 


67.0 

21.0 

6.0 

4.5 


140 
212 
248 
271 


212 
248 
271 


.706 
.742 
.750 
.770 


68.2 
58.9 
56.8 
51.9 


8 


Gasolene [Blount] 
Sp.gr. .711 
Be. 66.9 


66 

24 
6.5 
2.5 


140 
212 
248 
271 


212 
248 
271 


.700 
.731 
.741 

.762 . 


70 

61.6 
58.9 
53.8 


9 


Gasolene [Blount] 
Sp.gr. .715 
Be. 65.8 


59 

28.5 
7.0 
4.0 


145 
212 

248 
271 


212 
248 
271 


.701 
.736 
.750 
.765 


69.8 
60.2 
56.6 
53.0 


10 


Gasolene [Blount] 
Sp.gr. .712 
Be. 66.7 


62.0 

25.0 

7.0 

5.0 


145 
212 
248 
271 


212 

248 
271 


.699 
.730 
.742 
.758 


70.1 
61.8 

58.8 
54.8 


11 


Gasolene [Blount] 
Sp.gr. .710 
Be. 67.2 

1 


68 

22.5 
6.5 
2.0 


136 
212 
248 
271 


212 
248 
271 


.699 
.736 
.750 
.736 


70.1 
60.2 
56.6 
60.2 



852 



ENGINEERING THERMODYNAMICS 



Table CXI 1 1— Continued 
FRACTIONATION TESTS OF GASOLENES 





Class and Density of Original. 


Volumetric 
Per Cent 
Distilled. 


Temp, of Distillation. 


Density of 

Distillate, 

60° F. 


Density, 
Baum6. 


No. 


Deg. F. at 
Beginning. 


Deg. F. at 
End. 


12 


Gasolene [Blount] 
Sp.gr. .700 
Be. 70 


86.5 

11.5 
"Vb" 


133 
212 
248 
271 


212 
248 
271 


.692 

.739 


72.3 
59.5 


13 


Gasolene [Blount] 
Sp.gr. .718 
Be. 65 


59 

29 

8 

3 


145 
212 
248 
271 


212 

248 
271 


.704 

.742 
.755 
.768 


69 
58.8 
55.5 
52.5 


14 


Gasolene [Blount] 
Sp.gr. .717 
Be. 65.3 


64 
26 
6.5 

2.5 


149 
212 

248 
271 


212 
248 
271 


.705 
.740 
.754 
.770 


68.8 
59.4 
55.8 
51.7 


15 


Gasolene [Blount] 
Sp.gr. .717 
Be. 65.3 


68 

23 
5.5 
2.5 


149 
212 
248 
271 


212 
248 
271 


.705 
.743 
.755 
.773 


68.8 
58.6 
55.5 
51.2 


16 


Gasolene [Blount] 
Sp.gr. .717 
Be. 65.3 


67.5 

22 
5.5 
3.5 


143 
212 
248 
271 


212 
248 
271 


.706 
.742 

.758 
.770 


68 

58.8 
54.9 
51.8 


17 


Gasolene [Blount] 
Sp.gr. .715 
Be. 65.8 


58 

24 
9.5 
6.5 


136 
212 
248 
271 


212 

248 
271 


.700 
.733 
.749 
.770 


70 
61 
57 
51.8 


18 


Gasolene [Blount] 
Sp.gr. .705 
Be. 68.6 


73 
17.5 

5 

3 


131 
212 
248 
271 


212 

248 
271 


.697 
.736 
.751 

.768 


71 

60.2 
56.5 
52.5 


19 


Gasolene [Blount] 
Sp.gr. .705 
Be. 68.6 


74 

15.5 
5.0 
4.0 


140 
212 
248 
271 


212 

248 
271 


.696 
.736 
.745 
.764 


71.1 

60.3 
57.9 
53.2 


20 


Gasolene [Chambers] 
Sp.gr. .71 
Be. 67.18 


6.67 
6.66 
6.67 
6.67 
6.66 
6.67 
6.67 
6.66 
6.67 
6.67 
6.66 
6.67 
7.67 
5.66 
4.37 


148.8 

149.2 

167.0 

176 

176 

186.8 

197.6 

206.6 

212.0 

219.2 

226.4 

233.6 

248.0 

258.8 

284.0 


149.2 

167.0 

176.0 

176 

186.8 

197.6 

206.6 

212.0 

219.2 

226.4 

233.6 

248.0 

258.8 

284.0 

311 







TABLES 



853 



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854 



ENGINEERING THERMODYNAMICS 



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TABLES 



855 



Table CXV 
RATE OF FORMATION OF CO FROM C0 2 AND CARBON 











Volumetric Analysis. 








Temp. 
Deg. F. 


Time. 
Seconds. 












Form of Carbon. 


C0 2 


CO 


CO 


CO 


Authority. 








co 2 


co+co 2 




Fine, amorphous 


1472 


480 


13.6 


86.4 


6.43 


.864 




Charcoal, 2-5 mm 


1472 


480 


39.9 


60.1 


1.51 


.601 




Charcoal, hazel nut . . . 


1472 


480 


17.1 


82.9 


4.88 


.829 




Coke, 2-5 mm 


1472 


480 


79.1 


20.9 


.26 


.209 


Boudouard 


Coke, hazel nut 


1472 


480 


83.6 


16.4 


.20 


.164 




Gas carbon, 2-5 mm . . 


1472 


480 


80.1 


19.9 


.25 


.199 




Gas coke, hazel nut. . . 


1472 


480 


86.7 


13.3 


.15 


.133 






1472 


189 


49.7 


50.3 


1.01 


.503 






1472 


116 


49.6 


50.4 


1.01 


.504 






1472 


57 


48.2 


51.8 


1.07 


.518 






1472 


46 


47.8 


52.2 


1.09 


.522 




1. Charcoal, 5 mm... . 


1472 


24 


62.5 


37.5 


.60 


.375 


Clement 




1472 


16 


71.7 


28.3 


.40 


.283 






1472 


12 


75.5 


24.5 


.32 


.245 






1472 


2.7 


93.7 


6.3 


.067 


.063 






1472 


1.6 


96.1 


3.9 


.041 


.039 






1562 


123 


25.7 


74.3 


2.88 


.743 






1562 


54 


29.8 


70.2 


2.36 


.702 






1562 


24 


42.8 


57.2 


1.34 


.572 




2. Charcoal, 5 mm. . . . 


1562 
1562 


13 
9.3 


47.4 
70.3 


52.6 

29.7 


1.11 

.42 


.526 
.297 


Clement 




1562 


4.6 


70.3 


29.7 


.42 


.297 






1562 


3.7 


77.6 


22.4 


.29 


.224 






1562 


3.3 


77.5 


22.5 


.29 


.225 






1652 


64 


12.7 


87.3 


6.87 


.873 






1652 


44 


13.3 


86.7 


6.52 


.867 




3. Charcoal, 5 mm. . . . 


1652 
1652 


10 
4.3 


29.2 
50.2 


70.8 
49.8 


2.42 
.99 


.708 
.498 


Clement 




1652 


2.8 


68.9 


31.1 


.45 


.311 






1652 


2.2 


65.6 


34.4 


.52 


.344 






1697 


119 


5.3 


94.7 


17.9 


.947 






1697 


81 


6.7 


93.3 


13.9 


.933 




4. Charcoal, 5 mm. . . . 


1697 
1697 


12 

5.8 


15.2 

28.2 


84.8 
71.8 


5.57 
2.54 


.848 
.718 


Clement 




1697 


4.3 


35.8 


64.2 


1.79 


.642 






1697 


2.3 


62.5 


37.5 


.60 


.375 






1832 


70 


5.1 


94.9 


18.6 


.949 






1832 


18.6 


5.7 


94.3 


16.5 


.943 




5. Charcoal, 5 mm. . . . 


1832 


8.2 


9.7 


90.3 


9.3 


.903 


Clement 




1832 


3.7 


20.3 


79.7 


3.92 


.797 






1832 


2.3 


20.5 


79.5 


3.88 


.795 






2012 


36.5 


1.3 


98.7 


75.9 


.987 






2012 


10.4 


1.7 


98.3 


57.8 


.983 




Charcoal, 5 mm 


2012 


4.97 


1.9 


98.1 


51.6 


.981 


Clement 




2012 


3.6 


2.7 


97.3 


36.0 


.973 






2012 


1.9 


5.4 


94.6 


17.5 


.946 






1652 


142 


72.4 


27.6 


.382 


.276 




6. Coke 


1652 
1652 


80 
44 


86.9 
90.6 


13.1 
9.4 


.151 
.104 


.131 
.094 


Clement 




1652 


25 


94.3 


5.7 


.061 


.057 





856 



ENGINEERING THERMODYNAMICS 



Table CXV — Continued 
RATE OF FORMATION OF CO FROM C0 2 AND CARBON 





Temp. 
Deg. F. 


Time, 

Seconds. 


Volumetric Analysis. 




Form of Carbon. 


C0 2 


CO 


CO 

CO2 


CO 


Authority. 




CO +CO2 




6. Coke 


1652 
1652 
1652 


16 
9.6 
3.7 


95.1 
97.4 
99.2 


4.9 
2.6 

.8 


.051 
.027 
.008 


.049 
.026 
.008 


Clement 


7. Coke 


1832 
1832 
1832 
1832 
1832 
1832 
1832 
1832 


123 

80 

33 

19 
6.4 
4.1 
3.1 
2.0 


21.6 
35.6 
47.1 
68.0 
86.1 
88.5 
90.8 
93.7 


78.4 
64.4 
52.9 
32.0 
13.9 
11.5 
9.2 
6.3 


3.62 
1.81 
1.12 
.47 
.16 
.13 
.101 
.067 


.784 
.644 
.529 
.320 
.139 
.115 
.092 
.063 








8. Coke 


2012 
2012 
2012 
2012 
2012 
2012 
2012 
2012 
2012 
2012 


90 

30 

13 
6.7 
3.2 
1.8 
1.7 
1.6 
1.5 
.96 


2.9 
14.6 
33.9 
44.4 
68.3 
69.6 
76.0 
77.9 
78.6 
86.7 


97.1 
85.4 
66.1 
55.6 
31.7 
30.4 
24.0 
22.1 
21.4 
13.3 


33.6 
5.85 
1.95 
1.25 
.46 
.437 
.316 
.284 
.272 
.154 


.971 
.854 
.661 
.556 
.317 
.304 
.240 
.221 
.214 
.133 


Clement 






9. Coke 


2192 
2192 
2192 
2192 
2192 
2192 


19 

13 
8.3 
2.4 
1.6 
1.1 


1.1 
2.2 

4.7 
31.5 
56.1 
66.5 


98.9 
97.8 
95.3 
68.5 
43.9 
33.5 


89.7 

44.4 

20.2 
2.18 
.78 
.504 


.989 
.978 
.953 
.685 
.439 
.335 


Clement 






Coke '.. 


2372 
2372 
2372 
2372 


8.9 
4.1 
2.1 
1.1 


.1 

2.1 

6.8 

16.6 


99.9 
97.9 
93.2 
83.4 


999 
46.5 
13.7 
5.02 


.999 
.979 
.932 
.834 


Clement 






10. Anthracite 


2012 
2012 
2012 
2012 
2012 


34 
9.4 
5.4 
3.3 

2.4 


12.2 
39.9 
52.3 
69.8 
73.5 


87.8 
60.1 
47.7 
30.2 
26.5 


7.2 

1.5 
.91 
.43 
.36 


.878 
.601 
.477 
.302 
.265 


Clement 


11. Anthracite 


2192 
2192 
2192 
2192 
2192 


47 

10 
5.1 
2.8 
1.6 


.3 
14.4 

28.5 
57.7 
69.0 


99.7 
85.6 
71.5 
42.3 
31.0 


332.3 
5.95 
2.5 
.73 

.45 


.997 
.856 
.715 
.423 
.310 


Clement 


12. Anthracite 


2372 
2372 
2372 
2372 
2372 
2372 


12.4 
6.0 
3.6 
3.0 
1.91 
1.07 


.1 
3.5 
17.6 
19.1 
33.7 
49.7 


99.9 
96.5 
82.4 
80.9 
66.3 
50.3 


999 
27.6 
4.68 
4.23 
1.97 
1.01 


.999 
.965 
.824 
.809 
.663 
.503 


Clement 



TABLES 



857 



Table CXVI 
COMPOSITION OF WATER GAS 



No. 



9 

10 
11 
12 
13 

14 
15 

16 

17 

18 

19 



Description. 



Essen water gas, coke, Sexton 

Dellurck process water gas, Lewes 

No. 3 

Strong water gas, Moore 

Dellurck Process water gas, Lewes 

No. 1 

Average water gas, Lewes 

From anthracite before carburetting 

for illumination, O'Connor 

Dellurck Process water gas, Lewes 

No. 2 

Blue water gas, Morehead 

Water gas. Allen 

Uncarburetted water gas 

Uncarburetted water gas 

Essen water gas, coke, Thorpe 

Water gas before carburetting, 

Lowell 

Average water gas, Lewes 

Water gas before carburetting, 

average 

Water gas before carburetting, 

average 

Lowe water gas, anthracite, Thorpe. 
Water gas, anthracite, Loomis Petti- 
bone 

Water gas, bituminous coke, Loomis 

Pettibone 



Volumetric Analysis. 



54.52 



51.8 



47.97 



CO. 



31.86 

37.50 

35.88 

38 . 30 
40.08 

43.4 

39.95 
43.25 
42.89 
45.89 
35.93 
43.75 

43.2 
35.93 

42.75 

44.85 
42.10 

42.4 

32.6 



CH4. 



1.62 



4.11 



10 



.5 
.75 

1.05 
.31 

2.0 
1.05 

4.23 

4.41 

2.7 
2.9 



CC-2. 



12. 

4.08 
2.05 

4.73 
4.80 

3.5 

5.38 

3.0 

2.97 

3.87 

4.25 

2.71 

3.0 
4.25 

2.80 

4.45 
3.60 

3.5 

5.3 



02. 



00 



Ni. 



5.2 
4.33 

3.80 
3.13 

1.3 

3.36 

3.25 

3.74 

.71 

8.75 
4.00 

2.8 
9.95 

2.2 

.77 
9.80 

6.9 

16.8 



Ratios. 



CO 

C0 2 



9.2 
17.5 

8.1 
8.35 

12.4 

7.4 
14.4 
14.4 
11.8 

8.45 
16.1 

14.4 

8.45 

15.3 

10.1 
11.7 

12.1 

6.15 



CO 

CO + COi 



.90 
.95 

.89 



93 

.88 
935 
935 
92 
89 
94 

93 

89 

94 

91 
92 

92 

86 



858 



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TABLES 



861 



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cc 


CM 


-^ 


CM 


OS 


CM 


CO 


00 


CM 


00 CM 


o 


to 




LC 


00 


to 


OS 


CO 


00 


o 


to 


r^ 


a 


CN 


^ 


to 




c 


00 


"* 


o 


T* 


Tt< 


CO 


rJH 


Tt< 


CO 


cc 


TjH 




o 


T-H 


o 


CO 


«tf 


to 


00 


CO 


r^ 


c 


OS 


«o 


CO 




cc 


">* 


OS 


CO 


*-• 


T|H 


o 


N 


CO 


t> 


OS 


o 


t> 




r> 


00 


CO 


O 

T-H 


CO 


co 


o 


to 


CO 


c 

CN 


00 




tO 




CT 


T^ 


CO 


8 


© 


OS 


i> 


T-H 


to 


OS 


cc 


o 




tC 


00 


00 


Tt< 


t-H 


CO 


CO 


1-1 




T-H 


Tf 


o 




o 


CM 


OS 


CO 


to 


CO 


CM 


os 


OS 




to 


(N 


OS 




1 — 


00 


»o 


CM 


to 


1> 


TJH 


to 


CO 


C^ 


Th 


tC 


to 




'd 


CO 


to 


«* 


to 


CO 


^ 


o 


OS 


c 


1> 


r^ 


oo 




c 


00 


o 


CO 


00 


OS 


o 


o 




Tt 


to 


O 


to 




CC 


CM 


T-H 


CM 


CO 


00 


CO 


CM 


,_, 




00 




CM 




o 


to 


CO 


CO 


CO 




CO 


CO 


CM 


cc 


CO 


cc 


CO 




cc 


CO 


<N 


o 


t— i 


CO 


T— | 


CO 


OS 


fr- 


00 


LC 


CO 




(M 


o 


CM 


to 


co 


"* 


to 


CO 


OS 


ee 


CM 


cc 


CO 




l> 


CO 


(N 


CO 
CO 


i-H 


1-H 


T-H 


CM 


CM 


CC 


1> 


Tt- 


CM 




oc 


i-H 










<sf 








pj 






of 














a 








CD 














fc 




CM 


o3 








M 






03 






OS 

6 


o 

T— 

CD 

.-a c 


T-H 

6 


rH 

6 


T3 
a 


OS 
C 


1> 

6 




S c 




00 

6 


T3 





CM 

6 


.sf 
"3 

'3b 




of 

"3 

'3b 


faC 


*3 

'ft 


of 

"3 
'Sn 


CN 
C 


tC 

c 


'3 
'3b 


b£ 

O 




O 

02 


u 




-. ** 














u 






02 


> 


2 f" 


> 


> 


§- 


s> 


> 


C 


CO > 
«2 ^ 


*> a 


> 


o 



CN 


3 


+3 


*» 


■+3 


»J C 


■+= 


+3 




c 


a 


■+= 


C 




8 


8 


8 


^^ 


$ 


CD 


■+; 


-J ^3 


^ 


£ 


-M 


tz 


+3 


£ 


PQ 


£ 


£ 


ffl 


£ 


£ 


PC 


PQ 


M 


£ 


PQ 




s 



866 



ENGINEERING THERMODYNAMICS 

Table CXIX 
COMPOSITION OF OIL PRODUCER GAS 






Name. 


Volumetric Analysis, Per Cent. 


Ratio. 


B.T.U. per 

Cubic Foot. 


CO 


H 2 


CH 4 


CftB^ra 


2 


C0 2 


N 2 


CO 
CO2 


CO 


High. 


Low. 




CO +CO2 


Process of Interna- 
tional Amet. Co . 

Do 

Do 

Lowe process 

it a 
it it 


8.0 
8.6 

7.8 
7.3 
8.35 
6.0 


12.0 
10.0 
9.8 
47.4 
53.65 
46.0 


16.2 
7.0 
6.0 
28.6 
22.50 
26.0 


2.0 
4.2 
4.0 

10.0 
5.4 

10.30 


.2 
.3 
.4 
.2 
.4 
.3 


4.2 

5.4 

6.5 

2.0 

2.25 

3.0 


57.4 

64.4 

65.5 

4.5 

7.45 

8.4 


1.9 
1.6 
1.2 
3.7 
3.7 
2.0 


.66 
.61 
.55 

.78 ' 

.79 

.67 


275 

209 
192 
661 
543 
630 


249 
192 
176 

605 

487 
566 



Table CXX 
COMPOSITION OF POWDERED COAL, PRODUCER GAS 





Volumetric Analysis, Per Cent. 


Ratio. 


B.T.U. per 
Cubic Foot. 


Sample. 


CO 


H 2 


CH 4 


C ra H2 W 


2 


C0 2 


N 2 


CO 
C0 2 


CO 


High. 






CO +CO2 


Low. 


1 


15.85 


6.17 


4.09 




1.4 


9.2 


63.29 


1.7 


.63 


119 


Ill 


2 


13.52 


11.51 


5.17 




.3 


8.1 


61.40 


1.7 


.63 


140 


129 


3 


12.20 


10.50 


3.20 




.0 


7.6 


66.50 


1.6 


.62 


112 


103 


4 


18.2 


12.20 


2.1 


.1 


.1 


4.9 


62.40 


3.7 


.79 


128 


119 


5 


13.8 


10.4 


2.5 


.5 




8.0 


64.80 


1.7 


.63 


118 


109 



TABLES 



867 





p 

H 
PQ 

09 

9 


Per 

Cu.ft. 

32° F 

29.92" 

Hg. 


os HN00 00O^(N 

IOH0005NOMNM 

OCOOCOcOCOOi-HOO 

O00O5OOiOo3Oi» 


CO 

T-H IO 


94.53 
59.30 
88.09 

67.102 


59.15 
64.98 
66.67 


69.23 
100.8 
106.66 
107.2 

91.38 




73 

Pho 
Ph 


1258.6 

1469 

1176.4 

1931.7 

1196.6 

1219.17 

1244 . 1 

1211.7 

1209.5 


CO tO 

t-H OS 

T-H T-H 

CO CO 


1296 . 6 
735.9 

1278 

893.6 


680 

855.6 

918.2 


HH00ONH 

OS CO CO OS CO 

co to 00 

OS CO CO CO i-H 




p 

H 
O 


Per 

Cu.ft. 

32° F. 

29.92" 

Hg. 


CO rH CO 

i-HCoastocoiocor^co 

HHOStOOST-HOSHHt^oO 

cocooitor^r^iocOT" 

tH ,-H CO 


00 to 

OS CO 
t^ OS 

tO t-H 


659.6 

102.53 

277.13 

134.13 


110.43 
131.45 
144.78 


CO 

(OOJNNH 

CS^HNO 

to to CO i-H to 

i-H to t^ CO OS 

i-H tO OS 


'cfi 

W 
P 


Oh 


NOJH* 

OlCOCOCOi-HtOCOCOOO 
COOSCOiOOOOCOtOtO 

MOO^ohhcOO 1 *^ 

HtHHOONNlOO) 
lO(N(NNHHHH 


Tt< CO 

00 «* 
CO HH 
HH OS 


12888 
1273.8 
6344 

1928.9 


1397.4 
1849.9 
2292 


2276 . 9 
12100.4 
20519.7 
20255 
21301 


t-3 
< 
> 

o 
►J 


to 




o (2 


CO 

OOl HH CO T* 

OS O OS 00 to tooco 
00 • CO tO OS to CO 00 00 

• l> 

cor^cocoi-HT^r-icoto 

HrHN HH COCOCO 


to 

00 CO 

H* CO 

CO T-H 


18 . 5288 
12.4239 
22.89377 

14.381 


12.654 
14.074 
15.83022 


tO CO CO 
CO 00 00 O CO 
CO t^ to t-h rH 

t^ t^ CO CO CO 

i-H CO 


J. 

p 
H 

a St? 

* H 1 

«P3co 

o d 

CO << 


ft*-£ 

« ft ^ 

3 ° 


i-H CO to 

O CO Hfll NINO 
OONiONCOOOOON 

00"OHH05C003rHMOO 

t^O^t^OOi-HCO^cO 

ooooocoooo 


i-H CO 

CO to 

O i-H 


.05397 
. 08049 
.04368 

.069536 


.079021 

.07105 

.063169 


i-H 

to 

i-H 00 co 00 co 

O CO OS tHH tHH 
t>- i-H O t^ THH 

O O CO tHh 


< 


ftO k 
. ■ c3 


O H/i I> to -* CO CO O 

HHMlNiOHHCONM 

CO-HHt^TjHCOCOCO^HOJ 
CO 1-H 1-H l-H T-H i-H i-H 


CO CO 

00 os 
-tf 

t-H T-H 


8.9395 
.7309 
3.964 

1 . 1586 


1.059 
1.162 
1.4961 


1.5036 

9.06 
15.24 
15.079 
17.004 


ft£ ^ 


OJOJCO^cOiOOOCiO 
COCOtOCOI>OOOCOTti 

C0C0C5rt<cOiOiOcOTt< 

i-H r-H CO 


T-H CO 

O CO 
CO OS 
tHh CO 

tO i-H 


5.9774 

.7289 

2.1458 

.9989 


.9374 
1.023 
1.1711 


CO OS 

tO T4H CO 

ON^HOl 

co thh t> t>. co 

t-h tHH co 00 OS 

i-H to 00 


o 

M 

Ph 

to 
"m 

j*> 

"3 

C 
U 

'C 

(D 

s 

"o 


£ 


to CO ^ 

co to Tf 


T-H 

• 


2.2 
59.5 
9.27 

52.9 


66.69 

55.5 

42.5 


CO 

• 

T-H • • 

to 


6 



tO i-H 

i-H CO CO 


to to 

CO 


3.8 
10.0 
4.25 

5.5 


10.50 

8.2 
14.5 


OS 

CO • 1 • • 

' q6 '•'•'• '• 


6 


CO 
O CO CO 


i-H tO 


to • 


T* 




i-H i-H 


co »o 

CO • • • CO 


1-H 

H 

o 

Ph 
Ph 

u 


6 


rH OS CO 

: : : : : © ^ co ^ 


00 

CO rH 

T-H 








i-H 


d 


: : : : ; : ; : ; ; : : : : ; : ; : : : ; : 




■ • • O • • t-h Oi CO 

■ T-H • T-H CO T-H 


1> 

CO l-H 


os • 

CO • 




■ CO • 




P^ 

q 


w 



100 

31.35 

40.4 

27.1 


CO to 

TJH CO 

CO »o 


to 
i>- • c 

O : r- 

co 


CO 


CO 

to 


4.85 
98^2 


►h* 


M 


• O • ■ • • IO TJH CO 

• T-H • • • • CO t^ CO 

tO CO t« 


to 

CO 00 

HH i-H 


00 O tc 

r-H CO O 
CO H; 


co 
»o 

T-H 


2.81 
14.0 
29 


CO 

CO 


..... 


o 


. 








O CO t-h CO 



rH 00 t^ CO 




CO l-H 


co 

t-H IO O 

00 i> to 
CO CO CO 


CO 


14.34 

20.0 

12 


O to 
OS • • • CO 

d ' ■ • 

CO 






03 

O 

"o 

a) 

s 


Carbon monoxide. . 

Hydrogen 

Methane 

Ethylene 

Ethane 

Benzene 

Retort coal gas. . . . 
Coke-oven gas, rich 
Coke-oven gas, lean 
Coke-oven, gas, av- 
erage 


: & 

HJ 

o3 

a* 

1-2 

CD +3 
Sh -+J 

• al 

~ 3 


XT. 

eg 

faj 


w 
c3 
bfi 

<v 

o3 

-J fal 

) 03 cc 


1 



rH • 

P, ; 

CP oi 
Hg b£ 

o3 

5 P 3 

< 


Bituminous pro- 
ducer gas (up).. . 

Do. (down) 

Do. mond 

Lignite producer 
ffflS 


Alcohol 

Kerosene 

Natural gas, Kans . 



868 



ENGINEERING THERMODYNAMICS 



Table CXXII 
COMPOSITION OF BOILER FLUE GASES— (Volumetric) 





Stat. Boiler, Illinois Coal.U. S. Geological Survey. 


Locomotive Boiler, U. S. Geological Survey. 


Average 
of. 


Analysis. 


CO 
CO2 


CO 


Analysis. 


CO 

co 2 


CO 




C0 2 


2 


CO 


CO +CO2 


C0 2 


2 


CO 


CO +CO2 


4 


3.4 


17.5 











10.16 


8.49 


.13 


.0128 


.0126 


3 


3.7 


17.2 











11.10 


7.84 


.23 


.0207 


.0203 


5 


4.4 


16.3 











11.15 


7.52 


.20 


.0179 


.0176 


5 


5.0 


15.0 











11.45 


6.92 


.00 








5 


5.3 


14.7 


.1 


.0189 


.0185 


11.46 


7.49 


.10 


.00875 


.00865 


5 


5.9 


14.4 


.04 


.0068 


.00674 


11.50 


7.08 


.17 


.0148 


.0147 


6 


6.2 


14.1 


.03 


.00485 


.00482 


11.96 


7.00 


.23 


.0193 


.0189 


9 


6.4 


13.7 


.07 


.0109 


.0108 


11.96 


7.07 


.14 


.0117 


.0155 


16 


6.6 


13.0 


.10 


.0152 


.0149 


12.05 


6.93 


.15 


.0125 


.0123 


9 


6.8 


12.6 


.01 


.00147 


.0147 


12.20 


6.94 


.05 


.0041 


.0407 


14 


7.0 


12.8 


.06 


.0086 


.0085 


12.45 


5.87 


.22 


.0177 


.0174 


20 


7.2 


12.6 


.08 


.0111 


.011 


13.57 


4.49 


.20 


.0147 


.0145 


18 


7.4 


12.4 


.00 








13.87 


4.75 


.25 


.018 


.0177 


20 


7.6 


12.9 


.05 


.0066 


.00655 












14 


7.8 


12.1 


.03 


.00385 


.00375 












30 


8.0 


11.7 


.04 


.005 


.00498 












31 


8.2 


11.6 


.10 


.0122 


.012 












27 


8.4 


11.3 


.10 


.0119 


.01175 












16 


8.6 


11.1 


.10 


.0116 


.0115 












17 


8.8 


10.8 


.20 


.0228 


.0222 












19 


9.0 


10.7 


.10 


.0111 


.011 












14 


9.2 


10.4 


.10 


.0109 


.01075 












16 


9.4 


10.1 


.20 


.0213 


.0208 












10 


9.6 


9.9 


.20 


.0208 


.0204 












8 


9.8 


9.4 


.20 


.0204 


.02 












8 


10.0 


9.2 


.20 


.020 


.0196 












6 


10.2 


9.9 


.20 


.0196 


.0192 












8 


10.4 


8.9 


.5 


.048 


.046 












4 


10.8 


8.6 


.02 


.00185 


.00185 












3 


11.0 


8.8 


.36 


.0327 


.0317 












2 


11.1 


8.6 


.30 


.027 


.0253 












1 


11.4 


7.9 


.40 


.035 


.034 













TABLES 



869 



Table CXXIII 
LIMITS OF PROPORTION FOR EXPLOSIVE AIR-GAS MIXTURES 



Gas. 



Per Cent of Gas in the Mixture by Volume. 



Combining 
Proportion. 



When Air is in 
Excess. 



When Gas is in 
Excess. 



Authority. 



Carbon monoxide . 



Hydrogen . 



Water gas, theoretical 
il actual. . . . 



Coal gas . 



Boston illuminating gas 



Acetylene . 



Ethylene 
Methane . 



Ether... 
Benzene . 



29.6 
29.6 
29.6 
29.6 
29.6 
29.6 
29.6 
29.6 
29.6 



14.9 
14.9 
14.9 
14.9 



1 1 




Pentane 


2.6 


1 1 


2.6 


Gasoline 




86° Be 




" 71° Be..,.. 




" 65* Be 




Alcohol 


6.5 


(i 


6.5 


Blau oil gas 




Pintsch oil gas 




Ethane 





7.9 

7.9 
7.9 
7.9 
7.9 
6.5 
6.5 
9.5 
9.5 
9.5 
3.4 
3.4 
2.7 



16.5 
16.5 
13.0 
9.45 
9.45 
7.69 
5.00 
12.4 
12.4 
9.0 
3.8 
8.33 
7.9 
7.9 
5.3 
6.0 
6.7 
6.25 
6.67 
6.25 
6.67 
3.35 
3.35 
1.54 
2.96 
3.0 
4.1 
4.1 
6.1 
6.1 
5.0 
2.75 
2.75 
2.65 
2.65 
2.4 
2.4 
2.4 
2.5 
1.54 
1.54 
1.31 
3.95 
3.95 
4.0 



4.0 



74.95 
58.4 
75 

66.4 
54.4 
33.3 
72.0 
66.75 
54.3 
55.0 
16.7 
33.3 
19 

11.2 
16.7 
29 
20 

14.28 

25.0 

12.5 

20.0 

52.3 

49.0 

47.6 

66.7 

82.0 

14.6 

10.5 

12.8 

9.7 

13.0 

7.7 

5.0 

6.5 

3.9 

4.9 

2.5 

4.9 

2.4 

4.76 

4.76 

4.76 

13.65 

9.7 

8.0 

22.0 



Eitner 

Bunte 

Clowes 

Eitner 

Bunte 

M.I.T. 

Clowes 

Eitner 

Bunte 

Clowes 

M.I.T. 

M.I.T. 

Eitner 

Bunte 

Clerk 

Clowes 

Clerk 

Grover 

M.I.T. 

M.I.T. 

M.I.T. 

Eitner 

Bunte 

M.I.T. 

Clowes 

Eitner 

Bunte 

Eitner 

Bunte 

Clowes 

Eitner 

Bunte 

Eitner 

Bunte 

Eitner 

Bunte 

Eitner 

Bunte 

M.I.T. 

M.I.T. 

M.I.T. 

Eitner 

Bunte 

Hallock 

Lucke 

Clowes 



870 



ENGINEERING THERMODYNAMICS 



Table CXXIV 
RATE OF COMBUSTION OF COAL WITH DRAFT. 



["Conditions and Authority. 


Rate of Com- 
bustion, Lbs. 
Coal per Hour 
per Square Foot 


Draft in 

Inches of 

Water. 


Locomotive, Purdue, reported by Goss for two coals: 

(a) (b) 

Moisture 1 . 89 3 . 10 

Volatile 31.94 15.23 

Fixed carbon 57.71 72.75 

Ash 8.46 8.92 

Draught taken in smoke box, furnace draft less by loss due to 
tube resistance, which is from one and one-hall to three times 
the actual draft 


114 

108 

94 

69 

101 

99 

91 

91 

53 

82 

75 

50 

134 

131 

89 

74 

58 

36 


5.18 
5.15 
4.28 
3.09 
5.60 
3.85 
3.69 
4.37 
2.04 
4!56 
3.50 
2.25 
5.74 
5.65 
3.22 
3.10 
3.00 
1.25 




Ennis, buckwheat No. 1 


10 
15 
20 
25 


.30 

.45 

.70 

1.00 


Anthracite small coal 


Oneida, pea 

" buckwheat No. 1 

No. 2 

No. 3 

Eckley No. 3 


13.63 
13.58 
11.40 
11.34 
9.44 


.38 
.50 




.60 
1.04 
1.13 




Coree, buckwheat 

Stoker, buckwheat 

' ' rice 


19.80 

32.9 

28.0 


.34 

.58 
.92 










Full grate 


53.6 
63.4 
76.4 
93.4 


.40 
.59 


Torpedo boat reported by Lechner. 
Bit. coal. Locomotive type of 




.79 
1.37 


boiler 


Half grate 


66.7 
101.4 
113.0 


.78 
1.96 






2.27 


Whitham reporting Cambria run-of-mine bituminous coal under 
H.R.T. boiler. Ordinary conditions 


5 
8 
10 
12 
14 
15 
16 
18 
20 
22 
25 
28 
30 
34 
36 
40 


.04 
.11 
.13 
.17 
.19 
.20 
.21 
.23 
.24 
.26 
.27 
.29 
.30 
.32 
.33 
.36 



TABLES 



871 



<5l 
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• CHAPTER VI 

HEAT AND WORK. GENERAL RELATIONS BETWEEN HEAT AND WORK. 
THERMAL EFFICIENCY OF STEAM, GAS, AND COMPRESSED AIR 
ENGINES. FLOW OF EXPANSIVE FLUIDS. PERFORMANCE OF ME- 
CHANICAL REFRIGERATING SYSTEMS. 

1. General Heat and Work Relations. Thermal Cycles. Work and 
Efficiency Determination by Heat Differences and Ratios. Graphic Method 
of Temperature Entropy Heat Diagram. Whatever is known of the exist- 
ence of energy, its different manifestations or forms, the essential identity 
of the different forms, the possibility of converting one into another, and 
the laws of this transformation expressing quantitative relations between 
amounts of energy and changes of conditions of substances, has been all derived 
from very simple observations of the form and condition of substances, by a 
somewhat elaborate process of comparison and judgment, assisted by certain 
established mathematical principles. Experiments on substances suffering 
thermal changes are recorded directly in tables of observational data. The 
numerical values of any two series of readings are compared for the purpose 
of establishing, if possible, a general relation between them, a proceeding mate- 
rially assisted by plotting one against the other on cross-section paper yielding 
a curve. The various points will seldom lie all on the curve because of experi- 
mental error whether human or instrumental. The smooth curve passing 
through or near the greatest number of points is considered as representing 
the true conditions of relation, especially when repetition tests of outlying 
points bring them closer to the curve. Such a curve is a graphic representa- 
tion of a general relation between the two quantities involved and its equation 
is the mathematical expression for the law of relation. Two different laws so 
derived may, by similar methods, be combined with each other if any essential 
relation between the respective quantities really exists, and such a process will, 
if carried far enough, lead to more and more inclusive statements of the rela- 
tion between various quantities. Each generalization is a law, properly so 
called, whether summarizing one set of observations, or combining two sets, 
or inclusive of all those ever made by anybody at any time. These laws may 
be stated in three ways: 

(a) by equations; 

(6) by curves, graphically; 

(c) by words; 

874 



HEAT AND WORK 875 

and of these, the statement in words is least valuable because of the inad- 
equacy of language to express mathematical truths or to interpret physical 
mathematical facts. For solving numerical problems with actual substances, 
those laws which are least general and most concrete are most useful, because 
as a rule, they involve a lesser number of corrections and hypotheses than the 
broader generalizations which are reached only after the introduction of more 
and more assumptions, conditions and hypotheses. These various laws receive 
names when much used, the name sometimes being itself inclusive; all 
concerned with heat and work relations being thermodynamic laws, but the name 
may refer to a particular thing, such as the isothermal law for gases and the 
law of specific volume of saturated steam. In any case the law may be 
associated with an investigator's name, as in Boyle's or Avagadro's law, or 
they may have no name at all, appearing in books and papers simply as equa- 
tion x, or figure y, especially when not easily expressed in words or not fre- 
quently used. 

The relations of heat energy to work and the conditions for the transfor- 
mation of one into the other, have been found by experience to yield quite 
readily to the process of generalization, into more and more broad statements, 
crystallizing finally into two, under the masterly manipulation of great physicists 
and there has grown up a tendency to call these two statements alone, the laws 
of thermodynamics. Thermodynamics is defined by Rankine, who helped to 
establish it, as " the reduction of the laws according to which such (heat and 
work) phenomena take place, to a physical theory or corrected system of 
principles." 

Thermodynamics, as thus defined, is concerned with no numerical quantities, 
nor with any particular substance, nor for that matter with any actual sub- 
stance whatever, but it is a physical theory of energy in relation to matter and 
expressible by two all-inclusive laws known as the first and second laws of 
thermodynamics, and as such, is a branch of natural philosophy. These two 
laws are inclusive of individual laws of particular thermal relations, only in the 
sense that the concrete cases do not contradict them, or have served to create 
them, but not in the sense that any particular concrete fact can be derived from 
the two laws of thermodynamics; in most cases it cannot. 

Engineering is concerned with real substances and with numerical quan- 
tities, so that alone, the two laws of thermodynamic philosophy will not yield 
a solution of any practical numerical problem of design, or of analysis, of test 
performance of actual heat machines or thermal apparatus. To guide such 
numerical thermal computations dealing with actual substances and apparatus, 
which is the province of Engineering Thermodynamics, two differences must 
be noted ; first, numerical values for heat effects must be available for the various 
units involved, especially for that class known as physical constants, and second, 
their values must be known for actual substances or the materials of engineer- 
ing. These physical constants include, for example, such quantities as the 
coefficients of expansion, the specific heats, latent heats of fusion and vapor- 
ization, the ratio of the pressure volume product to absolute temperature, 



876 ENGINEERING THERMODYNAMICS 

and the exponent " s " in adiabatic expansion of gases and vapors. These 
so-called thermal constants must be accepted as given in authoritative physical 
tables until such time as they are themselves expressible by laws with numerical 
coefficients, when the special law replaces the table either wholly or in part, 
as illustrated more or less completely in Chapter IV. 

All the thermal constants are derivable by direct experiment, but in some 
cases, for example the specific heat of gases or superheated vapors at constant 
pressure and the latent heat of dry saturated vapors, the experimental difficul- 
ties are so great as to render indirect methods of determination preferable. 
From the general laws of heat and work a few important relations between 
these constants can be found, permitting of the determination of one of 
them more difficult to observe, from the values for others more easily measured, 
for example C v can be found from C p and R, or from C p and y, and the specific 
volume of dry saturated steam may be found by ClapeyrOn's equation. Not 
only are relations between the constants desirable for the determination of 
one from others, but also for purposes of checking observations made on differ- 
ent quantities, perhaps at different times and places, and by different experi- 
menters. Even though some of these coefficients of heat effect may not prove 
to be really constant, nearly all of them being variables in fact, they are still 
called constants for want of an equally short and descriptive name. These 
general heat and work relations are extremely simple in conception and easy 
to use, but when mathematically stated by a series of differential equations 
may lead to a most complex series of derived equations to express which com- 
pletely, would require perhaps a dozen volumes, but for engineering purposes 
this is not at all necessary or even desirable. 

Experiment has shown that the state of a body is related to its heat content, 
and that a given body with an unknown heat content at any particular state 
or time may receive heat, may expand, may be compressed, or may lose heat 
in all sorts of ways, but if at the end of all these processes it finally is brought 
back to its original state, it must then have the same heat content as it had in 
the beginning, even though the numerical value of this heat content cannot be 
measured. This is a most significant fact because it provides a means of dis- 
cussing heat taken in by the body, heat lost by it, and work done, each in 
its relation to the other. As energy is non-destructible, any work that has been 
done must be the exact equivalent of some heat that is not accounted for as 
heat, so that, assuming a body to pass through a series of thermal changes and 
to return to its original state, and therefore have the same heat content as before, 
it follows that the relations are as given in Eq. (885). 

(Heat added) = (Heat abstracted) + (Work done). . . . (885) 

To calculate the work done by a complete series of thermal changes it is only 
necessary to be able to calculate the heat added to the body or taken from it, 
during such a complete series thermal processes or algebraically as in Eq. (886) 

(Work done) = (Heat added) — (Heat abstracted). .... (886) 



HEAT AND WORK 877 

Defining any single thermal process as a thermal phase, such as evaporation 
at constant pressure, or expansion at constant temperature, and defining a 
complete series of thermal processes bringing the substance back to its original 
physical state as a thermal cycle, then a thermal cycle consists of a closed series 
of thermal phases, or on a diagram of changes the cycle is represented by a closed 
figure consisting of a number of separate connecting lines, each representing 
a phase. The work done by a thermal cycle is to be found directly, by sub- 
tracting from the algebraic sum of the heats added in each heat-adding phase, 
the corresponding algebraic sum of the heats abstracted in each heat-abstracting 
phase, so that the problem of finding the work done is resolved into 

(a) The establishment of a thermal cycle, representative of the processes; 
(6) The heat evaluation for each phase of that cycle. 

This work is, of course, done at the expense of heat added to the substance 
and the ratio of the amount transformed into work, to the amount supplied, is 
the thermal efficiency, which is algebraically defined by Eq. (887). 

/Thermal efficiency v Work d 

of transformation I = ^ — - — rT — . ) (a) 

V of heat into work / VHeat added/ 

(Heat added) — (Heat abstracted) 



(Heat added) * (5) 

/Heat abstracted\ , . 

\ Heat added / {c) 



. (887) 



To evaluate the various phase heats, positive and negative, graphic methods 
involving diagrams are of great assistance, but algebraic methods may also 
be used and for any particular problem, that method is used that saves most 
time, but judgment in the selection of method is acquired only by laborious 
experience in deriving solutions both ways. As algebraic methods are depend- 
ent directly on the relations of Chapter IV, it is only necessary to explain the 
graphic methods before proceeding to the detailed work of evaluating each phase 
by both methods. 

It was found convenient in the study of work as a function of pressure- 
volume changes, to use diagrams the coordinates of which were the two factors, 
pressure and volume, the product of which, represented by an area, indicated 
to scale the quantity of work, the diagram showing how the work varied 
with either volume change or pressure change. A similar diagram can be 
developed, equally if not more useful, for studying quantities of heat as func- 
tions of temperature changes, by taking as one coordinate, absolute tempera- 
ture, and for the other the heat per degree of absolute temperature, so that 
the area will indicate heat as the product of absolute temperature and heat 
per degree of temperature. As there is ordinarily little interest in the total 



878 ENGINEEEING THERMODYNAMICS 

heat content of a substance counting from the state of no heat, and as the 
question under examination is usually concerned only with the heat received 
or lost between two thermal conditions or between two temperatures, it is com- 
mon to take for the coordinates of the heat diagram absolute temperature for 
vertical distances and the heat gained or lost per degree of temperature starting 
anywhere, for horizontal distances. This horizontal distance, heat gained 
or lost per degree of absolute temperature was named by Rankine, the thermo- 
dynamic function, and by Clausius, the entropy and the latter and simpler 
name is now always used. Therefore, the diagram on which heat is repre- 
sented by areas is known as the temperature-entropy diagram, just as the 
old diagram on which work was represented by areas is known as the pressure- 
volume diagram, the coordinate names, vertical and horizontal, being used for 
the name of the diagram. As the pressure-volume diagram was also called 
the work diagram, so may the temperature-entropy diagram be similarly called 
the heat diagram. 

Horizontal distances on the temperature-entropy, or heat diagram are, from 
the above specifications of the diagram, derived quantities and not directly 
measurable physical quantities like volumes or temperatures. When a 
substance in condition A, receives or gives up heat to reach condition B, then the 
horizontal distance between the two corresponding diagram points is given 
by Eq. (888). 

(Entropy change from A to B) = — — — r— — ■ — '-^ — _, ; . . . ,, — (888) 
^ J fe ' /Absolute temperature F at which the\ v 

\ heat exchange took place / 



These horizontal distances or entropies can be laid off from an absolute zero 
of entropy or from any given value as a starting point, but as practical problems 
deal only with cyclic heat changes and not with total heat content, it is customary 
to plot only entropy differences in passing from one state to another, creating 
thus an artificial or arbitrary zero of entropy at the point where the heat content 
of the body is least. Entropies measured from this point are really differences 
in entropy between any point and that point where the body has least heat 
content. To avoid confusion the abscissa of any point will be set down as 
a difference. Introducing symbols, 

Let § a , §b, etc. = the absolute and indeterminate entropy of the substance 
in the condition represented by points A, B, etc.; 
(4> fc — 4> a ) = change in entropy, in changing condition from A to B, 
equal to the horizontal distance between the diagram points; 
T = absolute temperature F at which the heat is exchanged; 
Qa6 = the quantity of heat added to, or abstracted from the body 
in passing from state A to B. 



HEAT AND WORK 



879 



Then 



($& — <M = 



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It is evident that as substances may change temperature during a change 
of heat content, T will vary with Q, and as T is the temperature at which the heat 
is received, it is unknown, so that solution of Eq. (889) is possible, first, 



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Entropy 

Fig. 219. — Heat Diagram to Temperature-entropy Coordinates Representing Quantity of 
Heat by the Area under the Thermal Process Curve. 



by the use of an average temperature for the process, second, by what is prefer- 
able and equivalent, the substitution of a differential form, Eq. (890). 



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This formula Eq. (890) applies to all thermal processes in which the temperature 
changes with the heat content, such as heating solids, liquids, gases, and vapors, 
without change of state. When, however, the temperature does not change 
with heat reception or its loss, as for the boiling of liquids, condensing of vapor, 



880 



ENGINEERING THERMODYNAMICS 



freezing of liquids, melting of solids, and for isothermal expansion of gases or 
superheated vapors, then Eq. (889), the simpler form, applies directly. 

Referring to the diagram, Fig. 219, the point A, represents the original 
state, B the final state after receiving the quantity of heat Q ab , represented to 
scale by the area under the change of state curve, AB, down to the horizontal 
axis. It will be observed that heat added is laid off to the right and this is the 
usual way, accordingly heat abstracted is to be laid off to the left of the point 
at which it begins. This arrangement makes it possible to easily deal with 
cycles or complete thermal series of processes, as illustrated in Fig. 220, which 
represents the following four-phase cycle. 



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Curves. 



(1) Heat addition with rising temperature from A to B; 

(2) Heat addition at constant temperature from B to C; 

(3) Heat abstraction with falling temperature from C to D; 

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Each of these heats is represented separately in diagrams (6), (c), (e), (/), and 
the algebraic sum of all heats added, in diagram (d), of all heats abstracted 



HEAT AND WORK SSI 

in diagram (g), so that the heats added less the heats abstracted, is the dif- 
ference between the areas under the curves ABC of (d), and CD A of (g), and 
this is of course, the area enclosed by the cycle ABCD on the main dia- 
gram (a). 

It, therefore appears, that the evaluation of work done or the thermal 
efficiency of transformation becomes easy with such assistance as is rendered 
by the temperature-entropy diagram; the work is the area enclosed by the 
cycle diagram ABCD and the efficiency the ratio of enclosed area ABCD to 
all that area under the top boundary line ABCC'AA. 

Practical application of this method of study of efficiency of heat trans- 
formation into work requires a preliminary establishment of the character- 
istics of the separate phases, which in turn involves the physical constants 
discussed in Chapter IV, where certain relations between them were assumed 
that must be proved on rational grounds. 

Example 1. Evaluate the heat added to the substance in the change of condition 
represented by the line AB in Fig. 1, stating the results in heat units (B.T.U.) and 
in foot-pounds. 

The area of the figure below the line AB down to absolute zero temperature 
measured by the planimeter or by adding narrow vertical strips, w T as found 
to be 5.67 sq.ins. as originally plotted. On the diagram, 1 inch represented 
200° F. absolute in height and .2 entropy units horizontally, so that 1 square inch 
represents, 

200 X.2 =40 B.T.U. 

The entire area represents, therefore, 

40x5.67 =226.8 B.T.U. 

=226.8 X778 = 176,500 ft.-lbs. 

Prob. 1. In Fig. 2 evaluate by the heat added from A to B, and from B to C, the 
heat removed from C to D, and from D to A, and find the net work done by the cycle, 
first in B.T.U. and then reduced to foot-pounds. What is the thermal efficiency of 
the cycle? 

Prob. 2. A certain quantity of gas passes through a cycle in which it receives 1000 
B.T.U. of heat. The mechanical work of the cycle is 90,000 ft.-lbs. How much heat 
is removed during the cycle, and what is the thermal efficiency? 

Prob. 3. A steam engine has a thermal efficiency of 9.7 per cent. Each pound 
of steam entering the engine has 1050 B.T.U. more heat than the condensate leaving 
the condenser. How much heat must the condenser remove from each pound of 
steam? 

Prob. 4. If an engine were of 100 per cent thermal efficiency, how many B.T.U. 
must be supplied per hour per H.P.? How many, if the efficiency were 20 per cent? 
If the efficiency were 5 per cent? 



882 ENGINEERING THERMODYNAMICS 






Prob. 5. A hypothetical engine receives 1000 B.T.U. at a constant temperature 
of 900° F. and rejects heat at a constant temperature of 90° F. No heat is added or 
removed while the temperature is changing. What is the thermal efficiency? 

Prob. 6. A power plant using coal of 13,000 B.T.U. per lb., requires 2 lbs. of coal 
per hour per K.W. measured at the switchboard. What is the efficiency of the plant 
for converting heat of fuel into electrical energy? 

Prob. 7. If a pound of water be evaporated at a constant temperature of 212° F., 
the heat to completely vaporize it will be 970.4 B.T.U. What is the increase in its 
entropy? 

Prob. 8. To evaporate a pound of water at a temperature of 400° F. (247 lbs. per 
square inch absolute) requires 827.2 B.T.U. Find the increase of entropy due to 
evaporation. 

Prob. 9. Anhydrous ammonia evaporating at a temperature of 0° F. absorbs 555 
B.T.U. per pound. What is its increase in entropy due to evaporation? 

Prob. 10. Plot to scale the T§ diagrams for each of these numerical problems. 

Prob. 11. It has been proposed to apply the heat of condensation of steam after 
exhausting from a steam engine to the generation of ether vapor to work in an ether 
engine, the exhaust of which is to be condensed by water. Describe and sketch the 
steam and ether cycles showing what phases they have in common, what similar in 
kind, and what different, using T§ and PV coordinates. 

2. General Energy Equation between Heat Change, Intrinsic Energy 
Change, and Work Done. Derived Relations between Physical Constants 
for Gases and for Changes of State, Solid to Liquid, and Liquid to Vapor. As 

already stated, work done by a substance is always done at the expense of its 
heat, and while doing work the substance may be receiving heat or it may not. If 
no heat is being added at the time the work is being done, that work is derived 
directly from the heat content of the substance and as a consequence the heat 
content is lessened or the energy reduced by an amount exactly equivalent to 
the work done. Should heat be added while work is being done, the change 
in the energy content of the substance is the difference between the heat added 
and work simultaneously done. It has been the custom to call the whole 
energy content of a body its intrinsic energy, so that the above expression 
becomes for a case of heat addition, 



(Change in intrinsic energy) = (Heat added) — (Work done). . . (891) 

For complete thermal cycles, the body returns to the original state has the 
same intrinsic energy at the end as at the beginning, and the change in intrinsic 
is zero. During the cycle, heat may have been received at one time and lost 
at another, so that for the " heat added " term in Eq. (891) there must be sub- 
stituted a more general one representing the algebraic sum of all heats added, 
giving abstracted heats a negative sign, and similarly for the work term, there 
must be substituted one representing the algebraic sum of the works, work of 
compression being negative. Hence for a complete cycle, 



HEAT AND WORK 883 

(Algebraic sum of heats added) — (Algebraic sum of works done) =0, (a) 



or 

/Algebraic sum of 



gebraic sum of\ ,,, ,, ,. /TT , , ,, /NetworkV,. 

heats added ) = (Heat added) " (Heat abstracted ) = ( done. ) {b) , 



(892) 



It was on this form of the expression, that the work and efficiency methods 
described in the last section, were based; it is, however, a special case of a more 
general expression for any kind of thermal process whether a complete cycle 
or not and consisting of any number of phases whether overlapping or separate, 
which is derivable as follows: 

Let U = intrinsic energy of a body or its total energy of condition denned 

as the heat it has above a condition of absolute zero of heat, in 

B.T.U. 
" Q = the heat gained by a body in passing from any state to any other, 

which is the sum of all heat directly added less all abstracted 

as such, in B.T.U. 
IF = work done in passing from the first to the second state, which is 

the sum of all expansion less all compression work, or equal to 

the net work done, in ft.-lbs. ; 
J = Joule's equivalent of one B.T.U. in ft.-lbs. = 778 approximately 

or 777.52, more exactly. 

Then the more general expression for the relation between heat, work, and 
intrinsic energy, takes the following form, Eq. (893), which is one statement 
of the so-called first law of thermodynamics, 



dU = dQ-jdW. . . (893) 



This equation will be called for convenience the general energy equation 
for substances and is true for all sorts of thermal changes except those includ- 
ing as a part an irreversible process. In such cases the failure of the equilibrium 
of the state of the substance destroys the dependence of intrinsic energy on 
state alone which is otherwise true and assumed in the above equation. 

Applying this equation directly to perfect gases gives some useful relations 
between the physical constants. For such hypothetical substances which all 
actual gases closely approximate at some time, PV = RT, for 1 lb., in foot- 
pound units. The differential form of this is given by Eq. (894). 

d(PV)=PdV+VdP = RdT. (894) 



884 ENGINEEEING THEBMODYNAMICS 

When gases receive a little heat dQ, it is conceivable that it may be divided 
into two parts, one part raising temperature without doing work, and therefore 
given by C v dT B.T.U. for 1 lb., and a part transformed directly into work and 
equal to PdV foot-pounds, hence 

dQ = C v dT+jPdV. . (895) 

Combining Eqs. (893) and (895), 

■dU = CdT+jPdV-jdW (896) 

But as dW = PdV, 

dU = CJT (897) 

Interpretation of Eq. (897) yields the conclusion, that for perfect gases the intrin- 
sic energy depends on the absolute temperature only. Furthermore, the 
intrinsic energy of a perfect gas is equal to the product of its specific heat at 
constant volume into its absolute temperature plus a constant of integration, 
if this specific heat is not a function of temperature, which it probably is. 

Combining Eqs. (894) and (895) by elimination of dV gives 
dQ = C v dT+jdT-jVdP 

= (c,+j\dT~VdP (898) 

But the specific heat at constant pressure is given by Eq. (899), 

M§L ^ 

Whence by Eq. (898), 

or 

R = J(C P -C V ) (900) 

This Eq. (900) proves the assumption made previously that the gas constant 
R is equal to the difference between the two specific heats expressed in foot- 
pounds. 



HEAT AND WOKK 



885 



Substitution of this value for R in Eq. (898) gives another form to the heat 
expression, in terms of dT and dP variables, Eq. (901), 



dQ = C p dT-jVdP. 



Giving to dT in Eq. (895) its value from Eq. (894), 



dQ = ^d(PV)+\pdV. 



(901) 



(902) 



This can be put into an integral form, Eq. (903), directly, and the factor 
simplified by using the relation, 



R J(C p — Cv) T /Cj 



Heat added from state\ 
A to state B in foot- I 
pounds / 



(c;~~] iPbVb ~ PaVa)+ fa 



PdV (a) 



C 



XP b V b -P a Va) + (Work done) (6) 



(903) 



When gases expand they follow, as has already been shown, the general law 
PV S = K, and in so expanding the work done is given by Eq. (904). 



W 



/b pb 

PdV=K *£-^El(F,»--7.»-) . .. 

Substitution of this in Eq. (903) gives 

JQ b a=i^\(P b V b -P a Va)+(Y^i) (7* 1 --1V-') 



(904) 






'PaVa* 



c P 



(7 6 1 - s -F a 1 - s )+(^)(7 6 1 - s -F a 1 -«) 



(&-») 



§->) 



<PaV 



(Trf)^ 1 --^. 1 -) 



(905) 



886 



ENGINEEKING THERMODYNAMICS 



Hence 



o-) 



c v 



C v 



-1 



X( Work of expansion) .' . .. (906) 



To proceed further with the investigation of relations between physical con- 
stants requires a still more general energy equation involving more of them. 
One of the basic definitions of heat makes a quantity of heat proportional 
to the amount of its effect on pressure, on volume and on temperature, so that 
taking these quantities in pairs as independent variables, two fixing the third 
by the general property relations of matter, any heat increment may be equated 
to effect, by the three sets of algebraic relations of Eq. (907), 



dQ = adT+bdV, (a) 
= cdT+ddP, (b) 
= edP+fdV. (c) 



(907) 



The change in intrinsic energy is given by the following set, Eq. (908), corre- 
sponding to the above, in which the factors (a), (6), (c), etc., are unit thermal 
effects or physical constants. Substituting Eq. (907) in Eq. (893). 



dU = a dT+b dV-j P dV = adT+ (b-jj dV 



= cdT+ddP-jPdV 



dP+fdV-j PdV=e dP+ (f-j) dV 



(a) 

CO 

(c) 



. (908) 



These are, of course, partial differential equations and each must be a complete 
differential, that is, it must be possible to solve or integrate it. Therefore 
the differential coefficients of each coefficient with respect to the other variable, 
when its own variable is constant, must be equal to each other. 



dV 



(a) 



T=C 



/da \ d f h 

\dv) T ^c~df\ 



P" 

J J v=C 



(909) 



This is expressed symbolically in Eq. (909) and there are two others, similar 
but not set down because they are not to be used. 



HEAT AND WORK 887 

The change of heat per degree of absolute temperature has been defined 
as the entropy change as in Eq. (910) for the differential increments. 

di>=^. '•'■•.• ( 910) 



Combining Eq. (910) with the original general equation, Eq. (907), there results 
some new relations between constants, Eq. (911), and which constitute one form 
of statement of the second law of thermodynamics: 

d4> = |dr+£dF (a) 



= %dT+±dP (b) 
= ^dP+±dV (c) 



(911) 



These are also complete differentials so that the differential coefficient relations 
yield the following, Eq. (912), and two others, similar, that will be omitted 
to save space. 



Performing the operations indicated in Eqs. (909) and (912), they may be 
combined. 

From Eq. (909), 

w) t =c = W)v=c~JW)v=c •••••• ( 913 > 

From Eq. (912), 

l_(da\ I (db_\ _b_ 

T\dV/ T =c 



, c T\dT)v=c T 2 ' 



or 



(%) r .A$l-A (8 »» 



888 ENGINEERING THERMODYNAMICS 

Equating Eq. (913) to Eq. (914) the result is expressed by Eq. (915). 

I-HSL « 

This is a perfectly general expression, as no defining conditions or limiting 
hypotheses have yet been introduced except that of excluding non-reversible 
processes. Similar treatment will yield a long list of relations, omitted here 
because they may be found in works on the philosophy of thermodynamics 
and because they are not needed here. These particular ones are selected 
because they lead to values for specific heats which are defined mathematically 
by Eq. (916). 



MIL <«> 



(916) 



Referring back to Eq. (907) it is evident that, 

C v = a, and C p = c. 



One step has been taken in evaluating the constant (a), by establishing a value 
for the constant (6) in Eq. (915), and substituting this in Eq. (913), the com- 
plete evaluation of the constant a = C v , is given by Eq. (917), 



da\ = T/d 2 P^ 

dv) T =c~J\dT 2 J v =c 



whence 

'd 2 P 



t r 



rP2 dV+K' (917) 



Therefore the specific heat at constant volume for perfect gases is a function 
of absolute temperature only, or it is constant, because the constant of integra- 
tion K' must be constant or a function of temperature only, and it may be zero. 
By exactly similar reasoning, using other equations of the general group, it 
can be shown that a similar conclusion is applicable to the constant, c = C p , 
by Eq. (918), 



C = J 



fff 2 dP+K". (918) 



HEAT AND WORK 889 

These relations are derived here for two reasons: first, to show the general 
process of thermodynamic computation of relations between physical constants 
and second because this particular relation illustrates beautifully how utterly 
misleading some of the most laboriously reached of these conclusions may be. 
It was because of such a demonstration as this that for so long nobody even 
sought for a variation in the specific heat of superheated steam with pressure, 
which was ultimately found after engineering and not physical calculations 
showed that the old accepted values must be wrong. This does not indicate 
that conclusions based on such mathematical analyses will always be misleading 
because they are absolutely true for perfect gases, but they are to be adopted 
for real gases only with necessary restrictions based on how much departure 
there is by the real substance from the perfect gas. 

When a substance changes state, a similar analysis starting with the general 
energy equation yields some useful relations between the new constants involved 
for these processes. Starting with the general statement Eq. (895) or Eq. 
(907a) that heat added is equal to, something (a) multiplied by the tempera- 
ture change, added to the product of something else (b) and the volume change, 
it was found Eq. (915) b}^ various transformations involving the two laws of 
thermodynamics that the coefficient (b) was proportional to the product of abso- 
lute temperature into the rate of change of pressure with temperature which 
when inserted in Eq, (907a) gives Eq. (919), 



dQ = adT+j(^\dV (919) 



This equation may be applied to a change of state from solid to liquid and 
liquid to vapor at constant temperature, for which case it takes the form of Eq. 
(920), since dT = 0. 

d( i=-jQ dV (920) 



If the process be one of evaporation of 1 lb. and this be complete, then the heat 
added is the latent heat L, the final volume that of dry saturated vapor V v , 
and the initial volume that of liquid at the temperature of boiling V L , whence 
integration gives an expression for the physical constant, latent heat, Eq. (921), 



i(^) {Vv - Vl) (921) 



J \dT ; 



This is the Clapeyron equation previously used and now established, by methods 
used by Clausius and Clapeyron, who derived it independently. 



890 ENGINEERING THERMODYNAMICS 

For fusion it is believed that latent heats are constant for all pressures, in 
which case the change in the freezing- or melting-point with pressure is given 
in degrees per unit of pressure by Eq. (922), where V L = specific volume of liquid 
and V s the volume of solid. 



It is interesting to note that for ice and water, ice being lighter than water, the 
difference above is negative so that increase of pressure lowers rather than 
raises the melting- or freezing-point. Rewriting the equation in the following 
form, Eq. (923), for vapor-liquid changes another interesting conclusion becomes 
possible: 

JL =T% . (923) 



Vv-V L dT 

This indicates that latent heat in foot-pounds divided by the volume change 
in vaporization, which is in the nature of a pressure and may be regarded as 
the pressure resisting or opposing a change of state, is equal to the product of 
absolute temperature and the rate of change of vapor pressure with temper- 
ature. It has been remarked that this quantity given by Eq. (923) is nearly 
constant for several different vapors at the same pressure, so that each term 
is constant, Eq. (924) : 



JL =K 



Vv-V L 
or 

dP 



T <k= K 



Approx. when P is const (924) 



It is noted, however, that the constant K rises with pressure so that at any 
pressure 

T§=f(P), 

whence on integration, 

log* T = F(P)+K' (925) 

This Eq. (925) is the origin of form of some of the numerous equations for the 
relation between saturated vapor temperatures and pressures. 

When two different substances A and B follow similar laws, according to 



HEAT AND WORK 



891 



Eq. (924) then the boiling-points at equal pressures follow the law of constant 
ratio of Eq. (926), 



/ Boiling-point of one substance \ /rA s 

^r~^ • j. r m r~L = (Constant, at same pressure). 

\Boihng-point of another substance/ 



(926) 



It also follows that for substances having the same " K " or constant 
I— — lvalues, that latent heats are inversely proportional to the density 

\Ky— V l/ 

of vapors when V L is small compared to Vv and, therefore, that the generation 
of a cubic foot of any vapor in the group requires the same amount of heat. 

These applications of general laws to vapors are subject to the same remarks 
as were referred to the derivation of physical laws from mathematical analysis, 
that they are no more correct than the hypotheses on which they are based 
and in the world of derivation and deduction it is easy to forget the hypotheses 
and make a serious blunder. As Mellor remarks in his Chemical Statics and 
Dynamics, "No process of reasoning can establish a law of Nature. The elements 
of sameness — the law — must be actually discovered in the facts." 



Example 1. Air expands under a piston from a volume of 1 cu.ft. and a pressure 
of 300 lbs. per square inch absolute to a volume of 5 cu.ft. at a pressure of 40 lbs. per 
square inch absolute. Assuming that expansion follows the law of constant PV S , find 
the heat absorbed in B.T.U. Ratio of specific heats for air = 1.4. 

The work done from Eq. (904) is 



W 



PaV a 



log 



(F 6 1 - S -F a 1 - S ),ands = 



© 



log 



(ft) 



Substituting numerical values to find 



log 



300 
40 



s = 



log 5 



.875 
.699 



1.25. 



Therefore, 

rTr 300X1 t .. 
17 = 1^5 Xl44 

FromEq. (905) 



^1.251-1 



43,200 
-.25 



X (.667-1) =57,600 ft.-lbs. 



(Heat)&a = 



g- 1 - 25 






- I xw 



1 



"778 X 1.4 



1.25 



X57,600=27.7B.T.U. 



892 ENGINEERING THERMODYNAMICS 

Prob. 1. A pound of air at 60° F. and atmospheric pressure receives 500 heat-units 
at constant pressure. How much of this heat will be expended in raising the tem- 
perature of the air and how much in doing work? 

Prob. 2. Show that when air expands isothermally the work done is equal to the 
heat added. 

Prob. 3. A pound of air at 70° F. and a pressure of 12 lbs. per square inch absolute 
is compressed adiabatically until its volume is one-fourth of the original volume. One 
thousand heat-units are added at constant volume and the air then expands to the 
original pressure, whereupon it is cooled to the original temperature at constant pres- 
sure. Show numerically that the algebraic sum of heats added is equal to the work 
done. 

Prob. 4. A pound of gas having the specific heat at constant volume of .2, changes 
in volume from 100 cu.ft. to 150 cu.ft. while the pressure remains constant at 100 lbs. 
per square inch absolute. How much heat was added to the gas? 

Prob. 5. Prove that for a perfect gas R =J(C P — C V ) and show how much departure 
there is for air, methane, oxygen, nitrogen, carbon dioxide, steam. 

Prob. 6. One thousand heat-units are supplied to a pound of air at 60° F. and 
15 lbs. per square inch absolute pressure. What will be the temperature at the end 
of the heat reception and the work done, if (a) heat is received at constant volume, (b) 
if it is received at constant pressure, and (c) if it is received during expansion according 
to PV 1 - 1 -Const. 

Prob. 7. Show that for adiabatic compression the work done is CviT^ — T^J foot- 
pounds. 

JL 

Prob. 8. It has been stated that — — - is approximately constant. Taking the 

V v ~ Vl 

values of L, V v , and V L from the tables of Chapter IV, compare the results obtained 

for steam, ammonia, and carbon dioxide at pressures of 10, 25, 50, 75, and 100 lbs. 

per square inch absolute. 

Prob. 9. If air expands in a cylinder according constant to PV 1 ' 3 , find an expression 
for the heat gained or lost. 

Prob. 10. Determine whether heat is being gained or lost by ammonia vapor 
during compression when it is found that s has the value 1.2 and derive a quantitative 
expression for it. 

Prob. 11. Compare the change in intrinsic energy during vapor formation for one 
pound of steam, ammonia, and carbon dioxide at 32° F. 

Prob. 12. Derive from the tables a formula for vapor pressure in terms of tem- 
perature, of the form of Eq. (925) for steam, ammonia, and carbon dioxide. 

d 2 P 
Prob. 13. From the equation derived in Problem 12, find —^ and substituting in 

Eq. (917), find C v for steam, compare it with the values given in Marks and Davis, 
and explain the differences found. 

3. Quantitative Relations for Primary Thermal Phases, Algebraic, and 
Graphic, to PV and T3> Coordinates. Constancy of P, V, and T for Gases 
and Vapors, Wet, Dry, and Superheated. Heating of solids, liquids, gases 
and vapors without change of state takes place according to the fundamental 
relation between temperature change and quantity of heat, Eq. (927). 

(Heat quantity) = (Specific heat) X (Temperature change) X (Weight) . (927) 



HEAT AND WOEK 893 

If the specific heat is constant this is a most simple expression, but it has been 
shown that specific heat is variable and given generally by an equation of the 
form, Eq. (928), in which a, b, and c are constants, different for each substance. 

Specific heat = C = a+bT+cT 2 (928) 

In this equation, if b and c are very small, the specific heat it reduces to 
the form Eq. (929), 

Specific heat = C = a, (substantially) (929) 

From these equations the entropy change can be determined for all substances 
suffering heat change without change of state, from Eq. (910), as follows: 

^(Specific heat) .'.... (930 ) 



A A f d Q f 



but 



or 



Specific heat = -r^; 



Whence 



dQ = (Specific heat)dT; 
= (a+bT+cT 2 )dT. 

.2 ^2 ^2 

TdT 



4> 2 -<!>i=w a I ^-+b / dT+c I 



==wLlog e ^+b(T 2 -T 1 )+^(T2 2 -T 1 2 ). . . (931) 

This Eq. (931) is the entropy change for all heat gains or losses by solids, 
liquids, gases and vapors not changing state when the specific heat is a function 
of absolute temperature of the form given in Eq. (928), and reduces to the fol- 
lowing special forms, Eq. (932, a, b, c, d), for the condition stated. 

rp 

(For constant specific heat C) (4>2 — 4>i) =w C log e ^r («) 

For water if the specific heat be taken\ (<h 2 — d)i)=w log e — (b) 
as unity / e T\ 

/For gases and superheated vapors at\ ^ \ . (932) 

( constant volume with constant )(<t>2 — <h) =wC v log e — (c) 
\ specific heat ' 1 

/For gases and superheated vapors at\ ^ 2 

( constant pressure with constant )(<]>2 — <i>i) =wC p log e ^- (d) 

\ specific heat ' 1 



894 



ENGINEERING THERMODYNAMICS 



To illustrate the relation of these changes to each other there are plotted to 
PV and T$ coordinates in Fig. 221 the following cases each for 1 lb. of substance: 



(1) Heating of water from 32° F. to 400° F. with specific heat = l; 

(2) Heating of water from 32° F. to 400° F. with specific heat as given 

in the Marks & Davis Steam Tables using the columns called 
heat and entropy of liquid (above 32° F.) ; 

(3) Heating of air from 32° F. and 20 lbs. sq.in. to 500° F. at constant 

volume with constant specific heat = (.17); 

(4) Heating of air from 32° F. and 20 lbs. sq.in. to 500° F. at constant 

volume with variable specific heat, mean C v = .174 + .0000675(^—32) ; 

(5) Heating of air at constant 20 lbs. per square inch pressure from 

32° to 500° F. with constant specific heat C p = .239; 
'6) Heating of air at constant 20 lbs. per square inch pressure from 
32° F.to 500° with variable specific heat C p = .239+. 0000675 (£-32). 




1 












































250 


E 






















































































































































































































































































































































































































































, 












































CD 












































a 












































> 
































































































































































































0) 












































a 












































>° 
























































































oj 


r 










































a 


' 


^ 


c 


a> 




































O 


> 


Ai 


rC 


on 


ta 


it 


Pre 


SSI 


ire 


























<> 


\ 






v? 


-Vr 










4 








A' 


-D 












V] 


roi 


kd 


on 


eis 


ar 


;a unc 


er 


th 


X 


ur 


. e J 


B 



Entropy 



10 15 

Volume Cu. Ft. 



Fig. 221. 



-Changes in P, V, T, <£, Heat, and Work, for Constant Pressure and Constant Volume 
Phases of Liquids, Gases and Vapors, without Change of State. 



For constant pressure heating of a gas illustrated here by air, there will 
be an increase of volume depending on the temperature rise and work done, 
which latter is measured by the volume increase, but as there are two possible 
final temperatures for the same amount of heat, one for the constant and one for 
the larger variable specific heat, there will be two possible volumes increases 
and quantities of work done. In the example, the temperature has been fixed 
so that the heat quantities differ by the differences between the areas on the 
T$ diagram under the solid curve for constant and under the dotted curve 



HEAT AND WORK 



895 



for variable specific heat. These differences are so small and in the study of 
heat transformation into work so inconsequential that except when otherwise 
specifically stated, the constant value of specific heat will be used. This practice 
will save great labor as well and avoid the introduction of doubts which have 
real foundation, as to the accuracy of the determination of the variability laws 
so far announced. One special case of difference in practice is that for water 
and superheated steam, the entropy changes for which are calculated and form 
part of the steam tables under the heading " entropy of the liquid (above 
32° F.)" and "entropy of the steam," but in approximate calculations no serious 
error will be introduced by assuming constancy of specific heat for water and 
unity as its value, but the variability for the steam cannot be so safely ignored. 
Accordingly, with the above reservations the following relations of tabular 
Eq. (933) are found between the quantities for the gas heated at constant 
pressure from A to B., Fig. 221. 



Related Quantities for 
Constant Pressure Heating 


Formula 


Heat and Temp. 
Temp, and Volume 

Heat and Volume 

Work and Temp. 

Work and Heat 

Entropy and Temp. 


Qab =C v {Tb — Ta) 

V.-7.-V.(g-l) 

T . T . ir Qab R [Qab\ 
U ~ U - U C„T a -CAPa) 

W =Pa( V„ - V a ) =PaV a (y ~ U = B(T b - T a ) 

P a V a Qab R ~ 

W ~ t r "r Qab 

J- a v/ p \->p 
rp 

§b — §a = C p log e — 
J- a 



(a) 1 
(J>) 

(fi) 

(d) 
(e) 
(/) J 



(933) 



A similar set may be written down, Eq. (934), almost by inspection for the 
case of constant volume heating from A to C, Fig. 221, which, of course, causes a 
pressure rise, during which no work can be done, or W = 0. 



Related Quantities for Constant 
Volume Heating 


Formula 


Heat and Temperature 


Qac=C,(T e -T a ) 


Temperature and Pressure 


P £ -P.=P»(£-l) 


Heat and Pressure 


Pc ~ Pa=Pa \c)k) = c\fa) 


Entropy and Temperature 


<&c~ <J>a=C„ lOge — 
J a 



(a) 
(b) 

(c) 

id) 



(934) 



896 ENGINEERING THERMODYNAMICS 

The case presented by the water is peculiar inasmuch as it makes so little 
difference whether the water be heated at constant pressure or constant volume 
or at what pressure, its expansion being so small. For the purpose of illustra- 
tion it is assumed in the diagram that the water is at the same 20 lbs. pressure 
and that it is kept (a) at constant pressure, expanding from A' to D and (6) 
at constant volume, its pressure increasing from A' to E. In ordinary practical 
work no account is taken of the difference at all, first, because water in use is 
rarely heated at constant volume, and second, because the work of expansion 
is vanishingly small. 

Constancy of temperature characterizes all changes of state, solid to liquid 
and liquid to vapor, if the pressure be constant, and also is characteristic of the 
expansion of gases when the product of pressure and volume is constant, for 
in this case PV=RT= Constant For melting or boiling, then, the heat added 
is proportional to the weight of substance changed or the quality. 

Let L — latent heat of fusion or vaporization per pound; 
w = weight of substance present; 
x = fraction of total weight liquefied from the solid, or vaporized from 

the liquid, and to be called the quality; 
1— x = fraction of total weight remaining in the solid state in contact 

with liquid, or in the liquid state in contact with vapor. 

Hence for the changes of state at constant temperature and pressure the quantity 
of heat is the product of weight, quality and latent heat, Eq. (935), 

Q = wxL (935) 

The latent heat L is to be taken from tables of physical properties of substances 
but will not be the same at different pressures. The representation of the 
case of vaporization on the PV and T«I> diagrams, especially the latter, is usually 
associated with the heating of liquid which precedes vaporization and with 
the line of vapor superheat which succeeds vaporization if the heating is long 
continued. Accordingly the successive processes of heating liquid, evaporat- 
ing it, and superheating its vapor, are represented by three lines joining each 
other, for a single pressure, and by a family of lines if vaporization at all 
pressures are to be included. 

This is illustrated in Fig. 222 where, starting with liquid at 32° F. (A) heating 
at constant pressure to the boiling-point, B, will cause a slight expansion, shown 
A to B on the PV diagram, doing the work shown by the area ABB' A' A and 
requiring the heat shown by ABB' A' A on the T$ diagram, as already explained, 
but repeated here to assure continuity. The formation of vapor at the tem- 
perature represented by BB' on the T$ and at the corresponding pressure 
represented by BB' on the PV diagram, is represented by the line BC hori- 
zontally on both, until all liquid has been just vaporized which condition is 
denoted by the point D. Therefore, the area under the line of complete vapori 



HEAT AND WORK 



897 



zation BD and which is BDD'B'B on the T$ diagram, is a graphic representa- 
tion of the latent heat of vaporization to scale. SimilarLy, on the PV diagram 
the corresponding area BDD'B'B is the work of vaporization or P(V V — V L ) 
since AB is the specific volume of the liquid V L and AD the specific volume of 
the dry saturated vapor. Should the vaporization be not complete but stop at 
the point C, there will be present a mixture of liquid and vapor, to be called for 
want of a better name wet vapor. For such wet vapor as is represented by the 

point C the quality is given by = = x on the T$ diagram, and the heat used 

in making it at the boiling-point is represented by the area BCC'B'B and is equal 
to xL. On the PV diagram the point C for wet vapor is located in a similar 

way, BD corresponding to the quality x=^l, so that ===x, and the wet 

BD 

vapor will occupy a volume AC per pound of mixed vapor and liquid. 




p 


K 


B 










f> 




n 




r 


























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L 






























1 








i 






























i 




























6(1 


i 






























i 






























i 






























i 
i 






























i 




























40 


i 






























i 






























i 

















-f 















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Ati 


11 OS 


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N 



2 3 4 D' 5 

Volume in Cu. Ft. 



Fig. 222. — Changes in Pressure, Volume, Temperature, Entropy, Quality, Heat Added, and 
Work, done for Constant Temperature Heating of Liquid-vapor with Change of State. 



Tabular values for the increase of entropy for vaporization alone or for the 
entropy of dry saturated steam over liquid at 32°, which includes the liquid 
part, are very useful in solving problems with wet vapors by the above 
relation. 

Continued heating at the same pressure after the completion of vapori- 
zation, of course superheats the vapor along a line DE. If the whole process 
had taken place at some lower pressure it could be represented as in the dotted 
lines leading to the family of curves mentioned. It is not possible to express 
algebraically the relations between heat of vaporization or work and tempera- 
ture, except empirically and then only by an expression too complicated for use. 
Problems must, therefore, be solved by the use of the tabular values given at 
the end of Chapter IV. 

Gases receiving heat at constant temperature behave quite differently, 



898 



ENGINEERING THERMODYNAMICS 



and the relation between the quantities involved can be found directly from 
the general equation, Eq. (906) for heat and work of gases, here repeated: 



JQ, 



Cv 



ab : 



— s 



.C, \ 



X (Work of expansion) . 



If in the equation, s be given the value unity, which it must have for this case, 
the relation between heat and work for isothermal gas processes follows as 
given by Eq. (936) : 



Qab = j(Work of expansion) (a) 
=.jP«7a log. J? . (6) 



(936) 



Therefore when gases expand isothermally the heat added is equal to the work 
done in the same units or the work done in foot-pounds is equal to the heat added 
in foot-pounds. This is equivalent to saying that the internal energy remains 
constant and heat is converted directly and completely into work as fast as added 
during isothermal expansion or inversely for isothermal compression. To 
illustrate this case, the diagrams, Fig. 223, are plotted for the case of 1 lb. of 































































































T 


























































































oJ 800 














































A 








































b 


i 


< 




















































































































































































ta 














































03 

2 




























































11 


p 


it 




















ex 




















I IT 


















Q400 
















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1 










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a 


























































































H 







































































































































































































































Entropy 



t» 







































































































































































































































































A 
























































\ 


























































S 


























































s 


























































S 


























































•v 






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rk 






















































lu 




















































?00 




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pit.* 


















































Sc 


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- '4320 


















































H L 




















































































































































































A 


ta 


o 


P 


le 


■ic 


Lin 


e 














































































"1 
























































±L 







10 15 20 

Volume in Cu. Ft. 



Fig. 223. — Changes in Pressure, Volume, Temperature, Entropy, Heat added, and Work 
done, for Constant Temperature Heating of Gases and Vapors without Change of State. 



air at 150 lbs. per square inch absolute and temperature 300° F. expanding 
isothermally under the reception of 200 B.T.U.'s from A to B. The area under 
the curve or isothermal AB on the T$ diagram is equivalent to 200 B.T.U. and 



HEAT AND WORK 



899 



the area under the hyperbola AB on the PV diagram represents the mechanical 
equivalent of 200 B.T.U. or 155,504 ft.-lbs. For such cases as this, the following 
set of relations, Eq. (937), apply: 



Related Quantities for Constant 
Temperature Heating of Gases 


Formula 


Heat and Volumes 


n 1 V a 1R T . V a 
Qab =jPaV a log* — =-j~ T a log e y 


Heat and Pressures 


Co* =iPa7a log. g-^T. log. g 


Volumes and Pressures 


PaV a =P b V b 


Work and Heat 


Qab = -jWab 


Entropy and Temperature 


. , Qab Wab 



(a) 

(b) 
(c) 
id) 

w 



(937) 



|400 



-400, 



«1 



-200 A- 



— -=-n 



Atmosphen 



=^ 






1.0 
Entropy 



2.0 q> 



6> 400 



200 

















































































1 E 








































































































L 




















































1 






























- Co 


istai 


it Te 


mpe 


-atm 


e Li 


ae 








































x x 
















































A 


tmosi 


heri 


! Lin 


e 










A 








1 1. 





















15 
Volume in Cu. Ft. 



Fig. 224. — Constant Temperature or Isothermal Compression of Superheated Steam Reducing 
Superheat, and Subsequently Causing Partial or Complete Condensation. 



These same equations and general relations apply to superheated vapors provided 
they do not approach the region of wetness which they will do on isothermal 
compression if it is carried far enough. Condensation of vapors by isothermal 
compression is shown by the curve, Fig. 224, for the case of steam, on which 
the point A represents steam in the condition of high superheat at low pressure. 
Isothermal compression causes the change of state represented by the hyperbola 
AB on the PV and by the horizontal AB on the T$ diagram. When the point 



900 



ENGINEERING THERMODYNAMICS 



B is reached the vapor has acquired the pressure and volume occupied 
by dry saturated vapor so that further abstraction of heat at constant tempera- 
ture results in condensation from B to C without any further pressure change, 
and this would continue until all vapor had so disappeared, after which isothermal 
abstraction of heat would be impossible, because a liquid cannot give up heat 
without losing temperature. 

Constant- volume cooling of a vapor will cause a change of state just as does 
isothermal compression, as will also a constant-volume heating of a mixture 
of vapor and liquid, but constant-volume heating or cooling of a vapor within 
such limits as keep it always superheated follows the same laws as for gases. 
The departure from the gas laws for constant-volume changes of vapors is clearly 
shown by the T3> diagram in its relation to the PV as in Fig. 225. On the dia- 



T 

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Fig. 225. — Constant Volume Changes for Steam Showing Corresponding Changes of Quality. 

gram A B represents liquid heating to both T$ and PV coordinates, BC complete 
vaporization, so that the length of BC on the T<£ diagram is proportional to 
latent heat and on the PV diagram to the specific volume of dry saturated steam 
for the pressure. Similarly, GD represents the latent heat of condensation at a 
lower pressure to T& coordinates and the specific volume of low-pressure dry 
saturated steam to PV coordinates. Changes of temperature, pressure and vol- 
ume for steam continuously dry and saturated is represented by the satura- 
tion line CD. 

A condition of wetness of the higher pressure steam is indicated by the 
position of a point E on both diagrams and the following properties are defined 
by the relations of line lengths : 



Quality of steam at E = I = ) 

\bcJt* 



HEAT AND WORK 901 

/Wf 1 \ 
Wetness of steam at E = ( = ] =l — x: 

BC/t* 

/DEi\ 

Volume of steam-water mixture = (BE) PV = ( t^, ) X (BC)pv 

Heat of wet steam formation _xL_ /BE\ _ /Vol. of wet steam\ _ /BE\ 
Heat of dry steam formation L \BC/t<p \ Vol. of dry steamy \BC/pv 

These relations permit of the location of the line representing heat abstraction 
at constant volume and its heat evaluation by the T$> area under it. Of course, 
this phase is represented by a vertical EF to PV coordinates and the position 
of point F on the PV diagram indicates not only the volume the steam will 
have after heat abstraction to the low pressure, but the relation it bears to the 
volume of dry saturated steam at the same pressure. If (Vv—Vl)i is the high- 
pressure volume increase for dry saturated steam over its liquid, and (Vv— V£)2 
the same at the low pressure, then the original volume of the wet steam is 
(Vv— Vl)iXi-\-(1— xi)Vli and the final volume of the wet steam is 
(Vv— V L )2X2+(l— X2) Vl2 and as these are to be equal, 

(Vv-V L ) 1 x 1 + (l-x 1 )V L1 = (Vv-V L )2X2+(l-x 2 )V L 2. 

The volume of the liquid is, however, so small as compared to the volume of 
the vapor that it may be neglected and so will hereafter be neglected. 

Calling the specific volume of dry saturated steam V v to be taken from the 
tables, there results VVi^i = ^^2^2, or the product of quality and specific volume 
of dry vapors is constant. Points may be located along (EF) T $, by applying 
the equality between the ratio of partial to total latent heats on T<& and ratio 
of saturation to wet volume on PV coordinates. The heat abstracted is, of 
course, the area under FE or (FEE'F'F) and is best evaluated graphically by 
cross-section paper plotting when needed, which is not often. 

A series of such constant- volume lines is drawn over a T$ chart in Fig. 
226, which indicates the change in the character of the line at different degrees 
of wetness and its reversal of curvature from concave down to concave up, as 
it passes into the region of superheat, where it is the same as that for a gas. 



Example 1. Calculation and use of Diagram, Fig. 226, giving constant-volume 
for steam. To illustrate the method the location of the line of constant volume of 2 
cu.ft. will be traced. Let the first temperature be taken at 800° F. absolute for the 
first point A, corresponding to 340° F. From the steam tables dry saturated steam 
at 340° F. has a specific volume of 3.787 cu.ft., so that the quality when the volume 

2 

is 2 cu.ft. is =52.8 per cent. Therefore the entropy increase in making this 

6.7 vl 



902 



ENGINEERING THERMODYNAMICS 



(ureajg pa^aqjadng o% &\ddv %ou saop aj^os ojnssajj ) 

©iniosqy 'ui *bg asd 'sqi m aanssoaj 





































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(Jo) oan^uaoduio i o^niosq y 



HEAT AND WORK 903 

steam from 32° F. and at 340° F. is c^-fe = .528x1. 0984 +.4902 = 1.0703. Another 

point B is located by assuming a temperature ^=440° F. or !F&=900, for which 

<!>» — <i>32 = 1.5602 by the same method. 

To illustrate the use of the diagram in solving problems, suppose 1 lb. of wet 

atmospheric pressure steam, occupying 10 cu.ft. be enclosed in tank and heated to raise 

the pressure to 30 lbs. per square inch absolute, to find the final temperature, entropy 

and dryness. From 14.7 lbs. per square inch on the pressure scale project to point P 

on the constant volume line of 10 cu.ft. and follow this line to the point C for 30 lbs. 

per square inch absolute pressure. Projecting from C to D the absolute temperature 

is found to be 710° or t =250° F., and projecting from C to E the entropy § c — cj> 3: . = 1.332. 

CM 
The final quality -■— =72.4 per cent. 

OM 

Again, if heat be added to raise the temperature to 842° absolute the entropy is found 
by following the 10 cu.ft. line to the point K opposite the temperature, and projecting 
down from K to Q the entropy is found 4>fc — <t>3-2 = 1.724. The quality may be read off 
directly from the chart, Fig. 232, of the next section which carries lines of constant 
quality and which might be superimposed on this constant-volume chart. 

Prob. 1. Ten pounds of water are heated from 50° F. to 300° F. without change of 
state; what is the change of entropy? 

Prob. 2. What will be the entropy change for 100 lbs. of the following gases per 1000° 
temperature rise from 0° F. at constant pressure and at constant volume, assuming a 
constancy of specific heats as given in the tables of Chapter IV? Hydrogen, air, nitrogen, 
and oxygen. 

Prob. 3. Mallard & LeChatelier give for the specific heat of 2 , C p = (.303 +.00003 T). 
Compare the entropy change for 100 lbs. of -this gas from 0° to 1000° with this value 
of specific heat with the result of Problem 2. 

Prob. 4. A pound of air at 800° F. and a pressure of 150 lbs. per square inch gage 
expands at constant pressure until the temperature becomes 60° F. Find the volume 
at the end of expansion, the work done, and the change in entropy. A pound of air 
at 32° F. and 14.7 lbs. per square inch gage weighs .0807 lb. and the specific heat at 
constant pressure is .243, assumed constant. 

Prob. 5. If the same amount of heat be removed at constant volume from a pound 
of air at 150 lbs. per square inch gage and 800° F., as was removed in Problem 4 at 
constant pressure, what will be the pressure, entropy, and temperature change? The 
specific heat at constant volume is .17. 

Prob. 6. Water at 50° F. is changed to steam at 100 lbs. per square inch gage with 
5 per cent moisture. What is the entropy change? What would it be if ammonia had 
been the liquid instead of water? What would be the value if the vapor were super- 
heated 100° instead of being wet? 

Prob. 7. Draw a T$> and PV diagram for the following case. Ice is heated from 
0° F. to 32° F. and then melted, the water being heated to 212° and evaporated. The 
steam formed is then superheated 200° F. The specific heat of ice is .5 and the latent 
heat of fusion 142. 

Prob. 8. One pound of air receives 1000 B.T.U. while expanding isothermally from 
100 lbs. per square inch gage and 60° F. Draw the PV and the T$ diagrams for this case. 

Prob. 9. A pound of steam at 50 lbs. gage and 200° superheat is compressed isotherm- 
ally until it becomes 10 per cent wet. Show by a T& diagram the loss in entropy and 
work done. 

Prob. 10. A closed tank contains nothing but superheated steam at 200 lbs. per 



904 ENGINEERING THERMODYNAMICS 

square inch absolute at a temperature of 550° F. If the volume of the tank be 10 cu.ft. 
and it be cooled to 32° F., plot the T<$> and PV diagrams to scale and find the volumes 
of liquid and vapor at every 10 lbs. pressure and the corresponding heat abstracted 
for each 10 lbs. per square inch pressure reduction. 

4. Quantitative Relations for Secondary Thermal Phases. Adiabatics for 
Gases and Vapors. Constant Quality, Constant Total Heat, and Logarithmic 
Expansion Lines for Steam. Adiabatic expansion and compression for both gases 
and vapors, whether superheated or wet, are characterized or denned by constancy 
of entropy, which corresponds to zero heat change, and arc represented on the 
temperature entropy diagram as a straight vertical line for all cases. On the 
pressure volume diagram it is, of course, an exponential curve, PV S = Constant. 
It is possible now to find the value of s for both gases and vapors, including 
vapor liquid mixtures, best called wet vapors. Making the heat zero, in the 
general equation, 



Qdb —-J 



c 



.a 



i 



X (Work of Expansion) =0. 



It follows that since neither the work of expansion nor the specific heats 
are themselves zero, the s must be equal to the ratio of specific heats 
designated by y, 

C 

For adiabatic changes in gases, s = -^ = y .... (938) 



For exponential expansion of gases in general, and for adiabatic in particular, 
the following relations between the temperatures and pressures, or temperatures 
and volumes, are derived from the fundamental pressure volume relation, 

PaVa = P b V b s = Constant = P a V a V a s _1 = P b V b V b s ~\ 
Putting 

PaVa = RT a , and P b V b = RT b , 

there results the relation between temperatures and volumes, 

TaVa-^TtVj,- 1 . 



Also 

(Ta\^ = Vj = (Pa 
\T b ) V a \P b 

Therefore, 



s-l s-1 

T a P b s =T b P n s ' 



HEAT AND WORK 



905 



Collecting the series for convenience and giving to the adiabatic the special 
exponent all are given by the tabular equation, Eq. (939), 



Related Quantities 
for Exponential 
and Adiabatic 
Changes In Guses 
and Liry Vapors. 


Formula 




General Exponential Change. 


Adiabatic Change. 


P and V 


Pa my v a my 

P» \Va) ' Vj> \PaJ 


Pa 
Po 


/v b y V a /P b \T 

\Va) ' Vo \Pa) 


PandT 7 


Pa (TA^lTa (Pa\~ 

p b \rj 'T b \pj 


Pa 
Po 


(T a y=l T a (PaYV 

\To) J T b \P b ) 


V and? 7 


V a (T \^l T a (VA*- 1 
Vo \T a ) 'To \Vj 


V a 
V b 


(T yn T a my- 1 

\Ta) 'T b \Va) 



(a) 
(b) 



. (939) 



T 




























P 




























































150 

1 

.0 
«i 

S3 
& 

& ioo 

P, 

m 

jQ 

a 
2 

P< 50 





















































































































































750 


A 










s = 


1 






B 














A 




P 


>int 


A:- 


P = 


150; 


v= 


1.87 


;T= 


760 




































) 








B: 


" 


10 


" 


7.16 




760 










S&: 


/ ,, 


































C: 




40- 


" 


5.76 


; " 


608 






CO 


y 






> c 
































D: 


'• 


10 




',.25 




556 






H 




tt? 




































E: 


" 


10- 


" 


1.76 


; " 


513 




500 


r 






















































































































































J\ 


















































































































<? 




















250 








































V. 


<s\ 


Ntf 


V , 


s^V 




















































^> 


|E 


*Jd 


Tc 


— 


^ 


1^ 




















































































1 
1 


! 


i 




































— 


— 


— 


■ -(■ 


At 


a 


: 


ic L 


me 


-t- 


-t- 


-i-- 


— 




A 























1 


























1 


; 


I 






! 



.02 .01. .06 



.10 .12 <p 



7 V 



Entropy Volume in Cubic Feet 

Fig. 227. — Adiabatic and Exponential Changes for Gases and Comparison with Isothermal. 



To. illustrate the case of adiabatic processes and to bring out clearly the 
striking difference between the isothermals and adiabatics for gases on T§ 
diagrams in contrast to the small difference to PV coordinates, there is replotted 
in Fig. 227 the isothermal example of Fig. 223, and beside it is added an adiabatic 
for air, P7 1 * 41 = Constant, from A to E, between the same pressures as well as 
two general exponentials according to PV 12 = K from A to C, and PV 13 = K 



906 



ENGINEERING THERMODYNAMICS 



from A to D. The PV diagram shows but little variation in work done com- 
pared to the variation in heats required as shown on the T$ diagram where for 

s = 1 (Heat added) = (Work done) ; 

s = 1.2 (Heat added) = .51 X (Work done) ; 

s = 1 .3 (Heat added) = .27 X (Work done) ; 

s = 1.41 (Heat added) =0. 





i \ f i / i / 1 


v i _r 


7 X-t ¥ v 


~a u 


t 47 -A V 


4 t 4 


t XX 7— 


X 4\ 7- 


74 4 4 4 


9 V 7 \' 


tx zt4 -f - 


X t V 


4 X 4 -t t- X -7 


4 ' A 


t-A L V/ \ 7 


\7 TV 


4 7 V X -7 


Xt 7 t 


t -X -A X 7 


« in 


7% Zi X7 


3W ? 


e- 74 4 A -A 


74a V 


X- t vy 4 7 


-A-X a 


V y t"7 


3JL3AV ■ 




7 7+-£ ^ 


~/\_ /_\ y 


-jjLJx -4 


^ / r 7 7 i i 1 r 


7 t #YZ 




t-tJV% 


-^ V v- 15 


-ti T^fiv- 


/ \ /A Ica \y 


1 / A- \ 


A X-7 u s i 


6 t% t -r 4 


r "v ur vt i 


4 tjrf A 


S -^V * 4 4 


o / -/ /- /- ^>U- 


w^ -5ST ^ J - 


^ ' /-^ -J y 


-7L \ 4 U 


? 5 7j / l r ^ 


&J5 1 4 t 


§ tt-TJ- -/ ^ 7 


e 5: i u 4 u 


m lit/'' 


V V X 4 t 


u III/// 
P4 4-4--/ i— /- -^' 


V 3 4 u t 


-44 t/ Z 7 


S^ V V- t t 


4 - 77-/--/ 4 -7 


V 5 \ I 4 


ttfty y 


a V V } 1 


J^ty 7± 


S V £ V ] 




4-v ^4 V v- 


i / / / / $y 


V V V 4 t- 


3 ltn7i2L 


^ 4 V t 


77747-7 


^ V V w?— * 


7u-tt7 


Vv^?S 


JulC 7 


\w^J 


wv - 


^^TtA \- 






MmY 


^KvVvt 


_Jwy 


3044 


Ms 


^SsV 


Mr 


^s& 


12 3 4: 


5 ., 6 7 8 9 



u 



3.3 



Volume Ratio — 

i.8 v ' i.e 

Temperature Ratio i 



1*4 



1.2 



10 



Fig. 228. — Exponential Gas Changes. Relation between Initial and Final Ratios of Pressures, 
Volumes and Temperatures for Small Pressure Ratios. 



HEAT AND WOEK 



907 



This heat and work relation is given, of course, by the general equation just 
used, (Heat added) =j(- — t) X (Work done), and in which the coefficient has 
the following values for 

8*1; 1.2; 1.3; 1.41; 



62) 



r=V 



o. 



40 



35 



30 



0- |CL 
O 



rt 25 



20 



15 



X0 



^ 111 I -X 


.t t ± A A t 


I pf □ /I / 


j -/ w j i- -> A -X 


7 4 \L i V f %■ 


iiit. r_/ _ y x 


3 f fl J ±7 hV 


V- t_y l_I J/ 77 


\n i \f J rx ._ 


U r v z_l r 


7 t 7 7 r 71 Z 


S3 -vV- £ t x4 z 


_X7 7 r / T _7_ I , z 


1 f I If 17 1 / 


t-R-t 7 7 7 77 


tl X 7L 7t itz I 


77 A 7 7 7 t K 4 


7 7-X t Av 7 7E 4 


-77 7 5 IX 4 7 x i 


77 E X 77 7z I H 


777 7 W 4 77 77 J 


_J7X Z- V ^7 I 7 


77X 7 X- ? 4 4- 


77^ / 7 7 77 7 t , 


hk- k- h p V \ \ \ 


/" /'/ A A /i / \^> V. V V" r- 


ll / /■> \» « V* \ u " 

111/ / /Cl' \lP W> \^ \<" l</> 


tHT-/~ Z V A 7 7 t 


^fUiA^ fA \. X \. l 


T 77X17 7 X^T XT T 



40 



35 



30 



Q-loT 

o 



25 



15 



5 


10 


15 
Volume Ratio -I2 


20 


25 


30 












3.5 


3 


2.5 v > _ 
Temperature Ratio ■—• 


2 


1.5 


1 













Fig. 229. — Exponential Gas Changes. Relation between Initial and Final Ratios of Pre 
sures, Volumes and Temperatures for Larger Pressure Ratios. 



To facilitate calculations of P, V, T relations for expansions and compres- 
sion having various values of s, the curves, Figs. 228 and 229 have been plotted 

to a vertical scale of ( pM , and with a double horizontal scale for the correspond- 
ing I ^ ) and ( TrT ) • Each curve is for a different value of s, as marked on it. 



908 



ENGINEEEING THERMODYNAMICS 




HEAT AND WORK 909 

These are also given on logarithmic cross-section paper in Fig. 230 as arranged 
by Gunn, where all lines become straight, and to which an entropy scale is added. 
The simplicity of these relations for adiabatics of gases or superheated vapors 
not nearing the condensation condition, is in striking contrast to the complex- 
ity characterizing the adiabatics for vapors passing into the wet region. It 
was shown that isothermal compression of a superheated vapor would wet 
it, or inversely, isothermal expansion dry it or increase the superheat, and it 
now appears that adiabatic expansion of superheated vapors makes them lose 
superheat and become wet, or adiabatic compression dries them or increases their 
superheat, just the inverse of isothermal action. From the diagram, Fig. 231, on 
which OA is the liquid line and EJ is the saturation line to both T$ and PV 
coordinates it is clear that there are five typical cases of adiabatic change, 
either compression or expansion between any two pressures, depending on the 
relation of the adiabatic to the saturation and liquid lines. 
* First, AB passing through the state of all liquid at the high pressure and 
temperature, indicates that liquid atTthe boiling-point will evaporate or make 

vapor, by adiabatic expansion. The weight formed will be, (==) X (weight 

\OJ ) t* 

of substance), while its volume will be likewise, 1 = ) X (volume of dry satu- 

\0 J J t* 

rated vapor) = ( = ) X(OJ) PV , neglecting the liquid volume. 

Second, CD passing through a wet region at the high pressure also passes 
through the wet region at the low pressure, but the quality will not be the same 
at all points, though it may at two or more. The quality at the beginning 

and at the end ( = ) , while the initial volume before expansion is 



is . 

,AE T* \OJ/T4> 



A £\ X (AE) PV = (AC)p V , and the final volume is (=) X (OJ) PV = (OD) PV . 



\AE/T<l> \OJ/T<S> 

Third, EF passing through the state of dry saturated vapor at the high 
pressure, enters the region of wet vapor immediately for all lower pressures, 
proving that dry saturated vapors get wet by adiabatic expansion and, in addi- 
tion, providing means of evaluating the quality or moisture. Thus, for expan- 

/(~)jp\ 

sion, the initial quality is 1 and the volume (AE) PV , the final quality (= ) 

\0J/T4> 

/OF\ — 

and corresponding volume, (=) X (OJ) PV = (OF) PV , while the final moisture 

V J r# 



\OJJt* 



Fourth, GH passing into the superheat region at high, and into the wet 
region at low pressures, crossing the saturation line EJ on both diagrams at 
a point M. The high-pressure superheat is (T g — T e ) degrees, the low-pressure 

quality [=) and wetness, (==) . Assuming that superheated volumes 
\OJ /t$ _____ \OJ/t* 



910 



ENGINEERING THERMODYNAMICS 



V T T 
vary as absolute temperatures, then -=f=-=?- or, Vg = V e yfr } then the initial 

V e J- e J- e 
J 1 

volume before expansion is (AG)p V =(AE) PV X7jr, while the final volume is 

J- e 



OH\ 



X(OJ) PV = (OH) PV . 




12 15 18 21 24 

Volumes in Cubic Feet 



30 33 V 



Fig. 231. — Adiabatics for Steam showing Changes of Quality and Illustrating a Graphic 
Method of Finding "s" for Wet Steam. 



Fifth, IJ passing through the state of dry saturated vapor at the low pres- 
sure and indicating that dry saturated vapors immediately superheat on adia- 
batic compression, or that highly superheated vapors may become dry saturated 
and then wet by adiabatic expansion. 



HEAT AND WORK 



911 



Another case might have been added is represented by KL for adiabatic 
expansion entirely in the superheat region, but this is exactly the same as that 
previously considered for gases and has all its characteristics. 

The graphic construction of such diagrams is quite easy when the satura- 
tion line is used as a reference, it being plotted to both PV and T$? coordinates 
directly from the steam tables, because in the wet region the volume of wet 
vapor divided by the volume of dry saturated is equal to the quality of the 
steam, and in the superheat region the volume of superheated steam divided 
by the volume of dry saturated is equal to the ratio of actual steam temper- 
ature to saturation temperature absolute very nearly; the exact relations due 
to imperfection of steam gas are given in Marks and Davis Steam Tables. 
These principles applied to the diagrams give equal line ratios on both, so 
one can be transformed into the other, or a point fixed on one located cor- 
respondingly on the other. Thus for saturated steam 



Wet steam compared to dry saturated steam at 
the same pressure, Fig. 231. 



Similarly, for the superheated region the ratios are between verticals on the 
T$ and horizontals on the PV since the former represent absolute temperatures 
and the latter specific volumes; thus for 



SflN 


| - 


(SB\ 


SMJ 


T* 


\SM)pv 


~SQ\ 

SMJ 


T = 


\smJpv 


~SP\ 


\ 


fSP\ 


SMi 


'r$ 


\SM hv 



Superheated steam compared 
to dry saturated steam at 
the same pressure, Fig. 231. 



\EE'/t* \AE)pv \EE'/t* \AEJpv 
MG'/t* {sM/pv* \MG')t* \Sm)pv 



{ \jr Jt* \oj/pv 



From any of these adiabatics located on the PV diagram from the vertical 
line on the T& diagram, the exponent s can be determined by the measurement 
of simultaneous pressures and volumes at any pair of points and their sub- 
stitution in Eq. (14), Chapter I. Taking the two points E and F on Fig. 231, 
the exponent for adiabatic expansion of originally dry saturated steam can 
be found from 



log 



Sef = 



P f \0gP e -\0gP f 



><7 



log Vf— log V e 



912 



ENGINEERING THERMODYNAMICS 



Table CXXVI. 
VALUES OF "s" FOR ADIABATIC EXPANSION OF STEAM. 



A. Expansion of Water from 200 
Lbs. Abs. 



B. Expansion of Dry Saturated Steam from 
200 Lbs. Abs. 



Values of s for 10-lb. 


Values of s for Whole 


Values 


of s for 10-lb. 


Values of 8 for Whole 


Intervals 






Range. 




Intervals. 






Range. 




Pressure. 


Calcu- 


Cor- 


200 Lbs. 


Calcu- 


Cor- 


Range. 


Calcu- 


Cor- 


200 Lbs. 


Calcu- 


Cor- 


lated. 


rected. 


to 


lated. 


rected. 




lated. 


rected. 


to 


latol. 


rected. 


200-190 


.0987 


.1 


190 


.0987 


.100 


200-190 


1.132 


1.145 


190 


1.132 


1.143 


190-180 


.1435 


.141 


180 


.1175 


.118 


190-180 


1.153 


1.145 


180 


1.143 


1.143 


180-170 


.1847 


.182 


170 


.1348 


.135 


180-170 


1.142 


1.145 


170 


1.143 


1.143 


170-160 


.2304 


.223 


160 


.1519 


.153 


170-160 


1.148 


1.145 


160 


1.144 


1.143 


160-150 


.2671 


.264 


150 


.1682 


.168 


160-150 


1.138 


1.144 


150 


1.143 


1.143 


150-140 


.3069 


.305 


140 


.1843 


.184 


150-140 


1.128 


1.144 


140 


1.140 


1.143 


140-130 


.3509 


.346 


130 


.2007 


.202 


140-130 


1.150 


1.143 


130 


1.142 


1.142 


130-120 


.3911 


.387 


120 


.2172 


.218 


130-120 


1.130 


1.143 


120 


1.140 


1.142 


120-110 


.4304 


.428 


110 


.2341 


.235 


120-110 


1.135 


1.142 


110 


1.139 


1.142 


110-100 


.4738 


.470 


100 


.2517 


.252 


110-100 


1.137 


1.141 


100 


1.139 


1.141 


100- 90 


.5166 


.510 


90 


.2699 


.270 


100- 90 


1.148 


1.140 


90 


1.140 


1.140 


90- 80 


.5512 


.551 


80 


.2889 


.290 


90- 80 


1.126 


1.138 


80 


1.138 


1.139 


80- 70 


.5897 


.592 


70 


.3089 


.310 


80- 70 


1.144 


1.137 


70 


1.139 


1.139 


70- 60 


.6320 


.633 


60 


.3306 


.332 


70- 60 


1.138 


1.136 


60 


1.138 


1.138 


60- 50 


.6790 


.674 


50 


.3547 


.356 


60- 50 


1.125 


1.135 


50 


1.137 


1.137 


50- 40 


.7147 


.716 


40 


.3811 


.382 


50- 40 


1.143 


1 . 133 


40 


1.138 


1.136 


40- 30 


.7658 


.760 


30 


.4125 


.412 


40- 30 


1.131 


1.131 


30 


1.136 


1.135 


30- 20 


.8150 


.808 


20 


.4518 


.448 


30- 20 


1.131 


1.130 


20 


1.135 


•1 . 134 


20- 10 


.8718 


.870 


10 


.5085 


.504 


20- 10 


1.125 


1.128 


10 


1.133 


1.131 


10- 1 


1.0557 


1.042 


1 


.6381 


.638 


10- 1 


1.124 


1.126 


1 


1.124 


1.127 



























C. Expansion of Steam. Superheated throughout D. Expansion of Steam Initially Superheated 
Expansion, from 200 Lbs. Abs. and 540° Super- and Finally Wet, from 20" Lbs. Abs. and 150° 

heat. Superheat. 

(Note. — Crosses saturation line at 70 lbs. abs.) 



Values of s for .10-lb. 


Values of s for Whole 


Values of s for 10-lb. 


Values 


of s for Whole 


Intervals. 




Range. 




Intervals 






Range. 






Calcu- 


Cor- 


200 Lbs. 


Calcu- 


Cor- 


Range. 


Calcu- 


Cor- 


200 Lbs. 


Calcu- 


Cor- 


Pressure. 


lated. 


rected. 


to 


lated. 


rected. 


lated. 


rected. 


to 


lated. 


rected. 


200-190 


1.354 


1.342 


190 


1.354 


1.342 


200-190 


1.249 


1.334 


190 


1.249 


1.339 


190-180 


1.314 


1.342 


180 


1.333 


1.342 


190-180 


1.365 


1.332 


180 


1.306 


1.338 


180-170 


1.455 


1.342 


170 


1.374 


1.342 


180-170 


1.396 


1.330 


170 


1 . 336 


1.337 


170-160 


1 . 257 


1.342 


160 


1.340 


1.342 


170-160 


1.333 


1.327 


160 


1.336 


1.336 


160-150 


1.403 


1.341 


150 


1 . 354 


1.341 


160-150 


1.314 


1.324 


150 


1.331 


1.335 


150-140 


1.213 


1.341 


140 


1.323 


1.341 


150-140 


1.325 


1.321 


140 


1.330 


1.333 


140-130 


1.422 


1.341 


130 


1 . 340 . 


1.341 


140-130 


1.357 


1.316 


130 


1.334 


1.332 


130-120 


1.343 


1.340 


120 


1.340 


1.340 


130-120 


1.302 


1.312 


120 


1.329 


1.330 


120-110 


1.329 


1 . 340 


110 


1.339 


1.339 


120-110 


1.303 


1.306 


110 


1.325 


1.328 


110-100 


1 . 332 


1.339 


100 


1.338 


1.339 


110-100 


1.270 


1.300 


100 


1.317 


1.326 


100- 90 


1 . 338 


1.338 


90 


1.338 


1.338 


100- 90 


1.396 


1.292 


90 


1.328 


1.323 


90- 80 


1.287 


1.336 


80 


1.331 


1.336 


90- 80 


1.311 


1.283 


80 


1.325 


1.320 


80- 70 


1.331 1.335 


70 


1.331 


1.335 


80- 70 


1.337 


1.272 


70 


1.327 


1.316 


70- 60 


1.340 


1 . 334 


60 


1.332 


1.334 


70- 60 


1.230 


1.156 


60 


1.314 


1.304 


60- 50 


1.315 


1.332 


50 


1.330 


1 . 332 


60- 50 


1.150 


1.150 


50 


1.290 


1.289 


50- 40 


1.327 


1.330 


40 


1.329 


1.330 


50- 40 


1.144 


1.146 


40 


1.268 


1.270 


40- 30 


1.318 


1 . 327 


30 


1.328 


1.327 


40- 30 


1.138 


1.140 


30 


1.246 


1 . 250 


30- 20 


1.328 


1.325 


20 


1.328 


1.325 


30- 20 


1.093 


1.134 


20 


1.216 


1.226 


20- 10 


1.323 


1.322 


10 


1.327 


1.322 


20- 10 


1.157 


1.127 


10 


1.202 


1.200 














10- 1 


1.116 


1.120 


1 


1.163 


1.176 



Note. Irregularities in values of 5 have been corrected by plotting a smooth curve through calculated 
values, and taking corrected values from this curve. 



HEAT AND WORK 



913 



This gives a sort of overall value for s which is not uniform over the line EF, 
as other values from 



l og P c -log P p 
log Fp-loglV 



and s p f- 



log P,-log P f 
log 7,-log V,' 



for the same adiabatic using intermediate points, are not the same. 

To illustrate this point there have been prepared Tables CXXVI, A to D. 
Case B shows the results for dry saturated steam expanding in 10-lb. drops 
from 200 lbs. to 1 lb. per square inch absolute. After the first drop the steam 
is wet and becomes more wet for the next. Case A is for the case of initially 
all water, the percentage of steam increasing as expansion continues. In case 
C the steam is originally superheated to such a degree that it remains super- 
heated throughout the range of the expansion. In case D the original super- 
heat is small and at some point during the expansion the steam is momentarily 
dry and saturated and then becomes wet. 

Hot water shows by its adiabatic expansion values from, s = .10 to s = 1.042 
for the first and last 10 lb. interval, while for dry saturated steam the values 
are from s = 1.145 to s = 1.126. In neither case does the variation for the 
whole range agree with that for the ten pound interval corresponding. Steam 
always superheated shows values slightly decreasing from s = 1.342, to s = 1.322 
over the whole range, but when the saturation condition is passed there is a rapid 
change, as shown in the last table. 

Due to the incompleteness of the necessary data the results as calculated 
and as given in the second column of the table showed some inconsistencies. 
These former values were plotted and the values given in the third column of 
the table were read from a smooth curve drawn through the calculated results. 

Inspection of these tables shows conclusively the variability of s along one 
expansion line which is due to the condensation that is proceeding at a variable 
rate. This variability of s is also dependent on the pressure at which expansion 
begins and on the original amount of moisture or quality, and to show this, 
s has been determined for steam expanding from 250, 200, 100 and 50 lbs. to 
1 lb., from the initial and final volumes only and for original qualities of 100, 
90, 80, 50, 20 per cent, the results being set down in Table CXXVII. 

Table CXXVII 

VALUES OF "*" FOR ADIABATIC EXPANSION OF STEAM 

(Determined from initial and final volumes only) 



Pressure Range. 


Initial Quality. 


250° F. 
Superheat, 


100 o/o. 


90%. 


80%. 


50°/o. 


200/0. 


Water. 


From 250 lbs. to lib... . 
From 200 lbs. to 1 lb . . . 
From 100 lbs. to 1 lb. . . 
From 50 lbs. to 1 lb 


1.183 
1.185 
1.192 
1.206 


1.119 
1.124 
1.125 
1.124 


1.110 
1.117 
1.118 
1.118 


1.102 
1.107 
1.109 
1.111 


1.060 
1.066 
1.070 
1.073 


.961 
.966 
.969 
.963 


.662 
.638 
.565 
.490 



914 ENGINEERING THERMODYNAMICS 












If for any possible condition of expansion fixed by pressures and quality, 
an average value of s were known for the whole range, then work could be cal- 
culated by the PV methods of Chapters I and III, but even then the labor 
would be considerable and there would be necessary an almost infinite number 
of values of s. If there were no other better way of proceeding this would be 
satisfactory, but the entropy diagram and steam tables provide methods 
of exact calculation of such simplicity as to command admiration, and warrant 
the entire abandonment of all other methods of calculation of work of adiabatic 
expansion. As a matter of fact the thermal method, whether executed graphic- 
ally or by means of tabular values, involves practically no calculation at all, 
and it is now possible to read directly from a chart the work done for a complete 
cycle with adiabatic expansion so that labor is concentrated in the preparation 
of the chart itself. Indirectly the work of the expansion alone as a single 
phase may be found from the same data, as will be explained. 

For a complete cycle of water heating, evaporation at constant pressure, 
adiabatic expansion, and condensation at constant pressure, the work done is 
equal to the heat added less the heat abstracted, which, per pound of steam, 
is the difference between the total heats at the two different conditions before 
and after adiabatic expansion. Therefore, the work of such complete cycles 
is to be determined from the total heats of steam. Charts of the total heat 
to which are added lines of constant pressure, temperature and quality for the 
same entropy, are graphic means for reading off the work directly. 

For expansion alone as a single phase the work can be calculated from the 
total heats from the general relation, 

(Heat added) = (Change in intrinsic energy) + (Work done) . 

The expansion being adiabatic the work done will be the difference between 
the intrinsic energy before and after, with opposite sign, because the heat added 
algebraically, is zero. Therefore, 

Let Hi and #2 = Total heat, B.T.U. per pound dry saturated steam initial 
and final; 
qij Q2 and Li, L^ = Heats of liquid and latent heats, tabular values; 
Ui and ZJ2 = Internal energy at initial and final conditions; 
xi and. X2 = Quality or dryness fractions; 
Vvi and Vy 2 = Specific volume of dry saturated steam at high 

and low temperature, tabular values; 
Vl x and Vl 2 = Specific volume of liquid at high and low tem- 
perature which may generally be neglected. 



T^i 2= ( W ^l^?^ lt*± b ";-^ y ) =.7(^-^2). . . . (940) 



Then 

adiabatic expansion / 
But during the formation of steam from water 

(Change in internal energy) = (Heat added) — (Work done), 



HEAT AND WORK 915 

so that the internal energy of a pound of steam above 32° F. is the amount 
of heat it takes to make the steam starting with water at 32° F. less the work 

the steam does in coming into existence —XP(V V — V L ). Therefore, the work 

J 

done in adiabatic expansion is the difference between the total heats before 

and after, less the difference between the work of steam formation at the two 

states before and after in B.T.U. When steam is wet, 



Total heat = q 1 +x l L x B.T.U. (941) 

Vol. wet steam = xiF Fl +(l-x)F z , 1 Cu.ft. . .~~. . . (942) 
Therefore, 

(Work of steam formation) = Pi[xiVvi+(1—xiWli — Vli] 

=x 1 P 1 (V Vl -V Ll ) ft.-lbs (943) 



U 1 -U32 = qi+x l L 1 -jz 1 P 1 (V Vl -V L i) B.T.U. . (944) 



U2-Us2 = q2+X2L2-jx 2 P2(Vv 2 -V L2 ) B.T.U. . (945) 



Substituting in Eq. (940) the values of the intrinsic energy above 32° F., before 
and after adiabatic expansion as given by Eqs. (944) and (945) the work of expan- 
sion is given by Eq. (946), 

W 1 2 = J(qi-q2+x 1 L 1 -X2L2)-[x 1 P 1 (Vv-V Ll )-X2P2(V v -V L J]hAbs.(M^ 

Neglecting liquid volumes, this takes the form of Eq. (947) for the special case 
of expansion illustrated by the line from C to Q, Fig. 231, which is the case of 
constantly wet steam, 

Wcd = J(qa-qs+XcLe-x g L m )-(xcPcVv c -XaPaV V(l ). • . . (947) 

But the quality after expansion x a is a function of the quality before x c , which 
may be expressed in terms of entropies or the corresponding heats. 



916 ENGINEERING THERMODYNAMICS 

The entropy relations are given by Eq. (948) derived by the following steps : 



(<!>« - <M = (<i>c - M + (<!>« - <M ; 



hence 

Adding to each side the entropy difference from 32° to the point S, 

X a ($m - <i>s) + (cK - <|>32) = Xc(<$>e ~ <M + (§a ~ $32) .... (948) 






In words Eq. (948) may be stated that for points on the same adiabatic the sum 
of the entropy of the liquid from 32° to any temperature, and the product of the quality 
into the entropy of complete vaporization at that temperature is a constant. From 
Eq. (948) the value of x q , the final quality after adiabatic expansion can be found 
for use in Eq. (947) by using the tables of entropy for any initial quality of steam. 
A similar relation between condition after and that before adiabatic expan- 
sion can be expressed in terms of heats and temperatures. If the specific 
heat of water be taken as unity then from Eq. (948) this relation will be given 
by Eq. (949), 

x % =XQ k +loge Y a (949) 



In some steam tables, notably Peabody's, the value of x, the quality for a 
given entropy at any pressure and temperature, is given, and these values are 
extremely useful in such work as they eliminate the solution of Eq. (946) to Eq. 
(949), constancy of entropy characterizing adiabatic expansion, the tabular 
values solve such expansion problems directly. If corresponding qualities for 
equal entropies are not given directly in tables, which is the case with the 
Marks and Davis tables, it is possible to estimate without calculation, the 
quality at one temperature, for a condition defined by an entropy the same 
as for some other quality at another temperature, and an example will be 
given later to illustrate the procedure. 

Constant quality lines on the temperature entropy diagram are directly 
useful in this sort of problem work and permit of direct reading of the quality 

































HEAT AND WORK 




















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918 ENGINEERING THERMODYNAMICS 

at any temperature, after adiabatic expansion from another temperature and 
quality, and are especially useful in the region of superheat or for passing from 
superheat to wetness, which is the case in nearly all practical steam turbine 
problems. Such a set of lines is laid out on the chart, Fig. 232, graphically 
by the dividing each horizontal vaporization line proportionally and joining 
the one quarter, middle, three-quarter points and others located correspondingly 
between, in the wet region. In the region of superheat constancy of quality 
implies constancy of superheat, so that on each pressure line points Z and Z' 
are so located that ZY = Z'Y'. Changes of quality along adiabatics repre- 
sented by verticals, are given by the intersection of the vertical adiabatic 
with the quality curve. 

Constant total heat lines may be located directly on such a chart and serve 
to save the labor of computation of work, which is still appreciable even when 
quality changes during adiabatic expansion are known. 

When the end sought is work done by the whole cycle there are two forms 
for such charts, first, the ordinary temperature entropy chart with constant 
total heat lines located on it, and second, the total heat plotted as ordinates 
and entropy as abscissa?, which latter is known as the Mollier diagram. 

The steps by which points are located on the temperature entropy diagram 
for equal total heats illustrate a very important process in steam turbine work- 
ing. Steam expanding in a nozzle without friction suffers adiabatic expansion 
and the work of adiabatic expansion as just determined appears as kinetic 
energy of the jet of steam. If this steam velocity is reduced to zero, the kinetic 
energy will all be converted back into heat, which heat will be added to the steam 
at the low pressure, drying or superheating it in addition. This is precisely 
what happens in the throttling steam calorimeter for measuring steam quality, 
and was used by Grindley, Griessmann and Peake in determining the total heat 
of steam. It is evident that after such an action the steam will have the same 
total heat at the low as at the high pressure and from such equivalence, with 
measures of its quality, the heats could be found or at least checked. Between 
the stages of turbines part of the heat converted into kinetic energy in the 
nozzles is converted back into work, but the action so far as it goes is the same. 
These are important matters entirely aside from the more common one of 
direct reading of adiabatic expansion work from charts. 

In Fig. 233, let D represent a condition of wetness from which adiabatic 
expansion starts and proceeds to E with a loss of energy due to transformation 
into work, represented by the area CDEBC. If this energy be converted back 
into heat it will reduce the wetness by evaporation from condition E to F, the 
point F being located so that area EFGHE = area CDEBC. Another point 
K would be located so that area CDIJC = IKLHI. In this way a line DK 
can be drawn, on which the steam has always the same total heat, or along 
which it suffers no change in heat content, so that it is a constant internal energy 
line for steam. So long as the heat equivalent of expansion work is all con- 
sumed in drying wet steam the construction is comparatively easy, but if the 
steam is originally dry or nearly so it will become superheated, and if super- 



HEAT AND WORK 



919 



heated it will acquire more superheat. In fact for superheated steam, if it 
behaved like a perfect gas the internal energy would be a function of tempera- 
ture only and constant energy lines would be isothermals. It will be observed 
from the diagram that this is not the case. To illustrate the location of a point 
in the superheat region, consider an original condition represented by M to be 
the beginning of an adiabatic expansion ending at N, then the point will be 
located so that area CMAW = area NROPQN and drying has proceeded 
along NR, followed by superheating RO to the constant total heat line MSO. 




1.0 1.5 

Entropy 



2.5 



Fig. 233. — Constant Total Heat Lines for Steam, Method of Determination. 



When the points of such a diagram are replotted so that total heats above 
32° are ordinates and entropy from 32° are abscissa there will be a family of 
curves such as first plotted by Prof. Mollier and as represented in Fig. 234, 
known by his name as the Mollier diagram. On this chart the vertical distance 
from any pressure, temperature or quality, to any other, is the work done in 
heat units, by the whole cycle including an adiabatic expansion and can be 
marked off on a strip of paper and referred to the scale of heat to permit the 
work to be read directly, or the ordinate of the low can be subtracted from that 



920 



ENGINEEEING THERMODYNAMICS 



iri'a: '^hi^oi 




HEAT AND WORK 921 

of the high point. As this is so convenient for turbine work a scale of correspond- 
ing steam jet velocities is usually plotted beside that for total heats. A large 
scale chart of this sort is very necessary when many calculations of this sort 
are to be made and such may be plotted from the steam tables. 

Algebraic evaluation of the heat added during any expansion that is not 
adiabatic is often desirable, especially in dealing with cycles and their efficiency 
as heat converters into work. Unfortunately this is not always possible with 
precision, but approximations are possible for some cases, good enough for com- 
parative purposes. It is necessary for this work that the total heat be expressed 
as a function of temperature and this is possible only by algebraic expressions 
of the empiric sort and of complicated form derivable from the general 
expression, Eq. (950), 



(Heat added during expansion) = ( 7Uj> = I T-~, dT . 



(950) 



Integration of Eq. (950) is possible only when the differential coefficient ~ 

(l JL 

can be expressed as a function of T alone and by not too complex a function, 
if the final formula is to be of any practical value. Along the saturation curve 
the total heat of the steam is the sum of heat of liquid and latent heat at any 
given temperature above 32° F., so that the abscissa of any point on the T<$> dia- 
gram from 32° as an origin, will be given by Eq. (951), if C s is the specific heat of 
water, which may be taken as constant and equal to unity or as a function of T, 



>32 




(951) 



But latent heat can be expressed as a function of T approximately by Eq. (952), 

L = a+$T+yT 2 , 4 . (952) 

so that the entropy increase above 32° for saturated steam is a function of tem- 
perature given by Eq. (953), 



.^492 



T 

: - / C~ + ^+ 

492 

Corresponding to this integral expression is a differential form from which 
the differential coefficient of (j> with respect to T is given by Eq. (954), 

d§ C s a . 



922 ENGINEERING THERMODYNAMICS 

This takes the two following forms (a) and (6), Eq. (955), according as C s = l 
approximately, or is a series function of the temperature. 



For specific heat of water =1, -^ =— — -^+Y (a) 

Forspecific heat of water C s = a+bT+cT 2 , ^ = |+6 + cT-^+y (b) 



(955) 



Substituting the simpler form Eq. (955a), in Eq. (950), the heat added to keep 
steam dry and saturated as it expands will be given by Eq. (956), 



fMH' 



Heat added during expan- | 

sion from B to A along [ = / | 1_ " ^,+lT\dT (approx.) 
saturation line J 



= a log e J-(n-7 7 a)-|(n 2 -7 7 a 2 ). . (956) 



A more exact but still approximate result is obtained by substitution of Eq. 
(955 6), in Eq. (950), which gives Eq. (957). 



Heat added during expan- 
sion from B to A a 
saturation line 



pan- ] f Ta r a i 

long = / ya+(b+-r)T+cT*-^dT 



= alog e |- & -a(n-n)-^(n 2 -7 7 a 2 )-|-(n 3 -n 3 ). (957) 



Numerical problems cannot be solved without evaluation of these constants 
a, b, c, a, (J, y, which can be done approximately as follows: The Dieterici 
equation for the specific heat of water is 

C s = .9983 - .0000288(2 - 32) + .0000002133 (t - 32) 2 ; 
= .9994 - .00004242 - .0000002132 2 ; 
= .9738+.00015367 7 - .0000002132 72 ; 

whence 

a = .9738 ; 
b = .0001536; 
c =, 000000213. 



HEAT AND WORK 923 

Similarly, the Davis formula for the total heat of steam above 32° F. is 

# = 1150.3 + .3745(*-212) -.00055(*-212) 2 ; 
= 1046. 187 +.6077*-. 00055* 2 . 

From the specific heat of water the heat of liquid q becomes, 

q = .9994(7 - 32) - .0000424^ - 32) - .0000002 13* 2 (£ - 32) ; 
= l-.0007*-31.981-.0000356* 2 app. 

From these it follows that as L = H — q, 

L = 1078. 168 - .393* - .0005 U 2 ; 

^lsi.o+^er-.ooosiT 72 ; 

whence 

a = 1151.0; 
£ = .076; 
T = -.00051. 

Should the expansion proceed along any constant quality line in the wet 
region, the same method will apply, if, to the latent heat or to each term derived 
from it separately be applied the quality xasa multiplying factor. Hence 



Heat added during wet 
steam expansion from B 
to A for constant qual- 
ity x 



Or more correctly, 



= ^x\ogeY a -(T b -Ta)-^(T b 2 -T a 2 ). . (958) 



Heat added during wet steam expansion \ _ i Tl — n(T—T\ 
from B to A for constant quality x / ge T a { b a) 



—^(T^-Ta^-^TJ-T*). (959) 



It is even more difficult to express algebraically the heat added during expansion 
according to constant superheat, unless the specific heat be assumed constant, 
which is hardly proper, as it is known to be a function of both pressure and 



924 



ENGINEERING THERMODYNAMICS 















































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HEAT AND WORK 925 

temperature, which function is too complex to use here. If, however, it be 
desired to sacrifice accuracy to get a workable result, it may be taken as con- 
stant, and if C v be the mean specific heat of superheated steam for the .amount 
of superheat' t s degrees above the saturation temperature T e and T f , and constant 
for the expansion range, then the initial steam temperature will be T e -\-t s and 
final Tf+t s . For the heat of superheat C p t s B.T.U. per pound, the entropy 
above saturation will be 



Te 



' Te+ts dT 
C p \og e -~r^ ■■ I C v -y- 



which adds another term to the Eq. (954) similar to the first, and which on 
integration adds a corresponding term. As this case is of so little practical 
importance the result is omitted. 

Logarithmic expansion of steam, it has been stated, is the old assumption 
for cylinders, and it is of interest to see what changes in quality correspond 
to it. In Fig. 235 is plotted the saturation line ABODE to T$ and PV coordi- 
nates dotted. On the PV diagram are located a number of logarithmic 
lines for wet and superheated steam which it is found tend to cross the satura- 
tion line and so prove that evaporation takes place with logarithmic expansion 
of wet steam, and superheating follows. In fact for steam initially superheated 
the logarithmic expansion line is an isothermal and the superheat increases 
as the difference between saturation temperature for any pressure and the orig- 
inal temperature of the superheated steam. Thus, a logarithmic line A A' 
through A, a point of original dryness, is an isothermal as nearly as the super- 
heated steam approach to a perfect gas permits. It is located by the ratio 



I 



A'N \ = /A f O\ 
~EQJts \EOJpv 



Points in the wet region are located by proportionality of horizontal intercepts 
as for adiabatics. It is of no interest to compute algebraically the heat that 
must be added during expansion to keep it logarithmic, but for any case where 
it is needed, it can best be evaluated graphically by areas. 



Example 1. Calculation and use of Diagram, Fig. 232, lines of constant pressure and 
quality. Let it be assumed that the line of quality 80 per cent is to be located, starting 
with the pressure of 200 lbs. per square inch absolute, point A. From the steam tables 
£=381.9° F. or T a =841.9, the entropy of the liquid is .5437, of evaporation complete, 
1.0019, so that cJ)a-<J>32 = .8xl.0019+.5437 = 1.3452. To locate a point B in the super- 
heat region at the same pressure and for 100° of superheat, the steam tables are found 
to give directly <J>& — <J> 3 a = 1.6120. 



926 ENGINEERING THERMODYNAMICS 






As an example of the use of the diagram the following problem will serve. Steam 
at a pressure of 160 lbs. per square inch absolute, dry and saturated expands adiabatic- 
ally to atmospheric pressure and to some unknown quality to be found. From the 
point C representing the initial condition project vertically down to the pressure line 
14.7, at point D. By interpolation the quality is found to be 86.5 per cent, as point D 
lies between the two lines of 80 per cent and 90 per cent quality. 

Another example will illustrate the passage into the superheat region. Atmospheric 
exhaust steam at 20 lbs. per square inch absolute, is superheated 120° by a reheater 
and then expands adiabatically in an exhaust steam turbine to an absolute pressure of 
half a pound per square inch absolute, to find the final quality. The initial condition 
is represented by point E, from which projecting downward to the low-pressure line at 
H, lying between 80 per cent and 90 per cent, the quality is found by interpolation to 
be 88.4 per cent and the temperature by projecting to K, is T =540°. The corresponding 
volumes may be read off the chart Fig. 226, of the last section. 

Example 2. Method of calculating the Diagram, Fig. 235, for logarithmic expansion 
of steam. Assume for initial data, quality 50 per cent, initial pressure 60 lbs. and final 
pressure 14.7 lbs. sq.in. absolute. From the steam tables V r =3.585 cu.ft., T r =752.7°, 
<* = 4> 3 2 =1.0352. The point T is located as to pressure from the data, as to temperature, 
from the steam tables and as to volume, from the hypotheses, 

T _ P t Vt 60x3.585 

V t =— - = U7 = 14.62 rait. 

The quality at point T = ' =.546 and from the steam tables 4> — $32 =.3118, 

<j> e -<J> = 1.4447, whence 4*- <J> 3 2 = 1. 4447 X. 546 +.3 118 = 1.1006. 

The curve HCM is drawn between the same pressures, but from an initial 
quality of 95 per cent. From the steam tables 7^=6.812, ^ — ^32 = 1.5824 and 

V m =— — = — -~z — =27.8 cu.ft. But dry saturated steam at the low pressure has a 

specific volume of V e =26.79, so that the expanding steam has become superheated. The 

amount of superheat may be estimated from the Marks and Davis superheat tables by 

seeking the steam temperature for this pressure having a specific volume of 27.8, or it 

T 27 8 
may be estimated by the perfect gas law -=p = * . In this way it was found that 

1 z ^jO.79 

T m =695° and <j> m -<j> 3 2 = 1.7723. 

Prob. 1. A tank containing 500 cu.ft. of compressed air at 60° F. and twenty atmos- 
pheres pressure is drawn upon to supply a compressed-air engine. After the pressure 
has fallen to five atmospheres without heat exchange, what is (a) the air temperature, 
(b) the volume of atmospheric air measured at 32° F. withdrawn; (c) the work that 
has been done if all air withdrawn completely expanded; (d) the heat necessary to raise 
the remaining air to the original temperature? 

Prob. 2. If the working gases having y = 1.4 in a gas-engine cylinder expand according 
to the law PV lz = constant, the cylinder having a clearance of 15 per cent of its displace- 
ment, 20-inch bore and 30-in. stroke, initial pressure of 400 lbs. per square inch absolute 
and temperature of 2500° F., find (a) the final pressure and temperature, (b) the work 
done by expansion alone, and (c) the heat gained or lost during the expansion, (d) the 
net work of the cycle consisting of adiabatic compression from one atmosphere, combined 
with the above expansion and two constant-volume lines. 



HEAT AND WORK 927 

Prob. 3. Steam at 200 lbs. per square inch absolute and 150° of superheat adiabatic- 
ally expands to 100 lbs., then receives 100 B.T.U. at constant pressure and again expands 
to atmosphere. Find, (a) the quality after the first expansion and before and after the 
second, (6) the work done in each expansion, thermally, (c) the value of s for each expan- 
sion, {d) the work done in each expansion by PV methods compared with that found 
thermally. 

Prob. 4. Ammonia is being compressed adiabatically from 30 lbs. to 175 lbs. per 
square inch absolute. What quality must it have originally to have the final condition 
of, (a) all liquid, (b) 50 per cent vapor, (c) dry saturated vapor? (d) What will be 
its superheat if originally it were dry saturated? 

Prob. 5. Find the value of s for each of the cases of Problem 4 and determine the work 
of compression alone and of the whole cycle consisting of low-pressure evaporation 
preceding compression, high-pressure condensation and complete liquid cooling to low 
temperature, using both PV and T§ methods. 

Prob. 6. Plot to PV and T§ coordinates compression of 80 per cent quality ammonia 
and carbon dioxide from 0° F. to 70° F., according to the law PF=Const., and find (a) 
howjmuch heat must be added or abstracted per pound, by evaluation of Tfy areas and 
(6) the work done by both PV and thermal methods. 

Prob. 7. Steam in a cylinder has an initial pressure of 100 lbs. per square inch abso- 
lute and a quality of 90 per cent. It expands logarithmically to 30 lbs., is reheated to 
original temperature and then expands adiabatically in a turbine to 1 lb. per square inch 
absolute. Plot the T§ and PV diagrams and find (a) the heat added during logarithmic 
expansion and reheating, (6) the quality before and after adiabatic expansion, (d) the 
work done by expansion alone, in each stage checking calculations by diagram areas. 

Prob. 8. Using the thermal method entirely compare the work of admission, compres- 
sion and delivery for C0 2 and NH 3 compressors per pound of vapor originally of quality 
75 per cent, between —20° F. and 75° F. 

Prob. 9. On a T§ diagram plot the constant volume and constant quality lines 
for C0 2 and NH 3 , starting with superheated vapor at 300° F. and the pressure corre- 
sponding to saturated vapor at 75° F., ending the lines at 0° F. 

Prob. 10. For the steam cycle consisting of liquid heating, evaporation at 150 lbs. 
per square inch absolute, superheating 100°, adiabatic expansion to 2 ins., Hg absolute, 
and constant pressure condensation, (a) plot the PV and T<& diagrams, (b) find the work 
done by the five thermal methods and compare with the PV area, (c) modify the dia- 
gram by assuming adiabatic expansion to end at 50 lbs. per square inch absolute, closing 
with a constant volume line, and find the work so lost by T§ and PV methods. 

Prob. 11. If the steam of the first part of Problem 10 expanded through an orifice 
and impinged on a plate, what would be its final quality and velocity if (a) its final 
velocity were unchanged by impact, (6) reduced to half, (c) reduced to zero. 

5. Thermal Cycles Representative of Heat-engine Processes. Cyclic 
Efficiency. A Reference Standard for Engines and Fuel-burning Power 
Systems. Classification of Steam Cycles. In practice, heat conversion into 
work has narrowed down to a comparatively few processes out of the infinite 
number possible, considering all kinds of substances as possible heat carriers 
or work executors, and all possible states of each substance and the various 
modes of heat addition, abstraction, compression and expansion, that may 
constitute the thermal phases making up complete thermal cycles. The sub- 



928 ENGINEERING THERMODYNAMICS 

stances of importance are only two in general, first, water-steam, and second, 
a substantially perfect gas, which latter may be superheated steam or any other 
vapor, or a mixture of any number of so called permanent gases behaving as 
one. The processes through which water steam may pass, will, when studied 
yield certain principles of heat treatment of the fluid, fundamental to the 
steam-engine performance, while similarly basic ideas with regard to the gas 
engine follow from the same study of the processes through which gases 
may pass. There will result from such analysis of gas and vapor cycles, so 
called, a definite idea of the limiting possibility of the amount of conversion 
or maximum possible efficiency, and it will appear that the efficiency depends 
on how the heat is added and abstracted, how the substance changes state, 
how compression and expansion are related to each other and to the heating 
and cooling, how much heat is added per pound of substance, and the total 
range of variation of pressure, volume and temperature. In short, the efficiency 
can be found for any substance passing through any cycle and will be found 
to be not the same for all, sometimes quite high, often very low. Therefore, 
assuming, as is proper, that the object of engineers dealing with power generation 
is to convert as much heat into work as is possible or practicable, this cyclic anal- 
ysis will show what it is good to do, and what thermal actions must be avoided. 
It will, moreover, establish a numerical limit to the possible conversion or 
efficiency for any one series of processes that may be embodied in a machine, 
with which limit, the actual performance of a machine may be compared, thus 
giving a measure of the perfection of the mechanism as an executor of the cycle 
and showing how far improvements may be carried. For example, for given 
conditions of pressure and temperature, it may be found that a given steam 
engine is converting into work 20 per cent of the heat supplied to it in the form 
of steam, and this may look very low and great improvements appear to be 
possible, whereas on examination of the processes being carried out within the 
pressure and temperature limits imposed, it may be found that not more than 
25 v per cent efficiency is possible in the most perfect mechanism, in which case 

20 
the mechanism may be regarded as ~ = 80 per cent perfect and no amount 

Zo 

of ingenuity or expense can make it very much better than it is. This is a 
good illustration of the fact that the efficiency of a power generator is limited 
by two things, or is the product of two other efficiencies, first the efficiency 
of the thermal cycle being carried out by the substance and second, the efficiency 
of the mechanism as a cycle executor. The former is conveniently termed the 
cyclic or system efficiency and the second the mechanism efficiency, and this latter 
must not be confused with mechanical efficiency, which measures only mech- 
anical friction losses, whereas the former measures all energy losses in the 
mechanism not fundamental or necessary to the cycle itself. 

The great contribution of the cyclic analysis to engineering is, therefore, 
the establishment of standards of performance that serve as guides to improve- 
ment, telling clearly and unmistakably when one cycle or system must be 
abandoned in favor of another to attain a desired end, or where and how to 



HEAT AND WORK 929 

operate in the reduction of losses to improve a given piece of mechanism. With 
this as a criterion, it is evident that only those cycles that may be carried out 
by mechanism are worth studying, but a little caution is necessary here because 
someone may discover to-morrow how a given promising cycle, to-day considered 
a practical impossibility, may really be carried out and so a new power system 
be created. It is, therefore, of importance that this thermodynamic study of 
efficiency of conversion be so extended as to show what processes will yield 
high efficiencies, higher than the processes now considered good practice, so as 
to point the way for the mechanism designer that he may concentrate his energies 
on the clothing in metal of a process that promises better results than those 
in use. Cyclic analysis then has two criterions by which the worth while cycles 
are to be selected, first, their present practicability, and second, the high efficiency 
promised, realization of which depends on discovery of suitable mechanism 
to carry them out. No better illustration can be given than is supplied by 
the development c* the gas engine in the last half century. As originally built 
it was just a novelty, about the possibilities of which no one had any definite 
ideas: then thermodynamic study of its cycle showed two things: first, that 
its possible efficiency was very high, compared to the steam systems then in use, 
and second, that in all gas engines for efficient conversion, the cycle must include 
a compression of the working gases before heating. Both the higher possible 
efficiency with respect to steam and the necessity for precompression were known 
before their realization in gas engines of good construction and competitive size, 
and this knowledge was the real incentive to practical development. Not only is 
this so, but as higher efficiency began to be realized in gas engines than had been 
the rule with steam, so, as a direct consequence, was the improvement of the 
steam system stimulated, every loss carefully studied as never before, and means 
taken to reduce each, so that to-day in large-sized plants, especially with high 
pressure reciprocating and low-pressure turbine units, with superheated steam, 
feed-water heaters, economizers and high-vacuum condensing apparatus, the 
efficiency of the steam and gas system on the same coal fuel are substantially 
the same, and not less than twice as good as was considered satisfactory when 
the gas engine began its stimulating career. 

With this preliminary survey, the cycles for analytical study may be selected 
from among the great number of possibilities which are, of course, greater for 
gases than for steam, since in the latter case there is only one practical way 
of adding heat, that at constant pressure and temperature if the liquid heating 
and vapor superheating be excepted. It would perhaps be better to say that 
the bulk of the heat in the steam system or in fact any vapor system must neces- 
sarily be added at constant pressure and temperature. Heat abstraction 
likewise is essentially a constant-pressure and temperature process for vapor 
systems if the expansion is complete, and in the establishment of standards of 
possibility there is no reason for dwelling on any but complete expansion because 
this is always realizable in engines if really desired. Variation in vapor cycles 
one from the other can come in for consideration mainly as due, not to differences 
in modes of heat addition or abstraction directly, but rather from the differ- 



930 ENGINEERING THERMODYNAMICS 

ences in expansion between primary heating and heat abstraction, and in com- 
pression or its equivalent between heat abstraction and primary heating. The 
use of the term primary heating indicates that there may be secondary or minor 
heating in vapor cycles and this is a convenient way of distinguishing between 
the bulk of the heat received at constant temperature always to be considered 
as primary, and that received otherwise and to be considered as secondary or 
minor. To illustrate, the primary heat added will, of course, be the latent 
heat of vaporization of so much of the fluid as is vaporized if it is left wet. The 
secondary heat will be, (a) heat of liquid if liquid heating is part of the cycle; 
(b) heat of superheat if the vapor is superheated at the pressure of steam genera- 
tion before expansion begins; (c) heat added during expansion when it is not 
adiabatic, as for example constant quality logarithmic, or constant temperature 
for superheated vapor. 

With complete condensation of vapor as a result of heat abstraction at con- 
stant pressure and temperature, heating of liquid is necessarily one phase of 
the vapor cycles, but should the heat abstraction or condensation cease before 
condensation is complete and the remaining vapor be adiabatically compressed 
so that it just liquefies on reaching the high pressure and temperature which 
corresponds on a T$ diagram to a vertical line through the condition of high- 
temperature liquid, then there will be no liquid heating. 

All vapor cycles can be divided into two groups with respect to the mode of 
transition from condensation to vaporization phases ; the first group will include 
those that completely condense and so involve liquid heating, and the second 
those that substitute complete adiabatic compression for liquid heating. Of 
course, there may be cycles with part of one and the rest of the other process, 
but as their characteristics will lie between these two limits they need not be 
separately studied in a general investigation like this. 

Expansion may be adiabatic or may not, but if not, it must be according 
to some law for which there are algebraic relations, otherwise algebraic cyclic 
analysis is impossible and without this, principles are difficult to devise, though 
concrete problems may be solved. The only case of expansion that is of prac- 
tical interest and yet yields even approximately to algebraic analysis is that of 
constant quality, that is, for steam initially wet, with constant wetness during 
expansion, for steam initially dry and saturated, constantly so during expansion, 
and for steam initially superheated, with constant superheat during expansion. 
These cases of expansion for various constant qualities of steam are usually con- 
sidered as nearly representative of the possible action of steam in jacketed cylin- 
ders, and the heat added during expansion is supposedly representative of the heat 
lost by the jacket steam and gained by the cylinder steam at the rate required 
for constant quality. As a matter of fact it is very doubtful if jackets could 
so add heat to cylinder steam, and while they might give to it the total amount 
equivalent to constant-quality expansion it is pretty sure that the cylinder 
steam even if it received this amount of heat would not receive it at the rate 
corresponding to constant quality. To just about the same degree does the con- 
stant-quality expansion represent the skin friction heating effect of nozzles; 



HEAT AND WORK 931 

part of the work developed and appearing as kinetic energy is converted back 
into heat by velocity reduction. These cycles with constant-quality expansion 
are not of sufficient practical importance compared to those with adiabatic 
expansion to warrant equal treatment and while they will be indicated at the 
end of this section they will not be subjected to analysis. 

Adiabatic expansion is by all odds the most important case, because it is 
typical of heat addition completed before expansion begins, and heat abstrac- 
tion delayed until expansion is over. This is what would happen in cylinder 
engines if the walls were non-conductors of heat and did not absorb heat from 
the steam on steam admission or return it when the steam temperature dropped, 
and also what would happen in steam turbines if the nozzles were non-conduc- 
tors of heat and the surfaces in contact with the steam frictionless. Compari- 
son of cycles, otherwise the same, in which expansion is adiabatic with those 
in which quality is constant will show whether it is better to add all the heat 
before or continue addition during expansion and this will be taken up later. 
Of course, the adiabatic expansion may 

(a) Start with wet and end with wet steam; 

(6) Start with dry saturated and end with wet steam; 

(c) Start with superheated and end with wet steam; 

(d) Start with superheated and end with dry saturated steam; 
0) Start with superheated and end with superheated steam; 

so there may be five cases of adiabatic expansion cycles, each differing from 
the other fundamentally in the amount of heat added per pound of steam at the 
same pressure. 

From the preceding it appears to be possible to conveniently divide the steam 
cycles worthy of study into four, each with special cases for various degrees 
of initial quality or heat added per pound of steam at the high pressure, these 
four being the possible combinations of, (a) zero compression, (6) complete 
adiabatic compression; with (c) adiabatic expansion, and (d) constant quality 
expansion defining the typical phases, and represented by the corresponding 
diagrams of Fig. 236 to both T$> and PV coordinates. The relations of these 
various lines should be clear from the explanations given in the analysis of 
various phases and to make them clear to the eye, the saturation curve is 
extended across each diagram as a line of reference. 

. Referring to the PV diagram of Cycle I, it will be noted that vaporization 
B to C, or C' whether complete or incomplete, and including superheating C 
to C if there is any, is represented by the same sort of line as admission of steam 
to a cylinder, in which the initial volume V b is that of the liquid from which 
the amount of steam admitted is formed and which is so small as to be negli- 
gible in numerical work but is retained here to make the cycle operation clear. 
Similarly, the condensation line £>'", Z>", D', D to A, preceded by loss of super- 
heat D IV to D'" if there is any, is represented to PV coordinates, exactly as 
would be a constant-pressure exhaust. Therefore, although in cylinders there 



932 



ENGINEEBING THEEMODYNAMICS 



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HEAT AND WORK 933 

occur admission and exhaust, the efficiency of heat conversion can be studied 
by imagining as substitutes, constant-pressure heat addition or abstraction, 
which is equivalent to imagining the operations which take place in a whole 
plant from boiler-feed, steam-making, cylinder action, condensation in condens- 
ers, and return of condensate to boiler, to take place completely in the cylinder. 
There are no unwarranted assumptions involved in this substitution of equiva- 
lent thermal phases, as it is obvious that the cylinder will do the same work 
in either case and there will be necessary the same amount of heat to make 
it possible. However, as ordinarily carried out in separate apparatus these 
various processes involve some losses and some compensations, losses for example 
in pipe lines, and compensations from flue gas and exhaust steam heat. The 
compensations do not change the total heat required by the steam to put it 
in condition, but this amount is not all derived from fresh fuel when waste heats 
are partially returned. 

This same diagram represents in a similar way the action in a steam turbine 
plant in which the whole process is imagined as concentrated in the nozzles of 
the turbine, generation and superheating of steam represented by B to C, C ', C" , 
C IV does work represented by the area under this line to PV coordinates, and 
expansion does more work represented by the area under the expansion line, 
but in escaping from the nozzle this work is done only by pushing away some 
steam at the low pressure, which involves negative work under the low or back- 
pressure line or its thermal equivalent, the condensation line. The net work 
is the area enclosed by the PV diagram, as for cylinders, and is equal to the 
area enclosed by the T$ diagram of the equivalent thermal processes. 

All sorts of minor modifications of these cycles can be studied when special 
cases require it, as could also other cycles, for example, incomplete expansion 
followed by constant-volume release would be illustrated by Fig. 237 to T$ 
and PV coordinates for various degrees, ranging from no expansion at all to 
complete expansion. This diagram otherwise represents the conditions of 
Cycle I for dry steam and is inserted merely to call attention to the possibilities 
of cyclic modification, of which there are many others that will suggest them- 
selves to anyone. 

Some of these cycles adopted as representative and worthy of analytical 
study have received names associating them with some great man who studied 
them either first or most effectively. Thus the name Rankine is applied to the 
Cycle I, which is characterized by adiabatic expansion and no compression, and 
the name Carnot to those cases of Cycle II, that are without superheat and which 
are characterized by adiabatic compression and expansion, with heat added all 
at the high, and abstracted all at the low temperature. To other cases there 
is either no name applied or there is no agreement as to the name, so where 
confusion is likely to result from the application of a name to a cycle which 
elsewhere is applied to some others, the name will be omitted. There is abso- 
lute agreement as to the meaning of the Carnot cycle, and fairly general though 
not absolute acceptance of the name Rankine as signifying the cycle above 
defined, though some writers limit this name to the case of initially dry saturated 



934 



ENGINEERING THERMODYNAMICS 



steam, to which others apply the name Clausius. As the adiabatic-expansion 
and no-compression cycle is so closely representative of actual steam practice, 
it must be frequently referred to, and this will be called the Rankine cycle 
for wet, dry or superheated steam as a better term than Cycle I. In all other 
cases the cycle will be defined by its phases or number when discussed. 

The following analysis of these cycles will all be based on one pound of sub- 
stance of which varying amounts will be in the vapor and in the liquid states, 
and it will be assumed that when there is such a mixture it is of the same temperature 
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., 








































































































■ 


s 












































































1 




































f 


> 






<s 








|, 






\ 






























































1 












-- 


















Pi 






























































1 




1 














-^ 


~ 






































































• 






A 


i. 


losphwiu 


Line 










































































D" 


























l^ 2 

























































































































































































































































































































































































































































































































































Volumes in Cu. Ft. 



25 



Fig. 237. — Modification of Complete Expansion Steam Cycle by Incomplete Expansion 
with Constant Volume Closure Lines, 

collects next the walls and is cooler than the steam, and if superheated steam 
or gases are in cylinders that part near the walls will not have the same tem- 
perature as the rest of the mass, which is also the case in nozzles, as the heat of 
friction is imparted first to the skin fluid and may or may not be later communi- 
cated to the rest by conduction diffusion or mechanical mixture. 

Another assumption that is necessary is that of constancy of specific heat 
of water at unity, and of superheated steam at a mean value for the temperature 
range but not necessarily the same for different pressures or temperatures. 



Steam Cycle I (Rankine). 
First phase, from A 
volume). 



to B. Heating liquid (substantially constant 



HEAT AND WORK 935 

Second phase, from B to C. Heat addition at constant pressure, vapor- 
izing at constant temperature and possibly followed by superheating 
of vapor at rising temperature. 

Third phase, from C to D. Adiabatic expansion. 

Fourth phase, from D to A. Heat abstraction at constant pressure, 

condensation. 

Steam Cycle II (Carnot). 

First phase, from A to B. Adiabatic compression of vapor and liquid 
to all liquid. 

Second phase, from B to C. Heat addition at constant pressure, vapor- 
izing at constant temperature and possibly followed by superheating 
of vapor at rising temperature. 

Third phase from C to D. Adiabatic expansion. 

Fourth phase from D to A. Heat abstraction at constant pressure 
condensation. 

Steam Cycle III. 

First phase, from A to B. Heating liquid (substantially constant volume). 

Second phase, from B to C. Heat addition at constant pressure, vapor- 
izing at constant temperature and possibly followed by superheating 
of vapor at rising temperature. 

Third phase, from C to D. Expansion with constant quality. 

Fourth phase, from DtoA. Heat abstraction at constant pressure, con- 
densation. 

Steam Cycle IV. 

First phase, from A to B. Adiabatic compression of vapor and liquid 
to all liquid. 

Second phase, from B to C. Heat addition at constant pressure, 
vaporizing at constant temperature and possibly followed by super- 
heating of vapor at rising temperature. 

Third phase, from C to D. Expansion with constant quality. 

Fourth phase, from D to A. Heat abstraction at constant pressure, 
condensation. 

Prob. 1. Plot to scale to PV and T$ coordinates steam Cycle I modified by expan- 
sion in two stages, the first extending for half the temperature range, and with 
reheating in the receiver to the initial quality. 

Prob. 2. Plot to scale steam Cycle I, modified by logarithmic expansion. 

Prob. 3. Plot to scale steam Cycle I modified by logarithmic expansion in two 
stages as in Problem 1 but with reheating to initial temperature. 

Prob. 4. Plot to scale steam Cycle II modified as in Problem 1. 

Prob. 5. Plot to scale steam Cycle II modified by logarithmic expansion in the first 
stage for half the pressure range and cooling in the receiver to a quality of 90 per cent 
followed by adiabatic expansion in the last stage. 



936 



ENGINEERING THERMODYNAMICS 



Prob. 6. Modify Cycle II by two-stage compression adiabatic with intercooling 
and three-stage adiabatic expansion with reheating to original temperature. Inter- 
cooling to begin at half the temperature range and to completely condense the steam, 
reheating to be at I and § the temperature range and to just dry the steam. 

Prob. 7. Modify steam Cycles II and III by two-stage expansion, dividing the tem- 
perature range equally, the first to be at constant quality as in a jacketed cylinder and the 
second adiabatic as in a low-pressure turbine. 

Prob. 8. Modify steam Cycles I and II by stopping adiabatic expansion at half the 
pressure range, closing by a constant-volume line. 

Prob. 9. Plot to scale the following cycle: Water heated under pressure sufficient 
to prevent vaporization at 1000 lbs. per square inch absolute, the pressure being 100 
lbs. per square inch higher than saturated vapor pressure. Hot water allowed to escape 
through a valve to a receiver at 100 lbs. per square inch absolute, there brought to rest 
and adiabatically expanded to atmospheric pressure. Condensation from and at 200 c 
F. followed by return of water to high-pressure tank and there again heated. 

6. The Rankine Cycle. Work, Mean Effective Pressure, Jet Velocity, 
Water Rate, Heat Consumption and Efficiency of Steam Cycle I. Adiabatic 
Expansion, Constant-pressure Heat Addition and Abstraction, No Com- 
pression. 



Let 



Pc 




v c 




T c 


■ = - 


*Cc 




(4> c — <})32) , 





3i 
U 
hi 
Hi 



Qi = heat added in B.T.U. per lb. for whole cycle; 
$2 = heat abstracted in B.T.U. per lb. for whole cycle; 
IF = work in ft.-lb. per lb. for whole cycle: 

pressure volume temperature quality and entropy above 
32° F., for steam in condition represented by posi- 
tion of point C and with corresponding subscripts 
for other points. Pressures in lbs. sq. ft., volumes in 
cu.ft., temperature absolute Fahrenheit, quality in 
ratio of weight of dry saturated steam to total steam 
and water weight. 

heat of liquid, latent heat, heat of superheat and total 
heat above 32°, per lb. of dry steam, tabular value, 
at the high temperature and with subscript (2) for 
low temperature. 



$ c v' = constant mean specific heat of superheated steam from 



T e to 7>. 



Then since the change in intrinsic energy is zero for a complete cycle, the work 
done is equal to the heat change or the difference between heat put in and that 
abstracted for the whole cycle, which is algebraically expressed in Eq. (960) : 



W=J(Qi-Q 2 ) 



(960) 



HEAT AND WORK 



937 



As the efficiency of conversion of heat into work is the ratio of work done to 
heat supplied in the same units it is given by Eq. (961) : 



W = 0i-02 = 1 Q2 
JQi <2i Qi' 



(961), 



Both these expressions, Eq. (960) and (961), are perfectly general and true 
for any cycle, differences entering only when the heats supplied or abstracted 
are evaluated in accordance with the particular specifications for the cycle. 
For this cycle, Fig. 238, there are two forms of expression for the heat s 




p 




























































































































































































































































































































































































































































3 


C 


c 




;/ 


c 


'1 


C 


V 
























































10 






\ 


\ 


\ 


\ 




\ 


















y 




El 


Ih 


1 


n 


1 


in'al 


15 


» 






h 


e( 


1. 










\ 




















1 : 














\ 




v 


\ 






V 


















ially s 




' 


liea 

1 


te 


a 




,1 


li i;ilh 


s 


It 


11 


xted 




\ 




\ 












































\ 


\ 


L \ 


\ 




















T .it 


iallly s 






l< 


d 


fi 


nail 


' 


V(_ 


t. 
















s 


















\ 


















\ 




\ 


\ 


\ 






s 












8 


f 




1 . 






























> 


f\ 








































\ 






\ 








\ 










T 


lit 




a 






fi 




1 


I 


VL 


t. 












s\ 










s 


















, 




















\ 




^\ 












s 






li 






l3 


a 


Tlfl f 


n 


il 


y 


« 


et 
































































V 


\ 










































































































































It 


11 


>* 


>h 


ei 


1C 


L 


11 
















—- 






























U 






































L 










r c 






D 


in 





































































.5 1. 

Entropy 



10 15 20 25 

Volume in. Cubic ~Keet 



1.5 2, 

Fig. 238. — Rankine Steam Cycle. 



per pound, one, when superheat is initially present and the other when it is not, 
given as follows by Eq. (962), 



Q = q+xL = H — (1 — x)L f or saturated steam of quality x (a) 
Q = qJ r l J -\-} l = HJ r }i for superheated steam (6) 



. (962) 



These expressions will apply to all the special cases of this cycle which arise 
from different initial qualities and changes of quality due to adiabatic expan- 
sion and lettered (a), (b), (c), (d), (e) in the last Section. 

Case a. Diagram ABCD, Fig. 238, steam initially wet, finally more wet. 



Qi = qi — q2+x c Li, 
Q 2 = XdL 2 . 



938 ENGINEERING THERMODYNAMICS 

Therefore 

W = J(qi-q 2 +x c L 1 -x d L 2 ), ...... (963) 

E = l *§_ (964) 

qi-q2+XcL! v ; 

These two expressions Eqs. (963) and (964) for work and efficiency do not 
contain all independent variables because the quality x d is a function of the 
original quality x c , and of the extent of the adiabatic expansion. If, tables 
or charts are available, all these quantities may be read off and the answer 
obtained at this point. It is, however, desirable that this as well as similar 
expressions for other cycles be reduced to a function of temperature or other 
fundamental variable so that they may be compared without proceeding to 
numerical substitution. It has been shown in Section (4), Eq. (948) that for 
two points on the same adiabatic, the product of low-temperature quality, 
into the entropy of complete vaporization, is equal to the same for the high 
temperature added to the entropy change of the liquid from low temperature 
to high, or for this case, Eq. (965) and Eq. (966), 



X d (<i)d"> — §a)=Xc(§c> — <i>b) + (§b — §a) (965) 

^L 2 = ^ c Li + 7 7 a log e ^ (966) 

Substitution of Eq. (966) in Eq. (963) gives two forms, Eq. (967) and Eq. 
(968) for work, 

W = J^qi-q 2 +x c L 1 (l-^-T a log e ^ . .... (967) 
= j[(n- Ta) +x t U (l - y) - T a log e J] (approx.) 

= j[(T b -T a )(l+^-T a \og e ^ (approx.) . . . (968) 

The first form, Eq. (967), gives work, in terms of heats of liquid, temperatures, 
initial quality, and corresponding latent heat, and the second form Eq. (968) 
gives work in terms of the temperatures, the high-pressure latent heat and the 
initial steam quality on the assumption of unity as the specific heat of water. 
By putting the latent heat as a series function of temperature as in Eq. (952), 
Section 4, and substituting it in the last form for work, Eq. (968), a third form 
results, Eq. (969) in terms of initial steam quality quality and temperatures, 

W = J | (T b -T a )[l+x c ^+p+ y T b J-raloge^j (app.) . (969) 



HEAT AND WORK 



939 



The efficiency may also be put in terms of temperatures, initial latent heat 
and quality, or entirely in terms of temperatures and initial quality by similar 
substitution. The first form results from the substitution of Eq. (966) in Eq. 
(964), giving Eq. (970), which is exact, and Eq. (971), which is approximate, 
because the specific heat of water is assumed in it, to be unity. 



#=1 



1- 



Y^cLl + Ta \0gejT 

qi-q2-\-x c Li 



-jjTXcLi + Taloge T 
1 b J_a 

(Tt-TJ+XcL,! 



(app.) 



(970) 



(971) 



The first form Eq. (970) gives the efficiency in terms of liquid heats, tempera- 
tures, initial quality and corresponding latent heat, which reduce in Eq. (971) 
to temperatures, initial quality and corresponding latent heat, and in Eq. (972) 
by a final modification to temperatures and initial quality only, by introducing 
the temperature function of latent heat, 



E = \ 



z c (a+gn+Yn 2 ) + 7Uog e ^ 



(app.) 



(972) 



(n-^a)+^(a+&n+rn 2 ) 

In these equations for work and efficiency, 

a = 1151.0; 
g = .076; 
y= -.00051. 



Case b. Diagram ABA', Fig. 238. No steam initially present. If the initial 
condition is that of water at the boiling-point, x c = 0, and the three expressions 
for work and efficiency take the form Eqs. (973) and (974), 



W 



= j\qi-q 2 - T a log e pi from Eq. (967) (a) 

= j\T b -T a -T a log e ^ 1 from Eq. (968) or (969) (6) 



(973) 



E = l 



= 1 



Ta l0g e ^ 




from Eq. (970) (a) 

from Eq. (971) or (972) (6) 



(974) 



940 



ENGINEERING THERMODYNAMICS 



Case c. Diagram ABC'D'A, Fig. 238. Steam initially dry saturated. If 
the initial condition is that of dry saturated steam, x c = l, and the work and 
efficiency equation take the following special forms, Eqs. (975) and (976), 



W = j[q 1 -q 2 +L 1 (l-^-T a \oge^\ from Eq. (967) 
= j[(T b -T a )(l+^-T a log e ^ fromEq. (968) 

=/{(n-n)[i+(^+^+yn) 



(a) 
CO 



-T a \og e ^ fromEq. (969) (c) 

±a I 



E=l 



^Li+r B io & ^ 



qi — q2+Li 
T T 



1- 



T b -T a +U 

^(a+pn+ T n 2 )+r fl io ge |- 6 



"(n-r a )+( a +^n+Tn 2 ) 



from Eq. (970) 
from Eq. (971) 

from Eq. (972) 



(a) 
(&) 
(c) 



(975) 



(976) 



Case d. Diagram ABC CD" A, Fig. 238. Steam initially superheated and 
finally wet. Steam initially superheated to a degree indicated by condition 
C", that is, such as will allow it to become wet during expansion, will have dif- 
ferent work and efficiency equations, the general form of which is given in Eqs. 
(977) and (978), 

W = J(qi-q 2 +L 1 +h 1 -x d „L 2 ). ..... (977) 



E = l 



X d »L2 



qi-q2+Li+hi 



(978) 



As before, the final quality x a " is not an independent variable, but related to 
initial superheat by the equal entropy relations, 



and 



or 



(§d» — §a) = (<i>c" — <ba), 

x d " W" - 4>«) = (4v - 4v) + (civ - <M + (<j> 6 - <>«) , 



Xd»7jr = Sc'c» loge -Tfr+^r + lOge TyT- 
\ X d "L 2 = Sc'c"T a l0g e "T^ + TpLl + ^a log e jp 



(979) 






HEAT AND WORK 941 

This Eq. (979) will on substitution in Eqs. (977) and (978) give the work 
and the efficiency in terms of completely independent variables which may be 
reduced entirely to temperatures, as the only two. 

Case e. Diagram ABC'C' n D" r A, Fig. 238. Steam initially superheated 
and finally dry saturated. Steam with enough initial superheat to become 
just dry and saturated after expansion will have characteristics similar to the 
last case, but in which x d >> becomes x d >», and Xa»' = l, which yield the final result 
of Eqs. (980) and (981). 

W=J(qi-q 2 +L 1 +hi-L 2 ) 1 (980) 

E = \ ¥r-rr (981) 

In these equations the heat of superheat, hi=Sc>c'"(T C "> — T b ) is not an inde- 
pendent variable, being contingent on dryness without superheat after expansion, 
therefore, the temperature T c >» is fixed by the entropy relation that leads to 
Eq. (980) which, making £#« = 1, and T c » = Tc>" becomes Eq. (982), 

^ = £ cV ,,log e ^-+^+log e g (982) 



However, the two cases of small superheat just reviewed will give work 
per cycle and efficiency values between those for the two cases of, initially 
no superheat, and initially high superheat high enough to leave the steam 
superheated after expansion, so that the equations of the two cases of small 
superheat are omitted and those for high superheat set down. 

Casef. Diagram ABC C D D A. Fig. ,238. Steam constantly super- 
heated. When superheat is initially high enough to leave some superheat after 
expansion, however small the amount, the work and efficiency equations take 
the form of Eqs. (983) and (984). 

W = J(qi-q2+L 1 -L 2 +hi-h 2 ), (983) 

E = 1 L2 tr\ h (984) 

qi — Q2+Li+hi v J 

In these two equations the heats of superheat h 2 and h\ are not independent, 
but related by their temperatures through the equal entropy relations, 

or 

^^l0ge^-+J- = ^c^l0g e ^+^+l0g e ^. . . . (985) 
■L a * a I b ± b la 



942 ENGINEERING THERMODYNAMICS 

From this Eq. (985) the relation between T/v an d j> d iv can be found, which 
on substitution in Eq. (986), giving the final heat of superheat, in terms of 
the initial for substitution in Eqs. (983) and (984), will leave only independent 
variables. 

h - h \ Sa'"a IV (T d ' v -T a ) l 
h2 - h is,MT c iv-T b )\ ( 986 ) 

This is, to be sure, a very cumbersome transformation and even so is only 
approximate, but the exact expression is infinitely worse as a working equation 
It helps a little to assume a constant value for specific heat of superheated 
steam, S d '"£v = Sc'jv which gives Eq. (986), the form Eq. (987). 



kl T b 



T d iv 



T c iv 



(987) 



In this case it is not necessary to evaluate T c iv and T d *v from Eq. (969) 

(T d iv\ (T C IV \ 

but only the ratios ( — — ) and ( — — ) , but even this is bad enough as a time- 
consumer and will also be omitted to save space. 

These methods of work and efficiency determination furnish a thermal 
means of deriving some other important quantities, whether W and E are 
calculated from equations such as have been developed here, or from the 
shorter, more exact, and, therefore, more practical means, provided by the 
tables and charts of properties as related to entropy. 



Thus 

/Mean effective pressure in pounds\ _ / W 



j) (988) 



\ per square inch / \144X (vol. per lb. at low press 

If V2 be taken as the specific volume (tabular value) of dry saturated steam 
at the low pressure then this will have the two forms (a) and (6) of Eq. (989), 
which may be substituted in Eq. (988) for mean effective pressure. The 
second form is only approximate, as it assumes the superheated steam to behave 
as a perfect gas, while it does not, as may be seen by reference to Marks and 
Davis. 

(Vol. per lb. at low press) . = V2OC2 if wet finally (a) | 

Tr /Temp, of superheated\ . „ , , /n (989) 

= T2 Ump. of dry saturated /^ su P erheated < 6 > j 






HEAT AND WORK 943 

Again, since in round numbers the horse-power of 33,000 ft. -lbs. per minute 
is identically equivalent to 2545 B.T.U. per hour, the hourly heat consump- 
tion of the cycle per horse-power is given by Eq. (990). 

(B.T.U. per hr. per I.H.P.)=^F (990) 

£j 



Also if WR is the water rate or pounds steam per hour per horse-power 

(Cyclic wate rr ate)^, ^ B - T -^S^ h ^ H - P -) («) 

(B.T.U. supplied per hr. per I.H.P.) 
/B.T.U. per lb. st.\ /Heat of liquid at\ I (gem 

( at high press. ) — ( low press, above ) (6) 



(c) 



above 32° F. ^ 32° F. 

2545 



E(H 1 -q 2 ) 



Steam generated continuously at the high pressure and issuing from a perfect 

nozzle to a region of the low pressure would have a velocity measured by the 

work done on it by its own expansion. Since its increase in kinetic energy 

w 
from rest, is iMu 2 = —u 2 , the work done per pound of steam w, where w = l, 

will give to the pound of steam, the velocity in feet per second of Eq. (992). 

u = V2gXW = V$4:AXW (app.) (992) 

It is clear, therefore, that this maximum attainable jet velocity is directly 
proportional to the square root of the work done in foot-pounds or of the 
difference between the total heats per pound of steam at the conditions before 
and after adiabatic expansion. This fact prompts the paralleling of the 
heat scale on the total heat-entropy or Mollier diagram, Fig. 234, by a velocity 
scale computed as above. 

Attention has already been called to the possibility of evaluation of the 
work of cycles of this sort by means of the temperature-entropy and the 
total heat-entropy diagrams, and this is the only practical method for engineer- 
ing calculation, a fact that is appreciated only by solving problems first by the 
equation method and later by the tabular and chart methods. The Mollier 
diagram is constructed to read the differences (Q1 — Q2) directly so that work 
in heat units is only a question of reading the scale, and efficiency a mere question 
of slide-rule division. Moreover, the operations are precisely the same for steam 
of any initial quality, ranging from liquid water to the highest superheat, and 



944 



ENGINEERING THERMODYNAMICS 



changes in quality, superheat and volume are just as easily read off as the 
changes in total heat. 

As the Rankine cycle is the only steam cycle that reasonably approximates 
in representation the action of steam in cylinders and nozzles, and for which 
work and efficiency can be read off the entropy charts, this is an additional 
reason why it has become the standard of reference for steam engine and 
turbine performance among engineers. 

To make clear the great difference in labor involved in the solution of 
problems by tabular and by chart methods compared to the algebraic method, 
the following numerical examples are worked out, and it is important to 




1.5 q> 

Fig. 239. — Example of the Rankine Steam Cycle. 



1.0 
Entropy 



5 10 15 '30 25 V 

Volume in Cu. Ft. 



remember that the table and chart method is more accurate, as well as shorter, 
because it involves no assumption as to constancy of specific heats of water 
or steam, nor as to the accuracy of temperature relations with latent heat, 
all of which are necessary to secure even workable equations. In the curves 
at the end of this section are shown graphically the most important of the 
derived relations. 

Required the work per pound of steam, the thermal efficiency, heat con- 
sumption, cyclic water rate and mean effective pressure for the Rankine cycle 
between 100 lbs. per square inch absolute and one atmosphere for steam 
initially dry. This case is illustrated by Fig. 239. 



HEAT AND WORK 
First Method. By steam tables alone, without entropy data. 



945 



From the tables 



Hence 



Ti = 327.8° F. = 788° F. absolute (app.). 

7^ = 212° F. = 672° F. absolute (app.)- 

Heat of liquid (32° F. to 327.8° F.) =298 B.T.U. 

Heat of liquid (32° F. to 212° F.) = 180 B.T.U. 

Heat of liquid (212° F. to 327.8° F.) = 118 B.T.U. A to B, 

Fig. 239. 
Latent heat at 327.8° F. = 888 B.T.U. B to C, Fig. 239. 
Latent heat at 212° F. = 970.4 B.T.U., A to X, Fig. 239. 
Specific volume at 212° F. = 26.79 cu.ft. 

Qi = Heat added = 888 +118 = 1006 B.T.U. 

G 2 = Heat abstracted = final qualityX970.4 B.T.U. 



Final q uality = d£ = ^"4 a = X iZ |- a J 

<J>& — <j>a = l0g« 77^2 = .1625, 



* B -<>» = 78g = 1.1277 > 



and 



*«— 4>« = <J> d — 4>« = 1.1277+.1625 = 1.2902, 



970.4 -....„ 
<[>*-<[>« = -g^2~= 1.4447. 



Whence final quality = ~-rjj^ = -893, 
1.4447 



Q 2 = Heat abstracted = 970.4 X. 893 = 866.57, B.T.U. 

W = J(Qi- Q 2 ) = J(1006 - 866.57) = /(139.43) = 108476.5 f t.-lbs. 



1 3Q 43 
Efficiency = i^= 13.8%. 



Final volume of steam = .893X26.79 = 23.92 cu.ft. 



m.e. 



J(139.43) 108476.5 

P * 144 X. 893X26.79 144X23.92 dl,51bs - s ^ m ' 



946 ENGINEERING THERMODYNAMICS 



B.T.U. per hr.per I.H.P.=^=^= 18442 

hi .loo 



2545 
Water rate of the cycle = 9 Q .. An = 18.33 lbs., 

lo.oX-LvJUo 
Steam jet velocity = V64.4 X 108476.5 = 2643 ft. per second. 



Second Method. By ordinary entropy tables. 

From 32°F to 328°F entropy change for liquid = .4743 
From 32°F to 212°F entropy change for liquid = .3118 

From 212°F to 328°F entropy change for liquid = . 1625 

At 328 °F entropy change for vaporization = 1 . 1277 

From liquid at 212°F to dry steam at 328°F entropy change = 1 . 2902 

From liquid at 212°F to dry steam at 212°F entropy change = 1 . 4447 

Difference in entropy = .1545 

.1545 
Moisture after expansion = t~aT7^ = 10-7%, 

Quality after expansion = t-ttZ7 =89.3%, 

Heat supplied per lb. =(298 -180) +888 = 118 +888 = 1006 B.T.U. 
Heat abstracted per lb. = .893 X 970.4 = 866 . 6 B.T.U. 

Work = 139.4 B.T.U. 

Efficiency =|-g| = 13.8% 

The other quantities follow in the same way. 

Third Method. By Peabody's equal entropy table. 

For dry steam at 328° F. find (page 94) for quality = 1,(. 9994), the entropy 
= 1.60, and B.T.U. per pound above 32° F. = 1184.3. For wet steam at the 
same entropy at 212° F. read directly the quality = 89.15 per cent, and B.T.U. 
per pound = 1044.9 above 32°. 



Hence 



Heat supplied above 212° = 1184.5- 180.3 =1004.2 B.T.U. 

Heat abstracted above 212° = 1044.9 - 180.3 = 864.6 B.T.U. 

Work =(1004.2-864.6)= 139.6 B.T.U. 

Efficiency = ^| = 14%. 



HEAT AND WORK 947 

The other quantities follow in the same way. 

Fourth Method. Ordinary temperature entropy diagram with constant volume, 
quality and total heat lines. This method is essentially the same as Peabody's, 
as the final quality, initial and final heats are read off directly, work and 
efficiency are given by difference and ratio of these heats, and are slide-rule 
determinations. 

Fifth Method. Total heat entropy diagram. This method eliminates one 
step in finding work but is no shorter than the last for efficiencies. Work is 
given by the length of a vertical line joining initial and final conditions and is 
known without first finding the total heats. 

Sixth Method. By formulas. As this is merely a question of numerical 
substitution in equations which by inspection are clearly complex it is omitted 
from the text, but students are advised to do the work as an exercise. 

To show more clearly the relations between the various initial conditions 
and results derived from them, for this Rankine cycle, curves are plotted for some, 
taken in sets, as it is not convenient to show simultaneously all relations on 
a single curve. Thus Fig. 240 gives the thermal efficiency and shows how 
it varies with initial and back pressures and with the corresponding temperatures 
for steam initially dry and saturated. As heat consumption is in constant 
inverse proportion to the thermal efficiency, its scale is made to parallel that 
of thermal efficiency. Water rate being related to heat consumption through 
the high-pressure total heat per pound of steam above the terminal temper- 
ature, for the Rankine cycle, it is necessary to use the family of curves in the 
left-hand angle to pass from the vertical ordinate of heat consumption, on 
thermal efficiency, to the horizontal coordinate of corresponding water rate 
on the left. The resulting chart gives a direct solution of a problem on the 
relative value of two Rankine steam cycles for any pressure ranges when the 
steam is initially dry and, moreover, gives a numerical answer for the thermal 
efficiency, heat consumption and water rate. 

It is desirable to compare for given pressures the relative performance 
of the cycle with steam in various initial conditions of wetness or superheat, 
and to assist in this the curves of Fig. 241 are plotted according to a system 
similar to that used in Fig. 240, but with initial quality as the prime variable, 
but only for steam initially at 200 lbs. per sq. in. gage pressure, which maybe 
regarded as the high limit of present day general practice. 

These last two curves, Figs. 240 and 241, show the essential relations between 
the efficiency heat consumption or cyclic water rate, and the initial pressure 
or quality, but as capacity for work is just as important as efficiency of per- 
formance another pair of curves is plotted in Figs. 242 and 243, showing the 
work per pound of steam, and the corresponding maximum jet velocity for turbine 
nozzles or mean effective pressures for cylinder engines. In the curves, Fig. 
242, these two results fundamental to capacity for work are shown as functions 
of initial pressures for steam initially dry, and in the second case as functions 
of initial quality for 200 lbs. per square inch gage initial pressure, in both 



948 



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HEAT AND WORK 



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952 ENGINEERING THERMODYNAMICS 

cases for various back pressures. The very low mean effective pressure for 
low initial pressures such for example as in the region of one atmosphere, even 
with back pressures of only a fraction of a pound, is the very best dem- 
onstration of the value of modern low-pressure steam turbines against their 
only competitor, a low-pressure cylinder. Cylinder sizes must be inversely 
proportional to mean effective pressures, but there is no similar relation for 
turbines, the size of which depends on bucket speed and number of stages for 
a given output, so that the turbine is a logical substitute for cylinders in low- 
pressure ranges even when its mechanism losses exceed those of cylinders, 
as is often the case. 

Referring to the cyclic efficiency curves, Fig. 240, a fairly rapid rise with 
increase of initial pressure is found at first, becoming less as the pressure 
rises, but a most striking increase in efficiency with back-pressure reduction 
is demonstrated. Thus the gain from \ lb. to J lb. absolute is almost as 
great as from' 15 to 10 lbs., proving the superior value of low back over 
high back pressures from the efficiency standpoint. Initial quality has most 
influence when the steam is very wet, more wet than is ever used, as appears from 
Fig. 241, but the initial quality is of greater importance for low than for high 
back pressures shown by the slopes of the .25 lb. and 25.00 lb. curves near 
saturation especially. Initial superheat has a beneficial effect in all cases, 
but never large at any time. 

Example 1. Calculation and use of Diagram, Fig. 240, giving Rankine cycle efficiency, 
water rate, and heat consumption as junctions of initial pressure. The method used in 
plotting is illustrated by the location of point A, for assumed conditions of initially 
dry saturated steam at 100 lbs. per square inch gage pressure and 10 lbs. per square inch 
absolute back pressure. The efficiency is given directly by Eq. (976, b) as reference 
to the general T<& diagram, Fig. 238, shows these data to fall under Case c. Therefore 

E, = J h +T ^ ?) =1 / ggxS80 + 6 5 3. 2 lo4| |\ =16 ^ ^ cent> 
\ T b -T a +U J \ 797.9-653.2+880 / 

/ 

and the point A is located opposite this per cent on the efficiency scale. Passing to 
the left-hand angle of the diagram the point A' is located by Eq. (991, c), 

Water rate =m^) ^6 9( m8.7- 16 i.i) = 14 - 62 lbs - ^ LHR hour ' 

which fixes the abscissa of point A', the ordinate leing 16.94 per cent efficiency. 

To illustrate the use of the diagram, Fig. 240, the following problem will be graph- 
ically solved. Find the Rankine cycle efficiency, heat and steam, consumption for an 
initial pressure of 150 lbs. per square inch gage and dry saturated steam with a back 
pressure of 10 lbs. per square inch absolute. Starting at the initial pressure point B, 
proiect up to the 10-lb. back pressure curve point C, and then across to the efficiency 



HEAT AND WORK 953 

scale point D, reading there a thermal efficiency of 19.3 per cent and a heat consump- 
of 13 200 B T U per hour per I.H.P. Continuing across horizontally to the back pressure 
curve of 10 lbs. in the left-hand angle to point E and thence downward to the water- 
rate scale point F, the value 12.6 lbs. steam per hour per I.H.P. is read off directly. 

Example 2. Calculation and use of Diagram, Fig. 241, giving Rankme cycle performance 
as a function of initial quality. The determination of he position of the curve ABCD 
will illustrate the whole series. This represents the case of 2 lbs. per square inch 
absolute back pressure and 200 lbs. per square inch gage initial pressure. Point A 
represents an initial condition of all water; B, 50 per cent steam; C, dry saturated 
steam- D 200 degrees superheat. Reference is first made to the general T*> diagram 
to find under which case each set of data belongs so as to permit of the selection of 
the proper formula. The conditions represented by point A are those of Case b, and 
Eq. (974, 6) applies. Therefore 

, 847.9 
586.1 log e - — 

1— ^ = 17.34%. 

847.9-586.1 

which is the ordinate of point A, the abscissa of which is zero. Similarly a condition 
of 50 per cent quality falls under Case a, and Eq. (971) applies. Therefore, 




Eb = 1 ~ \ V» - T a +x c U I " 1 ~ \ 847.9 -586.1 +(.5X838) 



847.9\ \ 

-1=25.68%. 



Dry saturated steam falls under Case c, and Eq. (976, b) applies, so that 

( T «L + T log £\ /& 838 ) +586.1 log.g^\ 

y-LH-r B log. r J V§47^_^ 5§5i =27.66%. 

Ec = 1 \~Y^T^+U~/ \ 847.9-586.1+838 / 

An initial superheat of 200° falls under Case d, and Eq. (978) applies, and 
__XdU___ x _ ^ (approx.) 

Referring to the Mollier diagram, Fig. 234, Section .9 as the easiest way, ^ = .851, 
Which could of course have been calculated from Eq. (979). Substituting it, 

»_! _ ^i^ 021 -=28.24%. 

^ _X "847.9-586.1 +838+110.5 



954 ENGINEEEING THEEMODYNAMICS 



The 



In the left-hand angle the corresponding curve to be located is A'B'C'D'. 
ordinates of these points are the same as for A, B, C, and D, so it is only the abscissa 
that is to be determined as follows from Eq. (991, c) : 



= 56.1 lbs. I.H.P. hr. 



(WR)a' 


2545 


2545 


(T b 


-T a )E (847.9-586.1) X.1734 


{WR)v 




2545 


2545 


(T b 


-Ta+xME 


(847.9 -586.1 +.5X838) X.2568 


(WR) C ' 




2545 


2545 


(TV 


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(WR) d > 




2545 


2545 



= 14.55 



= 8.37 



7.45 



To illustrate the use of tne chart, Fig. 241, find the thermal efficiency, heat and 
"steam consumption, for the Rankine cycle, when steam is 90 per cent initially dry at 
200 lbs. per square inch gage pressure, and the back pressure 15 lbs. per square inch 
absolute. From the scale of quality at 90 per cent, point E, project up to point F on 
15-lb. curve, and then horizontally to point G at 18.98 per cent thermal efficiency and 
13,400 B.T.U. per hour per I.H.P. heat consumption. Continue across to H and down 
to K, reading the water rate value 14.4 lbs. of steam per hour per I.H.P. on the bottom 
scale. 

Example 3. Calculation and use of diagram, Fig. 242, giving work, jet velocity, and mean 
effective pressure, as functions of initial pressure for the Rankine cycle. Assume for point A 
an initial condition of dry saturated steam at 150 lbs. per square inch gage and 5 lbs. 
per square inch absolute back pressure. Then the cycle work will be given by Eq. 
(975, b) under Case c, so that 

W a =J [(T b - T a j (l +^j - T» l0g e |?J 

=778 [(825.9 -622.3) (l +f§|) -622.3 log, |||] =185,700 ft.-lbs. 

Passing to the left the abscissa of point A' is to be found and this is given by 
Eq. (988) as 

, , W 185,700 _ __ 

(m.e.p.) =-— = 144x6(U9 =21.32 lbs. per square inch. 

In the above the value of F 2 is found from the specific volume of dry saturated steam 
and final quality, or better, taken directly from the Mollier diagram. The velocity 
scale paralleling that for work is calculated directly from Eq. (992) and when 



W =250,000, u = V64.14 x250,000 =4010 ft. per second. 



HEAT AND WORK 955 

To illustrate the use of the diagram, Fig. 243, find the jet velocity, work per pound 
of steam, and mean effective pressure for the Rankine cycle for steam at 75 lbs. initial 
pressure gage, dry and saturated expanding to 10 lbs. absolute. Project up from 
point B to point C and across to point F where there is read, work done = 121,000 ft.-lbs. 
per pound of steam. Continuing across to D and down to E, (m.e.p.) =24.8 lbs. per square 
inch, or continuing CD across to G the jet velocity is 2790 ft. per second. 

Example 4. Calculation and use of Diagram, Fig. 242, giving for the Rankine cycle, 
work, jet velocity, and mean effective pressure as functions of initial quality. Curve ABCD 
is fixed by the four points, A for all water initially; B for quality 75 per cent; C for 100 
per cent, and D for 250 degrees superheat, so each point belongs to different case of the 
Rankine cycle and is given by a different though corresponding equation. 

From Eq. (973, b), Case b. 

W a =j(r b ^T a -T a log e y) =778^847.9 -662.3 -622.3 log e |||) 

= 25,750 ft.-lbs. 
From Eq. (968c), Case a. 

W b = J [(T b - T a ) (l +^fj - T a log e g] 

= 778^(847.9 -622.3) ^+- 7 |^- 8 ) -622.3 logc|||] =156,000 ft.-lbs. 
From Eq. (975, b), Case c. 

W c =J^T b -T a )(l+^-T a \og e ^ 

=778 [(847.9 -622.3) (l+J^) -622.3 log e |||] =200,000 ft.-lbs. 

From Eq. (977), Case d. 

W d =J{T b - T a +Za -x d U +h) 

= 778[847.9- 622.3 +838 -(.899x1000.3) +135.2] =231,200 ft.-lbs. 

To illustrate the use of the diagram, Fig. 243, find work, jet veloicty, and mean effec- 
tive pressure, for the Rankine cycle when initial pressure is 200 lbs. per square inch 
gage, 50° superheat and back pressure' 1 lb. per square inch absolute. Projecting up 
from point E to F and across to G, read, work =272,000 ft.-lbs., velocity =4190 ft. per 
second, and stopping on the 1-lb. curve at point A' the mean pressure 7.4 lbs. per square 
inch is read directly below at K* 



956 ENGINEERING THERMODYNAMICS 

Prob. 1. A locomotive engine with steam at 250 lbs. per square inch absolute initial 
pressure and 90 per cent quality, exhausting to atmosphere, uses 24 lbs. of steam per 
hour per I.H.P. What fraction of the Rankine cycle is being realized? 

Prob. 2. A low-pressure steam turbine supplied with atmospheric steam 75 per cent 
quality, exhausting to a 28-in. vacuum at 3000 ft. elevation, is developing 70 per 
cent of the Rankine cycle heat. What is its heat consumption per hour, its water 
rate, thermal efficiency, and work per pound of steam? 

Prob. 3. A combination low-pressure turbine and high-pressure piston engine operates 
on 60 per cent of the Rankine cycle efficiency in the former and 70 per cent in the 
latter. The initial pressure is 190 lbs. per square inch absolute with 100° superheat, 
receiver quality 80 per cent, and pressure 30 lbs., and back pressure J lb. per square 
inch absolute. What is the combined thermal efficiency and water rate? 

Prob. 4. What would be the jet velocity of the turbine of Problem 2 if single stage 
and all losses took place in the nozzles. 

Prob. 5. If a five-pressure stage turbine suffered no heat losses whatever and each 
stage were to develop the same energy and jet velocity from 190 lbs. and 100° 
superheat to 1 lb. absolute, (a) what would be the pressure in each stage, (6) the jet 
velocity, (c) the initial quality? (d) Plot a curve through these quality points. 

Prob. 6. Compare the thermal efficiencies of two turbines working between the 
pressures of Problem 5, one with 100 per cent initial quality and the other with 200° 
superheat, each developing 65 per cent of the Rankine cycle. What will be the differences 
in the respective heat consumptions and water rates. 

Prob. 7. What mean effective pressure would be developed by a Rankine cycle for 
the data of Problem 6 and how much cooling water between 45° F. and 70° F. would 
be required by a condenser per hour per cyclic horse-power. 

Prob. 8. The value of a reheating receiver is to be judged by comparing the Rankine 
cycle efficiency with and without it for the following data. Initial pressure 200 lbs. 
per square inch absolute, 50° superheat, receiver pressure 40 lbs., and back pressure 
2 lbs. per square inch absolute. Find the thermal efficiency for (a) one-stage adiabatic 
expansion ; (b) two-stage adiabatic expansion with receiver reheat to original temperature ; 
(c) express the value of the reheat in any proper terms. 

Prob. 9. Which case of the Rankine cj^cle corresponds to the following data: 

Initial pressure Back pressure 

lbs. sq. in. absolute. Initial quality. lb. sq. in. absolute. 

{a) 200 200° superheat 20.0 

(6) 190 150 10.0 

(c) 180 100 5.0 

(d) 170 50 4.0 

(e) 160 dry saturated 2.0 
(/) 150 90 per cent , 1.0 
(g) 140 80 2.0 
(h) 130 70 4.0 
(i) 120 80 5.0 
(i) 110 90 10.0 
(k) 100 • dry saturated 20.0 

Prob. 10. Find for any one or all cases of Problem 9, the heat added and abstracted, 
work per pound, mean effective pressure, thermal efficiency, heat consumption, and 
cyclic water rate by each of the five methods of the text or any one. 



HEAT AND WORK 



957 



7. The Carnot Steam Cycle and Derivatives. Work, Mean Effective Pressure, 
Water Rate, Heat Consumption and Efficiency of Steam Cycle II, Adiabatic 
Expansion and Compression, Constant Pressure Heat Addition and Abstrac- 
tion. To this cycle illustrated in Fig. 244 the same general Eqs. (960) and (961) 
of Sec. 6 apply, and need not be reproduced. It is in the evaluation of the heat 
supplied and abstracted that all differences between this and the Rankine Cycle 
arise. In this case there is no heating of the liquid, so the heat supplied to or 
abstracted from the cycle will have a different relation to the total heat per pound 
of steam. Steam in condition C has received from the beginning of the cycle 
some part of the latent heat at the high pressure, so that the heat supplied is 
given by the product of quality into latent heat and this is equal to the total 

























































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Fig. 244. — Carnot Steam Cycle and Derivatives. 

heat of the wet steam above 32° less the heat of the liquid. Hence if Hi is 
the total heat of the steam in the initial condition whether wet or dry, it is 
given by Eq. (993), 



Qi = Hi — q± = xL\ for steam initially wet (a) 

— Li for steam initially dry saturated (&) 

= Li+/ii for steam initially superheated (c) 



(993) 



The heat abstracted is, however, differently related to the total heat per pound, 
because the steam is never fully condensed at the low temperature in this cycle, 
but is reduced to water by steam compression at the end, therefore only part of 
its low-pressure latent heat is ever abstracted. Calling the low-pressure quality 
at which abstraction begins by X2, and that at which it ends by xj, then the heat 



958 ENGINEERING THERMODYNAMICS 



per pound of steam not abstracted is #2'Z/2+<?2, and the amount abstracted 
is the total heat above 32° for the condition at which abstraction begins less 
the above quantity, symbolically by Eq. (994). 



Q2 = H2 — X2L2 — q2 = (x2 — X2 f )L2 for wet steam (a) 

= (l—X2 f )L2 for dry saturated steam (6) 
— (l—X2 f )L2+h2 for superheated steam (c) 



. (994) 



Each of the cases arising from various positions of the adiabatic expansion 
line due to different initial quality will be taken up separately for evaluation. 

Case a. Steam initially wet, ABCD, Fig. 244, Carnot cycle. 

W = J(Qi- Q 2 ) = J[x c Li - (x d - x a )L 2 ] (995) 

E=l _9l = l _ {Xa-Xa)L2 (9%) 

Qi x c Li 

In these two equations, Eq. (995) for work, and Eq. (996) for efficiency, 
there are two dependent variables x d and x a which are related high pressure 
qualities at C and B by conditions of equal entropy, Eq. (997) and Eq. (998) : 

(<l>a-<i>a') = (<!>&- <M, Or -~- = \0geY a - 

Therefore z a L 2 = T a log e ^ (997) 

J- a 
AISO (4>d— 4>a') = «>c— <I>a') = (4>c— 4>&) + ( 4>6 ~ 4>a') , ° r -f~ = ~7p Hoge yT, 



or 



xJv^XcLx + Talofr^ (998) 

i b i a 



Whence combining Eq. (997) with (998) to evaluate the quality difference 
term of Eq. (995) and Eq. (996), this becomes Eq. (999): 

(x d - x a )L 2 = tj^XcLi + T a log e -^—T a log e -£ = ~-x c Li. . . (999) 

J- b ± a ± a ±b 



HEAT AND WORK 959 

Substitution of Eq. (999) in (Eq. (995) gives Eq. (1000), which is the work 
done by the cycle. 

W = j[x c L 1 -^x c L 1 ^=Jx c L l (l-^) (1000) 

Substitution in Eq. (996) gives the efficiency by Eq. (1001), 

This is a most remarkable case in several ways, and is one example of the 
famous Carnot cycle. It is characterized by reception of all heat at the high 
temperature and abstraction of all that is abstracted at the low temperature, 
and it will be noted, that the efficiency depends on the temperatures only and not 
at all on the quality of the steam. 

^ It can be demonstrated that between two temperatures, maximum and 
minimum, it is not possible to convert more heat into work or to get a higher 
thermal efficiency for any substance than is possible for this Carnot cycle. 
It is, therefore, a standard of comparison of value, because no system, whether 
using gases or vapors as media in an engine mechanism, however perfect, 
could yield a higher efficiency than such a Carnot cycle. 

Its value as a standard of comparison is, however often overrated, because 
it represents an unattainable high value for efficiency, whereas other standards 
like that of the Rankine for steam are attainable in proportion as the losses 
are located and reduced to zero. The difference between the attainable max 
imum for steam efficiency of the Rankine and the unattainable maximum 
of the Carnot, is not so great as is the corresponding difference for gas engine 
standards of reference. In the case of gas engine efficiencies there is a very 
great difference between the efficiency of the Carnot for the temperatures used 
and the efficiency of the most closely representative attainable ideal cycle 
for the particular method of working, and this difference is significant. 

The demonstration of the proposition that efficiency of the Carnot cycle 
is a maximum for the temperature range is very simple, by the aid of the T<S> 
diagram ABCD, Fig. 244, which is a rectangle, whose height is the temperature 
range, whose base the entropy range, and the enclosed area of which is the 
work done. It is evident that between two temperatures of working the max- 
imum work area of any cycle will be that of the rectangle embracing and 
enclosing all points and this is, of course, the Carnot cycle. 

Case b. Steam initially dry saturated, ABCD', Fig. 244, Carnot cycle 
For this case x c becomes a* = l, and x a becomes x#, whence 

W = J(Q 1 -Q 2 )=J[L 1 -( Xd ,-x a )L 2 ], ..'.;.. (1002) 



960 ENGINEERING THERMODYNAMICS 

and 

E=l _fx^a\ L2 (1003) 



But 



and 



T 

x a L 2 = T a \oge-jr, 



Xd L2 = TtrLl + T a loge yf • 
1 h ± a 



Therefore 



tx d .-x a )U = ^U ••■.. (W04) 



T, 



Substitution of Eq. (1004) in Eqs. (1002) and (1003) gives work by Eq. (1005), 
and efficiency by Eq. (1006). 

W = JU{1-^) (1005) 

. -RM^y • • (ioo6) 

This is another case of the Carnot cycle, that for dry steam, as the last was 
that for wet steam, and the cyclic work is of course greater. Comparing the 
work per pound in the two cases it is found to be directly proportional to the 
quality of the steam initially. 

Case c. Steam initially superheated and finally wet {ABC CD A), Fig. 244. 
To the latent heat at the high pressure is now to be added the heat of 
superheat to get the heat supplied, while the heat abstracted has the same 
form as before, so that work is now given by Eq. (1007), and efficiency by Eq. 
(1008); 

W = J(Qi-Q2) 

- JlLi+fe - (x d „-x a )L2} = J[Li+Sc>ATc» - T b ) - (xr-zjl*], (1007) 



and 



._ 1 (x d ,.-x a )L 2 _. (x d .>-x a )L 2 ....... (1008) i 

E=l — U+hi Li+S*AT*.-T>y 



HEAT AND WOEK 961 

In these two equations, the two dependent variables %a» and x a are related 
to initial quality or superheat by the expansion conditions. As before 



But 



whence 



T 

X a L 2 = T a log e 7^. 
1 a 

(§d» — <t>aO — (&» — <iv) + (4>c — 4>&) + (4>6 "" <Iv) , 

Xd"L/2 a 1 *c" , Z/l . , Tfy 

— ^— =b C 'c» l0g e y" +Y^~ ge Y' 

x#.L 2 =SwT a log, ^f+^Li + Ta log, p. 

1 b 1 b la 

.: (fc.-JBjii-flw.r.loB.^+Jii (1009) 

Substitution of Eq. (1009) in Eq. (1007) and (1008) gives Eq. (1010) for 
work and Eq. (1011) for efficiency. 

W = jh 1 +h 1 -S c ^T a \oge^+^Li^ • • (1010) 

S^Ta l0g e ^+|^1 (1011) 

E = l T , /> T b 

Li+hi 

These expressions do not reduce to the simple form found for the Carnot 
cycle, from which this case is a departure. This case is less efficient, indicating 
that superheat heat is of less value than latent heat in this cycle when judged 
by the temperature limits, which are now T C " and T a . However, for the same 
pressures, initial and final, this cycle is more efficient than the Carnot and as 
in practical machines it is pressure limits that control, it is a proof of the 
superior value of superheat in ordinary working over saturated steam. This 
was also the case with the Rankine cycle as was shown. 

Case d. Steam initially superheated and finally dry saturated, ABC C D A, 
Fig. 244. By substituting in the last case x d " = x d '" = l, the work and efficiency 
Eqs. (1012) and (1013) for this case follow at once, but hi, the heat of initial 
superheat, has a particular value, that which yields after expansion dry 
saturated steam. Hence 

W = J^L 1 +h 1 +S c ^T a \og e ^+^L 1 ^, .... (1012) 



E = l- 



Sc'c't'Taloge^r + ^L! 
1 b 1 b 

Li+fo 



(1013) 



962 ENGINEERING THERMODYNAMICS 

Case e. Steam continually superheated, ABC I C IV D JV D'"A, Fig. 244. The worl 
and efficiency equations in their first form can be set down by inspection, 
Eqs. (1014) and (1015): 



W = J[Li+hi-(l-x a )L 2 -h2], (1014) 

and 

E=l _ (l-x a )L2+h 2 _ 

In these equations the heats of superheat are given by hi=S c >c IV (Tc*v—T b ) 
and, h2 = Sd»> a iy (T d IV — T a ), and the amounts of superheat are related by 
the necessity for equal entropy. Therefore (§<f v —$a) = ($c? v — $b), which is 



equivalent to 



S d ,,, d IV\0g e ^+^ = S c , c Iv\0g e -^+^ + \0g e ^. . . (1016) 
-La -La Ij, lb l a 



This relation between superheats, Eq. (1016), together with 



T 

x a L2 = T a \og e7 ^, 

J- a 

will, when substituted in Eqs. (1014) and (1015) give expressions for work and 
efficiency which are somewhat too complex to be very useful. 

Unlike the Rankine cycle, this one cannot be solved by the temperature 
entropy and Mollier diagram charts, though the tables and diagrams are of 
value in finding individual quantities. For example, on the T§ diagram 
containing the various characteristic lines, the final quality or superheat can 
be read off directly, thus avoiding the equations; likewise heats of superheat 
for any temperature can be read off by passing along a constant-pressure line 
from any high temperature and total heat, to the lower temperature of satura- 
tion for the same pressure and its total heat. The difference between these 
total heats is the heat of superheat. There are many other short-cut ways of 
picking off various quantities from the diagram that cannot be detailed here 
but must be learned by experience. The really important fact to emphasize 
is that while the T§ and Mollier diagrams do help in finding separate quantities 
they do not yield the answer directly. 

Dividing any of these work expressions by the volume range will give the 
mean effective pressure as in all other cases, but it is to be noted that the 
volume range is exactly the same as for the Rankine Cycle I, so that the mean 



HEAT AND WORK 



963 



effective pressure will be given by Eq. (1017), using in it the value of low- 
pressure volume, Eq. (1018) : 



(m.e.p.) = 



W 



144 X vol. per lb. at low pressure 



Jibs, sq.in. . . . (1017) 

/ 



(Vol. per] 

lb. at low > = ^2^2, if wet finally 
pressure J 



(a) 



= V / Tem P- of superheated steam X J ^ P ^eated i (b) 
\ Temp, of saturated steam / ' 1 fi ji j 



finally 



(1018) 



B.T.U. 
per hr. 
perl.H.P. 



2545 . 



E 



in general 



Z!&P. for saturated steam wet or dry 



(a) 



(1019) 



Cyclic wa- ] 
ter rate 
= WR J 



2545 



r for super- 



E[(latent heat) + (heat of superheat)] heated (a 

I steam J 

2545 



#[ (latent heat) X (quality)] 



rq , for wet steam 



(6) 



(1020) 



It might seem as if steam jet velocity could be computed for this cycle 
as for the Rankine, but as a matter of fact such would be at present a purely 
academic proceeding because it is difficult to conceive of continuous steam 
generation necessary to nozzle operation with a cycle that requires adiabatic 
compression of wet low-pressure steam. This would be regarded to-day as 
an impracticable procedure though, of course, it is within the range of possi- 
bility and may be brought to a probability at any time by invention. To meet 
such a contingency and fill out the set of derived quantities, the jet velocity 
may be set down by the same general expression Eq. (1021); 



V = V2gXW = VUAXW, (app), 



(1021) 



There is added a set of four curves similar to those derived for the Rankine 
cycle, two, Figs. 245 and 246, giving thermal efficiency and the quantities derived 
from it, and two more, Figs. 247 and 248, giving the work per pound of steam 
and its derivatives. One curve of each set gives the result as a function of 
pressures for initially dry steam and the other as a function of quality for a 
given initial pressure. Comparison of this cycle with the Rankine is possible 



964 



ENGINEEBING THERMODYNAMICS 




HEAT AND WORK 



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ENGINEEEING THEKMODYNAMICS 




HEAT AND WOKK 



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968 ENGINEERING THERMODYNAMICS 

by taking off points from these curves for any given data and solving simi- 
larly for the same data by the Rankine curves. 

Inspection of the efficiency as a function of initial pressure, Fig. 245, shows 
the variation of this cycle to be of the same general character as for the Rankine 
cycle curves, Fig. 240, but usually higher numerically. Thus, dry saturated 
steam expanding from 100 lbs. per square inch gage to 1 lb. per square inch 
absolute yields 29.8 per cent for the Carnot, as against 26.8 per cent for the 
Rankine, and the values for 200 lbs. to 1 lb. are 33.6 per cent for the Carnot, 
and 30.1 per cent for the Rankine cycles. However, efficiency plotted as a 
function of initial quality or superheat, brings out a striking difference, 
comparing this cycle, Fig. 246, with the Rankine cycle, Fig. 241. For this 
case the efficiency is independent of quality up to saturation, whereas for the 
Rankine it increases with quality, so that the Rankine is somewhat less effi- 
cient for dry steam and very much less for very wet steam. In the super- 
heat region the character of the curves for the two cycles 'is the same, both 
showing a small gain by superheat, and at about the. same rates. 

Example 1. Calculation and use of the Diagram, Fig. 245, giving for the Carnot cycle, 
the efficiency heat and steam consumption, as functions of initial pressure. Two 
points A on the right and A' on the left will be located to show the method used in 
plotting. Assume initially dry saturated steam at 100 lbs. per square inch gage and 
10 lbs. per square inch absolute back pressure. Then by Eq. (1006) as these conditions 
fall under Case b, 

Ea = 1 - jr = 1 -^9 = 18.13 per cent. 
Passing across to A' its abscissa is given by Eq. (1020, b), 

( ^'=lf = ira80 = 15 - 951bsperhoUrperLH - P - 

To illustrate the use of the diagram, Fig. 245, solve the problem: For the Carnot 
cycle with dry saturated steam between 150 lbs. per square inch gage and 10 lbs. 
absolute find the thermal efficiency, heat, and steam consumption. From point B pass 
up to C and across to D, reading efficiency =21.1 per cent, and heat consumption 
12,060 B.T.U. per hour per I.H.P. Passing horizontally to E and down to F' the 
water rate of 13.9 lbs. per hour per I.H.P. may be read off directly. 

Example 2. Calculation and use of the Diagram, Fig. 246, giving for the Carnot cycle and 
its derivatives, the efficiency, heat, and steam consumption, as functions of initial quality. 
The location of the curve ABCD will illustrate the calculation of the series, point A, 
representing the condition of all water initially; B, 50 per cent steam; C, dry saturated 
steam; and D, 200° superheat all for 200 lbs. per sq.in. gage initial pressure and 2 lbs. 
per sq.in. absolute back pressure. Each falls under a special case of the cycle, all 
apparently has the same thermal efficiency as B and C, but really it is an impossible 
being Carnot cycles, except the last, where superheat is present. The first point A 



HEAT AND WORK 969 

condition because without some steam the cycle cannot exist, though with an infinitely 
small amount it can. 

FromEq. (1001), E a =E b =E c = 1-^ = 1 -f?^=30.8 per cent. 

1 & o47.9 

T^WT^nnn^F i fa"-*-)*" , (.851 -.2159) x 1021 

From Eq. (1008), ft-1- ^ +fc =1 m+110Ji =31.62 per cent, 

in which xa" and x a , the two qualities, are determined by the steam tables or from the 
Mollier diagram. Passing to the left-hand angle the water rates for points A', B' , 
C't and D' are given by the formuals, Eq. (1020) : 

,__ 2545 2545 ... 

(WR)a> = & t = ~~tt~ (indeterminate). 

2545 
{WR)v = .3088x.5xS38 =19 ' 67 lbs " per hour P er IHP " 

2545 
^^-5088^838 = 9 ' 838 " " " 

(WR)g.= 2545 - ^ =851 „ u 

^WK) d E(Li+hi) 3i. 6 2(838+110.5) 8 ' 0i 

Lines of constant dryness are drawn across this set and to show how these are found 
the point M on the 60 per cent quality fine will be located for 200 lbs. initial gage and 
5 lbs. absolute back pressure. From Eq. (1006) the ordinate or efficiency is found and from 
Eq. (1020, b) the abscissa or water rate, whence 

T 622 3 

^ = 1-^ = 1-^=26.61 per cent. 

2545 2545 

(Tffl) - = E^ = J 661X.6X838 = 19 -° 2 Ibs " ^ hoUr P er LHP - 

The use of the diagram requires no further explanation after what has been given 
for the Rankine cycle, nor is it necessary to explain the other diagrams completing 
this set to correspond to the set of four for the Rankine cycle. 

Prob. 1. For the data of Problem 1, Section 6, what fraction of the Carnot cycle 
efficiency is being developed? 

Prob. 2. For the same back pressure as in Problem 2, Section 6, what initial pressure 
for the Carnot cycle would yield the same efficiency? 

Prob. 3. Compare the Rankine and Carnot derivative cycle efficiencies for 190 lbs. 
initial, | lb. back pressures per square inch absolute with 100° superheat. 



970 ENGINEEEING THEEMODYNAMICS 

Prob. 4. With the operating data of Problem 3, a steam engine shows by test a 
thermal efficiency of 18 per cent, what per cent of the Rankine and of the Carnot 
derivative cycles is being developed. 

Prob. 5. Solve for the data of Problem 8, Section 6, the corresponding case with the 
Carnot derivative as the standard of reference. 

Prob. 6. Which case of the Carnot cycle or its derivatives corresponds to each set 
of data of Problem 9, Section 6. 

Prob. 7. Find for the data referred to in the last problem the performance of the 
corresponding Carnot cycle or its derivative. 

8. Gas Cycles Representative of Ideal Processes and Standards of Refer- 
ence for Gas Engines. While comparatively few thermal processes may be 
regarded as important as representatives of what may happen in trans- 
forming heat into work by steam as a medium, it is not so when perfect or 
reasonably perfect gases become the medium. Gases may receive their heat 
at constant pressure, constant volume or constant temperature, whereas steam 
must receive nearly all of its heat at constant pressure and temperature, as 
heats of liquid and superheat, when also involved, are both small parts 
of the total. Gases may, moreover, be expanded isothermally with the 
product of pressure and volume constant, or adiabatically with the product 
of pressure and the gamma power of the volume constant; moreover, compres- 
sion by either of these laws may or may not precede heat addition in various 
ways peculiar to it. It is possible to construct for gases a very great number 
of cycles, much greater than for vapors within the range of reasonable pos- 
sibility of execution, but when the criterion of easy execution by mechanism 
that shall be very simple is imposed, the number drops at once. In selecting 
those worth examination there are three conditions to be satisfied by the 
result. First, the cycles selected shall most closely represent processes now 
being executed in practical machines so that the cycles may serve as standards 
of reference for comparison with actual performance, which comparison shall 
show how perfectly the mechanism is carrying out the imposed process. 
Second, the selected cycles should include some that might reasonably be carried 
out especially if they promise high returns in efficiency, controllability, mean 
effective pressures or any other desired end not yet attained otherwise. Third, 
they should include those that yield general guiding principles, however 
academic 6he cycles may be or however impracticable. As a matter of fact 
comparatively few cycles, but more than it was found necessary to study when 
steam was the medium, will serve these ends, the same cycle falling at times 
in two or all three of the above groups. 

Consideration of gas cycles involves differences in the bases of comparison 
from those that were satisfactory with steam by reason of the differences in 
limiting conditions, With the steam it is the pressures that constitute the 
fundamental limits to results attainable, boiler pressures as high and con- 
denser pressures as low as are commercially economical are used and the 
temperatures that correspond are mere accidental incidents. The one exception 
is introduced by superheated steam with a rise of temperature at the same 



HEAT AND WORK 971 

high initial pressure, and this is of little importance as a departure because 
the temperatures used seldom exceed 200° superheat and the heat of super- 
heat 10 per cent of the total. Thus, with steam the amount of work that will 
be done per pound, depending as it does on the amount of heat per pound, 
is fixed within fairly narrow limits by the physical properties of steam itself, 
far more than by the various phase combinations that may form different 
steam cycles because these latter are so few. 

Gas cycles are not only more numerous even when the limit of practicability 
is imposed than vapor cycles, but the amount of work per pound of gas, depend- 
ing on the efficiency of the cycle and the amount of heat added to the gas per 
pound, is not limited by the physical properties of gases. Gases may be 
caused to take up as much heat as is convenient to give them, but steam may 
not carry more than a definite amount fixed by the safe and economical high 
pressure. There is, however, a practical limit to the heat per pound of gas, 
at least at present, which curiously enough is of about the same order of magni- 
tude as for steam. Gases may receive their heat from an external source, 
in which case the engines embodying the cycle are termed external combus- 
tion engines and in this case the amount of heat a pound of gas may receive 
is limited by the temperature to which it is safe to heat metals, somewhere 
about 1100° F. as a high limit and the temperature of cooling water as a low 
limit, these limits corresponding roughly to about 200 B.T.U. per pound. 
Thus, external combustion gas cycles may receive only about one-fifth the 
amount of heat per pound of medium as may steam, which is roundly 1000 
B.T.U. for ordinary conditions. This low limit is partly responsible for the 
abandonment of external combustion gas engines in favor of internal combustion, 
in which explosive mixtures are made with fuel and air, or in which air is caused 
to support non-explosive combustion of fuel in closed pressure chambers. 
With internal combustion as a mode of heating the heat per pound of working 
fluid will depend on the thermo-chemical reaction equations, which fix both 
the weight of air needed to burn a pound of fuel, and the heat of combustion 
per pound of fuel, which together fix the heat of combustion per pound of 
working gases. This for air-fuel mixtures is in round numbers about 1000 
B.T.U. per pound or about the same as for steam, but for oxygen-fuel mixtures 
it would be about five times as much. 

Gas cycles may receive, therefore, a definite amount of heat per pound, 
limited at present to about 1000 B.T.U. per pound of gases maximum, so that 
the problem is, to decide how the greatest part may be transformed into work 
or which of all the various cycles will give the most work when receiving this 
amount of heat. "This is not the only question, however, because one cycle will 
require larger volumes of gas to do the same work than another, requiring that 
a piston engine embodying this larger volume cycle be larger. Another cycle 
will require the gas to rise to a very high pressure to do the given amount of 
work and its engine will have to be very strong and heavy to resist these high 
pressures. Therefore, gas cycle efficiency must be studied, not alone, but in 
conjunction with corresponding pressure and volume ranges which are infinitely 



972 ENGINEERING THERMODYNAMICS 



•I 



more varied than for steam. The basis of comparison should in all gas cycles 
be the amount of heat received per pound of working gases rather than the 
pressures which impose the corresponding limit on steam cycles together 
with corresponding temperatures of boiling and condensation. 

With this preliminary survey of the relation of gas to steam cycle limi- 
tations, the next step is to examine some proposals and engine constructions 
embodying gas cycles to serve as a basis for judging practicability, as a pre- 
liminary to the selection of the cycles for analytical study. 

One of the first schemes for operating a gas engine was employed by Brown, 
who was no doubt influenced by the plan of operation used in the very old 
steam engines, known as atmospheric engines. These old steam engines 
drew into the cylinder a charge of steam at atmospheric pressure and then 
injected water to condense it, the resulting vacuum allowing the superior 
atmospheric pressure to perform the working return stroke. Brown paralleled 
this by burning gas in a large flame at his suction port, filling the cylinder 
with hot gases of combustion during the out stroke, at atmospheric pressure. 
He then injected water which more or less suddenly chilled the gases, causing 
a pressure drop at constant volume and with this vacuum the return stroke 
began. This, by reason of the considerable residual volume of the cooled 
gases, was necessarily a compression stroke and by reason of the slowness 
of operation, large weights of metal and of injected water, was approximately 
an isothermal compression, the non-gaseous bodies taking up the heats of 
compression of the gases as fast as liberated. Thus, the cycle of the Brown 
engine pretty closely approximates, (a) heating at constant pressure; (6) cooling 
at constant volume to original temperature; (c) heat abstraction at constant 
temperature to original volume and pressure. There were many modifications 
of this, constituting other so called atmospheric cycles, but as all were essentially 
inefficient and required large displacement volumes to do only a little work 
as measured by low mean effective pressures, they were all abandoned, as 
were also the atmospheric steam engines in favor of others working at initial 
pressures above atmosphere. 

The next practical proposal embodied in an engine was that of Lenoir 
about fifty years ago and from which modern gas engines may be said to date, 
so that steam engines of the practical sort are about one hundred years older 
than gas engines of the modern kind. Lenoir drew into his cylinder an explo- 
sive gaseous mixture and at about one-third stroke exploded it by an electric 
spark. This caused a rapid rise of pressure, ideally a constant-volume heating, 
and was followed by adiabatic expansion for the rest of the stroke. If too much 
mixture had not been taken in, this expansion might be complete and end 
at atmosphere at the end of the stroke, otherwise there would be a terminal 
drop at release. Neglecting the charging stroke that ideally neither does 
or uses up work, the cycle would be, (a) constant-volume heating, beginning 
at atmosphere; (6) adiabatic expansion to atmosphere; (c) constant-pressure 
cooling to original volume, this last being the thermal equivalent of exhaust 
as for steam. 



HEAT AND WORK 973 

This same series of operations was embodied in the Otto and Langen free- 
piston engines that succeeded the Lenoir, but with one modification. The 
piston engaged the shaft by a clutch on the down stroke only, charging, 
explosion and expansion occupying the up stroke, the cylinder being vertical, 
single acting and open upward. This construction allowed the piston to rise 
as a free projectile till stopped by the vacuum created in the cylinder behind 
it by overexpansion. Returning slowly by its own weight, the low-pressure 
gases in the water-jacketed cylinder were compressed nearly isothermally 
to atmosphere. Hence the Lenoir cycle is here extended below atmosphere 
adiabatically with an isothermal return. Both these cycles and others of the 
same general sort are termed non-compression cycles because the gases are 
not compressed before heating, and are inefficient compared to others that 
have preliminary compression, as will be shown, and this is why they were 
abandoned. They did, however, demonstrate the superior advantages of 
internal combustion over external, about the same time that external com- 
bustion engines had demonstrated the superiority of compression over non- 
compression cycles, though they were not purely such, as some of their heal) 
was exchanged during compression. 

These external combustion cycles, involving precompression, are best 
illustrated by the Stirling and Ericsson engines. The Stirling used an enclosed 
mass of air so arranged that it could be brought in contact with a hot part 
of the chamber and a cold part alternately, and likewise caused to pass through 
a regenerator or large mass of porous or divided solid matter like wire gauze 
packed together. In passing through this regenerator the air gave up heat 
if hotter itself, which was the case when it flowed from the hot to the cool 
chamber in one direction, and took up heat when it was cooler on the return. 
The various parts of the mechanism were so arranged that the flow through 
the regenerator took place at constant gas volume, which could be accom- 
plished by two pistons moving synchronously in opposite directions, the regen- 
erator being between the cylinders. Expansion is accomplished by one piston 
moving out faster than the other and compression inversely. During com- 
pression the gas is in the cold cylinder which keeps its temperature about 
constant; ideally it would be so, and after isothermal compression the air transfer 
takes place through the regenerator which adds heat to the gas at constant 
volume. Expansion now takes place in the hot cylinder, the air absorbing heat 
from the fire, ideally at constant temperature, and this is followed by a 
constant-volume return flow through the regenerator, during which the gas 
is cooled to its original temperature and the regenerator is heated. The cycle 
is, therefore, (a) constant-temperature compression; (6) heating at constant 
volume; (c) expansion at constant temperature; (d) cooling at constant volume. 
The regenerator heat is, however, algebraically equal to zero at the limit 
and the only heat derived from the fire is that for isothermal expansion. 

Ericsson arranged his pistons differently so far as their relative motions 
are concerned and secured two constant-pressure phases between the two 
isothermals, by making one piston move out faster than the other during 



974 ENGINEEEING THERMODYNAMICS 

heating and in one form of engine there was substituted a transfer piston that 
did no work at all, leaving but one piston, acted on by gas pressures. The 
Ericsson cycle was ideally, (a) constant-temperature compression; (6) con- 
stant-pressure heating; (c) constant-temperature expansion; (d) constant- 
pressure cooling. 

Embodying the advantages of internal combustion, which are, (a) more 
heat per pound of gases, and (6) faster working because of the removal of con- 
duction limitations to heat absorption by the gas, and in addition introducing 
the fundamentally necessary precompression before heating and after heating, 
utilizing the strong transforming power of adiabatic expansion, Otto embodied 
in his engine a new cycle which has come to be the standard of nearly all 
modern engines. This Otto cycle is, (a) adiabatic compression; (6) explosion 
or heating at constant volume; (c) adiabatic expansion to the original volume; 
(d) exhaust at constant volume which is equivalent thermally to constant- 
volume cooling to original pressure. This was modified by Atkinson, who 
expanded to. more than the original volume, and various attempts at com- 
pounding were made subsequently to carry the expansion to atmospheric 
pressure instead of limiting it to the original volume before compression as 
did Otto. r j 

Contemporaneously with Otto in Germany, Brayton developed an engine 
in America, using non-explosive but internal combustion which operated on 
another cycle, that bore to that of Otto the same relation that the Ericsson 
did to the Stirling. Brayton substituted constant pressure for Otto's constant- 
volume phases between the two adiabatics. His cycle, also known by other 
names such as Joule, consists of, (a) adiabatic compression; (b) constant-pressure 
heating by internal combustion; (c) adiabatic expansion to atmosphere if 
the cylinder is big enough for the amount of heating employed; (d) constant- 
pressure cooling as the thermal equivalent of atmospheric exhaust. Later 
Diesel utilized the same cycle in a single cylinder, Brayton having used two, 
which structural change necessitated a modification of cycle, as the expansion 
in one cylinder can proceed only to the original volume, and the Diesel cycle 
becomes, (a) adiabatic compression; (b) constant-pressure heating by internal 
combustion; (c) adiabatic expansion to original volume; (d) constant-volume 
cooling as the equivalent of exhaust, to atmospheric pressure. 

Following the success of the steam turbine, many proposals have come 
forward for the operation of gas turbines involving, (a) compression of air 
adiabatically and delivery into and through a combustion chamber for, 
(b) heating at constant-pressure followed by, (c) adiabatic expansion in 
turbine nozzles instead of in cylinders, to atmospheric pressure and which 
must be followed to complete the thermal cycle by, (d) constant-pressure 
cooling at atmosphere, which is, of course, the Brayton cycle as used in 
cylinders. 

This review while decidedly incomplete will serve to indicate the desirability 
of selecting for study the following types of gas cycles: 



HEAT AND WORK 975 

(1) An atmospheric cycle, to show its essential inefficiency and low mean 

effective pressure 

(2) Non-compression cycles, as a basis of comparison with compression cycles 

to prove the value of precompression. 

(3) The external combustion compression cycles, to demonstrate the value of 

precompression for gas cycles, to illustrate the action of a regenerator 
and as a basis of comparison of isothermal with adiabatic compression 
cycles. 

(4) Constant-volume heating, adiabatic precompression cycles, as typical of 

most modern engines and supplying a standard of reference for them. 

(5) Constant-pressure heating, adiabatic precompression cycles, as typical 

of the rest of the modern engines and of the gas turbine proposal, and 
supplying a standard of reference for them. 

(6) Constant-temperature heating, adiabatic precompression cycle, to complete 

the list of possibilities and demonstrate a most valuable, general law, 
viz., that the efficiency of all cycles of four phases in which the com- 
pression and expansion lines follow the same law and are included 
between a pair of similar heating and cooling lines is the same, and 
equal to that of the general, Carnot, cycle, the standard for both gas 
and vapor cycles. 

In the diagram, Fig. 249, these selected cycles are illustrated to PV and 
T$ coordinates. The PV diagram is of greater value than the T$ in this analysis 
of gas cycles, each point of which must be calculated, before the heats, work 
and efficiency, the volume and pressure range with quantities derived therefrom, 
may be determined. In order that there shall be some proportionality between 
these cyclic diagrams, so necessary for a fair visual comparison, they are all 
plotted to the same scale on the following assumptions of initial condition, 
one atmosphere pressure, absolute temperature 492° F. app. = 32° F., volume 
12.38 cu.ft. per pound, that for air. To the pound of gas in each cycle is added 
1000 B.T.U. after a compression of 7 atmospheres absolute for compression 
cycles, except for the two external combustion cases, which are allowed 200 
B.T.U. and 3 atmospheres compression absolute. Another modification or 
exception necessary is that of Cycle VIII, which has isothermal heating followed 
by adiabatic expansion. With 1000 B.T.U. added isothermally at 10 atmos- 
pheres pressure initially, the pressure would fall below atmosphere before 
adiabatic expansion could begin, and the final volume would become several 
thousand cubic feet. This cycle has accordingly been modified by allowing 
only 200 B.T.U. per pound. 

In all cases the specific heat of the gases is taken as constant and the 
working fluid is assumed to be air; any other procedure leads to unmanageable 
formulas, and errors involved in the practice can be allowed for at the same 
time as errors from other sources, in one or two inclusive correction factors 
when it is necessary to pass from these ideal to real engine conditions. 



976 



ENGINEEKING THERMODYNAMICS 



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Fig. 249.— Typical Gas Cycles. 



HEAT AND WOEK 977 

The cycles illustrated are described as follows: 

Brown Atmospheric. Cycle I. 

First phase from A to B. Heating at constant atmospheric pressure. 
Second phase from B to C. Cooling at constant volume. 
Third phase from C to A. Isothermal compression. 

Lenoir. Cycle II. 

First phase from B to C. Heating at constant volume. 
Second phase from C to D. Adiabatic expansion. 
Third phase from D to B. Cooling at constant pressure. 

Otto and Langen. Cycle III. 

First phase from B to C. -Heating at constant volume. 
Second phase from C to D. Adiabatic expansion. 
Third phase from D to B. Isothermal compression. 

Stirling. Cycle IV. 

First phase from A to B. Isothermal compression. 
Second phase from B to C. Heating at constant volume. 
Third phase from C to D. Isothermal expansion. 
Fourth phase from D to A. Cooling at constant volume. 

Ericsson. Cycle V. 

First phase from A to B. Isothermal compression. 
Second phase fron B to C. Heating at constant pressure. 
Third phase from C to D. Isothermal expansion. 
Fourth phase from D to A. Cooling at constant pressure. 

Otto. Cycle VI. 

First phase from A to B. Adiabatic compression. 
Second phase from B to C. Heating at constant volume. 
Third phase from C to D. Adiabatic expansion. 
Fourth phase from D to A. Cooling at constant volume. 

Brayton. Cycle VII. 

First phase from A to B. Adiabatic compression. 
Second phase from B to C. Heating at constant pressure. 
Third phase from C to D. Adiabatic expansion. 
Fourth phase from D to A. Cooling at constant pressure. 

Carnot. Cycle VIII. 

First phase from A to B. Adiabatic compression. 
Second phase from B to C. Isothermal expansion. 
Third phase from C to D. Adiabatic expansion. 
Fourth phase from D to A. Isothermal compression. 



97B ENGINEERING THERMODYNAMICS 

Prob. 1. Modify the Brown atmospheric Cycle I, by substituting adiabatic for 
isothermal compression and plot between the same temperature limits. 

Prob. 2. Plot the modification of the Lenoir Cycle II, introduced by assuming 
constant-pressure heating to follow constant volume, each equally sharing the whole 
heat added. 

Prob. 3. Modify the Otto and Langen Cycle III by substituting adiabatic compres- 
sion and isothermal expansion. 

Prob. 4. Plot to scale a modification of the Stirling Cycle IV, due to substitution 
of adiabatic compression and complete isothermal expansion to original pressure. 

Prob. 5. Modify the Ericsson Cycle V by substituting adiabatic compression to 
maximum temperature. 

Prob. 6. If the Otto-Atkinson Cycle VI were executed in two cylinders, one carrying 
out the Otto part and the other the rest of the extended expansion, plot their indicator 
diagrams to the same stroke, base, and correct pressure scale. 

Prob. 7. Modify the Otto cycle for two-stage adiabatic compression with perfect 
intercooling and both with and without final perfect cooling. 

Prob. 8. Assume the Brayton Cycle VII to be carried out in four cylinders, two 
belonging to a two-stage adiabatic compressor with perfect intercooling and two 
to a compound steam type of engine with perfect reheating, each without clearance, 
and plot the cycle. 

Prob. 9. For all cylinders having 5 per cent clearance each, plot to the same stroke 
base the indicator cards using correct pressure scales. 

Prob. 10. Plot a combination of Diesel and Carnot cycles in which after adiabatic 
compression half the total heat added is received at constant pressure and the other 
half at constant temperature. * 

9. Non-compression Gas Cycles, Brown, Lenoir, Otto and Langen. Work, 
Mean Effective Pressure, Volume and Pressure Ranges, Efficiency, Heat and 
Gas Consumption. 

Brown. Cycle I 

Let Qi = heat supplied in B.T.U. per pound gas; 
11 $2 = heat abstracted in B.T.U. per pound gas; 
1 1 C p and C v = specific heats of gas at constant pressure and volume and 

assumed constant; 
" r v and r P = volume and pressure ranges respectively = (maximum) — 

(minimum) ; 
11 H = B.T.U. per cubic foot gas as supplied. 
Then referring to the diagram of the cycle, Fig. 250, 

W=J(Qi-Q 2 ), -." - (1022) 

and 

E=l-% ..'.."..- (1023) 

Qi = C p (T b -T a ), o o . . (1024) 

Q 2 = C v (T b -T c )+jP a V a \0geY' • • • ° • ( 1025 ) 



HEAT AND WOEK 



979 



Evaluation of these heats and determination of all derived quantities requires 
first the finding of the pressures, volumes and temperatures at each point 
in terms of the initial conditions, P a , V a , T a , and Qi, as imposed. 

Point A : 



P b = P G 



Qi 



T b =T a +^ = T a Y, 



Qi 



v,=v a f=v a [\^^\ 



(a) 
(&) 
(c) 



(1026) 



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Entropy 

Fig. 250— Brown Gas Cycle. Cycle I. 



.6 



The point C is located, at the intersection of a constant volume line from B 
and an isothermal from A, hence P a V a = P C V C = P C V^ Hence for 



Point C: 



v c =v b =v a 1+ 



-/ c — J- ay 
c Vc " . . Ql F 



(&) 

(c) 



1+ 



Oi 



(1027) 



980 



ENGINEERING THERMODYNAMICS 



Knowing thus the coordinates of all points in terms of the initial conditions 
and heat supplied, any desired quantities can be set down at once. Thus, 



-i<h + § r . 108,(1+^ 



Also 



W = J{Qi-Q 2 )=JQ 1 { 1~) -RT. log, 



*- 1 -S- 1 -r^ k *( 1 - + ^v)''--- ■ • • (1030) 






rp^Pa-P^P, 



1- 



1+ 



Qi 






. . (1031) 



(1032) 



T^=4[l-^log e (l-c|r)]. • • • 0033) 



144tv 144 



Heat consumption, B.T.U. per hour per I.H.P. 



2545 
E 3 



Gas consumption, Cu.ft. gas per hour per I.H.P. = 



2545 
EH' 



. . (1034) 
. . (1035) 



Lenoir. Cycle II 



Referring to the diagram of the cycle BCD, Fig. 251, the points are to 
be evaluated as before. 

Case a. Complete expansion to atmospheric pressure.^ 



Point C: 



V c =Vi 



Q] 



T c =T b +f = T b X 



(a)} 
'(b) 



Qi 



P c = P b ^ = P b l l+^r ) =P b X. (c) 



. . (1036) 



HEAT AND WOEK 



981 



in which 



X=l + 



C v T b 



Point D: 



Pd — Pb, 

V d = V c (j^ y =V b Xy 



}--l 



(a) | 
(b) 

(c) 



r • 



(1037) 



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S 7-5 
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t k 


^ 






t i - ~t 


A 


J 


L 


A 


i 


tiz 




V 


\ 


V 


V 


"s 6' 






p j j 1 p 1 "— ' 3 


X M.I 1 ,' 



45 60 

Volume in Ou. Ft.. 



T 


c 




j 




7 








T 




i- 




t 




J 




7 


.a 


f 


< 


7 - 


u 


j_ _ 


,a 






7 - iV 


w 


/ < D 




\ j T 


u 






/ : /: d 






a 


7 ' / 


"* 


A - _// 


Q, 


A- / / 


s 


/ //' 


















yS\ t J^' 


1 


1000 ^^^_J 


J, 






492 gf- 




b r 




j _-_ 


_ _-. 



75 y 



Entropy 



4 <p 



Fig, 251. — Lenoir Gas Cycle. Cycle II. 



Q2 = C p (T d -T b )=C p T b (Xy-l). 



(1038) 



W=J(Qi-Q 2 )=J[Qi-C p T b (Xy-l)].. . . (1039) 



Q: 



Cpi b 



E = l-^ = l-^(Xy-l), ...... (1040) 



Qj 







r r =»F«-7,= 7i(X'-l). 
r,-P.-ft-ft(Z-l). 



(1041) 
(1042) 



982 



ENGINEERING THERMODYNAMICS 



(m.e.p.) 

B.T.U. per hr. per I.H.P. = 
Cu.ft. gas per hr. per I.H.P. = 



W = J[Q!-C P T b (Xy -1)] 

144r F 1 

1447 6 (X"r-l) 



2545 
E ' 

2545 
EH' 



> (1043) 



(1044) 
(1045) 



Case b. Incomplete expansion with constant-volume cooling to atmosphere 
Fig. 251, BCD'E. 

Up to the point C this cycle is the same as the last, but noints D' and E 
are fixed by the limited expansion volumes available. 



Ye 

Let Ry = ratio of final to initial volume, or — = R v . 

Vb 



Point B'\ 


V d , = V b R v , (a) 




Pd " Ic \V d ) ~Rv*- RS {h) 




w-$y*2?-* <■> 


Points: 


V e =V d , = V b R v , (a) 




P e = P b , (b) 



Whence 



T e = T b -=r = TbRy, (c) 

Vb 



Q2 = C.(T d ,-Te)+C p (T e -T b ) 



^A^Ty-l 



(1046) 



(1047) 



(1048) 



W 



=J r (Qi-Q 2 )=j[Qi-CX^(^- 7 -l)-C 2 ,n(^-l)] (1049) 



^ = 1 -| = 1 -^(^- 1 )-¥^- 1) - ' ' (1 ° 50) 



r v =V e -V b = V b {R v -l), 



(1051) 



HEAT AND WORK 
r P = P c -P b =P b (X-l) 



(m.e.p.) = 



144n 



j\Qi-C v T^ v (^- y -l^-C P T b (R v -l)\ 



B.T.U.per hr.l ^2545 
per I.H.P. I " : E ' 



fCu.ft. gas perl _2545 
ihr. per I.H.P. I ~ EH ' 



983 
(1052) 

(1053) 
(1054) 
(1055) 



Otto and Langen. Cycle III 

Case a. Complete expansion to initial temperature. Fig. 252, BCDB. 

The point C will be as located for the two previous cycles, but point D. is 
at the intersection of an adiabatic through C and an isothermal through B, 
therefore, 

7, 



and 



Hence 



Point D: 



PaV d = P b V b , or P d = P b jf, 

V d 



PaV a y = P c V c y , or P d = P c (J^ y 



p b \vj ' 



and V d =V t 



(ft)""' 



T d = T t 



d— J- D 



(a)} 



V d =VX l (b) 



Pd = 



l 



to 



(1056) 



Q 2 = jP>V b log, p? = -p log e (Z)-v -i = C v T b log e X, . (1057) 



W=J(Qi-Q2)=J(Qi-C v T b \o% e X), . ' . 



(1058) 



984 



ENGINEERING THERMODYNAMICS 



E=1 _Q2 = 1 _C v T b log e X 



Q] 



Oi 



(1059) 



rF= 7 d -F 6 = F & (XT-i-l) ; 



(1060) 



3 6 



*^ 



r P = P-P d = P b X ^- = PJX V), • • (1061) 

Xv- 1 \ Xy- 1 



{m . e . v ,w m= cM S jc lt 



144/y 



1447 6 (Z^" 1 -1) 
T 

6000 

^ 5000 



^ 4000 



3000 



| 2000 



1000 



50 100 150 200 250 300 V 

Volume in Cu. Ft. 



V=7118 492 

-Pi . 





(1062) 







































Cj 


















































































































































































— 


~ 


D' 










V' 


v 










B^ 






** 












D 
1 





















.10 .20 .30 .40 <p 
Entropy 



Fig. 252.— Otto and Langen Gas Cycle. Cycle III. 



B.T.U. per hr. per LH.P.= 



2545 

E ' 



(1063) 



/ Cu.ft. gas per hr. \ 2545 „™^ 

{ per I H. P. } =-M> (1064) 



Case b. Expansion to temperature lower than that at atmospheric pressure but 
higher than the initial, with constant-pressure cooling to complete the cycle BCD'EB, 
Fig. 252. 

To locate the point E, which may be anywhere between B and the expan- 
sion line, it is most convenient to imagine the expansion proceeding to some 

p 
fraction of the initial one atmosphere pressure, accordingly let m = — , so that 

if expansion continues to half an atmosphere, m=2, 



HEAT AND WORK 



985 



Point D' 



id'- — , 

m 



(a) 



Point E: 



V*=V.(j£) y =V t (Xm)y, (6) 

7-1 j 



whence 



Pe = Pt 



T e =T^ n (Xm)y 

Pe 



(a) 
(6) 
(c) 



Q 2 =-fPeVc l0ge p^ + C^r,- n) 

RT b/ 



Jm 



(Xm)v loge (m)+C p T b 



(Xm)~ 



m 



W = J(Qi-Q 2 )=JQi 



RT b 

m 



(Xm)y log e (m) 






# = 1-^ = 1 



(Im)T log e (m) + 



L m J 

7 p n[(Xm)7 1 
Qi L m J' 



(1065) 



(1066) 



(1067) 



(1068) 

(1069) 
(1070) 



(m.e.p.) = 



rv = V d ,-V b = V b [(Xm)y -I], 

rp = Pc-P* = P*(x-^V (1071) 

] 



W 

1447y 



1 r JL 

JQi~^r(Xm) y log, (m)+JC p n[^^ 7 



m 

^_ 

1447 & [(Xm)^-l| 

[B.T.U. per 1 2545 
l hr.perl.H.P.J # > 

■Cu.ft.gafi i 2545 



, (1072) 
(1073) 



r^u.tt.gafi i 2545 /in ^, 

^ perhr. per^ = ^7T, (1074) 

I LH.P. J ^# 



986 



ENGINEERING THERMODYNAMICS 



ZT H 


| | ■ 


L 


~L X 


Y 


t - 1 X 


T 


t J 


T 


4 ' 8 


j 


4 X - 


\ 




P 


I 1 


Y 


t F- 


T 




[J 


1 F 1 


\ 




£4 


-j£t £4 


bg\ 


©1 isX 


%\o 


v^ \ CO rt \<c S 


-jst 


\ 1 ^ 


jftt 


»-m\ --hi 


& 


IK 41 


J^ 


-IV 6 


c 


SV- - L -^-8 


5 


X- t g 


*>^r- 


X t 


^\ • 1 


It 4 


sSL 


+3 4 


t^ 


t L j X 8 


+v 


I \ I 8 


v it X 


I V t 


s ± 


i A 4 




1 -\ A 


v 


4 ■ -S V- 8 


^v 


I \ t * 


*v 


4 \ A 


\ 


1 \ ' 


^s^ 


T \ A 


^^ 






! "^V \ ®* 




""^-i ^ ^ 


• X 




;uao J9d ut jJouaprga iBuuatfj 


' ' ^^--^ <=>"<S, 


S tr- §5 iO 


s o! & § -- — £ — _:v^ 


g= — f— = C- ^ 


-8- ° 8 8 


§ — — §z:z:~— = ii==:= 


^^ = --qi_— — — S:^:;-^.^ ©J - 8j— g 


t --. "~ 


— . -X""~ — m^"^^ 5 ^^^^ ^ 


T ~~" ■—— — - 


1 """"""^^ " s= ^*^=,^ s ' s ^ s;; s'n. <=> 


Oh — 


" = — «■ J ~^^. ^^v,\ 0\_ ^ ** 


w ~"" B= ^--. 


""-^ *■*":*. ^V. ^"^S^-X^J 


a3 "^ — » 


^^»^ ^^ i\$\\^\i> 




V^ N v fy\ ^KA^V o 


o *^'"~' > ^- 


^H *5\. n\. ** v* V \ \ 


w ^^ 


\ *k ^XI^Wii 






"Pi ^"^ 


§k ^ A4U - 


p 


-4>^ -&± ^-Ti^ffl 


H 


4^k J&^js *v&\l 


■pq- 


J&, ?A ^Y ffflfl 




% -$*__ S 44H st 




% #V t-4-K 




^ 3 tul 




$v f 4 tffl 




4&*- 4 4 tfflts* 




A > 1 s < 




■ \ \ I - 




3X 4 t it 




V 4 4 lts f 




A 4 4 jflt rt - 




XV i-4 it ; 




X t L J 








i \ 1 "- 1 




V 4 1 il 




4 L-lh 




i 1 s 








4 ti ,: 




\ 111 



o ^ 



HEAT AND WORK 



987 




988 ENGINEERING THERMODYNAMICS 

The curves, Figs. 253 and 254, give efficiencies and mean effective pressures, 
and show the relation between the important quantities for the complete 
cycle in each case. They prove the inferiority of the Brown atmospheric 
cycle compared to the other two by reason, not only of its low efficiency, but 
also its small mean effective pressure, which requires large cylinders. For 
efficiency the Otto andLangen complete expansion cycle is superior to the Lenoir, 
but its mean effective pressure is very much too low to compete. Its should 
be noted that while the mean effective pressure of such cycles as the Lenoir 
is the work divided by the volume range, the mean effective pressure to be 
expected from the execution of even the perfect cycle in cylinders must be the 
work divided by the whole or maximum volume, because that part of the 
displacement up to explosion is used in getting the charge into the cylinder, 
so that the cylinder mean effective pressures will be smaller than those for 
the cycles in the ratio of volume range to maximum volume. Comparison 
of the performance of all these non-compression cycles will show them to be 
beneath consideration in comparison with compression cycles, to be analyzed 
next, in both efficiency and mean effective pressure. All these non-compression 
cycles show an increase of efficiency with heat supplied, so that for small heats 
supplied which would correspond to weak explosive mixtures or much excess 
air in any case of internal combustion, the efficiency would be less than for no 
excess air. This also is in striking contrast to the compression cycles, the 
efficiency of which is not only better but independent of the amount of heat 
supplied. 

Referring to the efficiencies, Fig. 253, all three cycles show rising efficiency 
with increase of heat added, the Brown being lowest with the value of 20.8 
per cent for 1000 B.T.U. per pound of gases, with Lenoir at 38.4 per cent and 
Otto and Langen at 78.4 per cent for the same heat supplied. For 200 B.T.U. 
per pound these three cyclic efficiencies fall off considerably, having the values 
10.8 per cent, 18 per cent, and 48.4 per cent. These are all, however, high 
values and stimulate interest in what will be attainable in cycles with compression, 
which, it was pointed out, promise still higher lesults. It must be remembered, 
however, that in executing these gas cycles in engines realization has fallen 
far short of cyclic promise, due to excessive heat losses in the mechanism itself. 

Example 1. Calculation of Diagram of Brown gas cycle, Cycle I, Fig. 250. 
P a = l atmosphere. 
V a = 12.39 cu.ft. (one lb. of pir). 
T a =492°F. absolute. 
Qi = 1000 B.T.U. added per pound of air. 

Pb=Pa = l atm. 



Assumed data ■ 



Point B 



T b = T a +Q-= 492 +^ = 4766° F. absolute. 
n = F^=^Xl2.39=117.8 cu.ft. 
- <k, = C, log, p = .239 log. ~ = .538. 



HEAT AND WORK 



989 



Point C 



V c = F 6 = 117.8 cu.ft. 

T c = T a =492° F. absolute. 

_ PaV a 1X12.39 

P c =-rr-= =.105 atm. 



F c 117.? 

T 
<!>& - 4>c = C log c ^ = .17 log, 

-* c 

cj> c -4>a = .538-.383=.155. 



4676 



= .383. 



Example 2. Calculation of Diagram of Lenoir gas cycle, Cycle II, Fig. 251. Data 
assumed as per Example 1. 



Point C 



7 C== 7 6 = 12.39 cu.ft. 

7^ = ^+^=492 +^=6374° F. absolute. 

T 6S74 
P - P -n = l92" = 12 - 96atm - 

d> c - d» = C P log, |- c = .17 log e ^ = .436. 



Point D 



Pd=Pb=l atm. 

/P V 713 
Va = Ve \p) =12.39(12.96) 713 =77 cu.ft. 



T d = T c 



6374 



J-) 

12.96/ 



.286 



3056° F. absolute. 



.436. 



Point D' 



F d , =45 cu.ft. (assumed) 



Pa,=Pc 



= 12.96 



r,-r.^ =6374 

- 4>& = cj> c — 4>& = .436. 



12.39 
45 

12.39 



= 2.13 atm. 



=3803° F. absolute. 



Point E 



V e = V d , =45 cuit. 
P e =P b =l atm. 

n = TV C^ =3803 X^ = !785° F. absolute. 
- <fe = C„ log, ^ = .239 log e ^ = .308. 



990 



ENGINEERING THERMODYNAMICS 



Example 3. Calculation of Diagram of Otto and Langen gas cycle, Cycle III, Fig. 252. 
Data assumed as before in Example 1 and point C same as for Lenoir cycle of Example 2: 



Point D 



Point E 



Point D' 



T d = n=492°F. absolute 
V d 

V b 12.39 



•.sr-HSfr- 



7118 cuit. 



Pd ~ Pb V d 7118 

■ §d — §b — §c — 4>& = .436, 



.0017 atm. 



7 C =45 cu.ft. (assumed). 
P e =P 6 = latm. 



T e = T b -=mx 



45 



12.39 



= 1786° F. abs. 



T 1786 

4* - <fe = C„ log c ^ = .239 log, — = .30' 



T d , = T e = 1786° F. absolute. 
*-T-®"->- (if 



=291.7 cuit. 



-P<Z' =Pe TT - 



45 



7* 291.7 
4>c — $& = .436, 



.154 atm. 



Example 4. Calculation and use of Diagram, Fig. 253, giving for the non-compression 
gas cycles Brown, Lenoir, and Otto and Langen, the thermal efficiency, heat, and gas con- 
sumption as a function of heat supplied. One point will be calculated for each, A for 
Brown, B for Lenoir, and C for Otto and Langen, all for 1000 B.T.U. added per pound 
of working gases assumed to be air, initially at one atmosphere pressure, and 32° F. 



FromEq. (1030), E a = l 



1 

Y 


RT a . 


(■ 


1 


53.3 X492 



1.4 778X1000 



lo&( 



1 + 



1000 \ 



239X492 



) =21.1 per cent. 



Fro mEq .(1040), Ss = l^[( 1+ ^) 7 -l] 

, .239X492IV 1000 V 713 ,1 „„ „ 
- 1 — l000-K 1+ I^i92J - 1 ] =38.3 per cent. 



HEAT AND WORK 



991 



C v Th 



From Eq. (1059), E c = 1 - ~ log, 1 + 
Hi \ \j 






1 .17X492. / , 1000 \ noa 
= 1 -^00^ 1Oge ( 1+ l7^2) = 7 ™^^' 



The location of one point D in the left-hand angle will serve to illustrate all the 
rest. From Eq. (1035) for the' Brown cycle, and assuming the fuel gas to have 200 
B.T.U. per cubic foot: 

Cubic feet fuel gas per hour per H.P. = -^ttt = — — ^rz =42.41 cu.ft. 



EH .30x200 



To illustrate the use of the diagram find for a Lenoir cycle receiving 800 B.T.U. 
per pound of working gases the thermal efficiency, heat consumption, and cubic feet of 
300 B.T.U. per cubic foot fuel gas per hour per I.H.P. From the 800 point E pass 
vertically to point F on the Lenoir curve and thence horizontally to G on the efficiency 
scale, reading 35.2 per cent and heat consumption, 7250 B.T.U. per hour per I.H.P. 
Passing across to the 300 B.T.U. calorific power curve at H and down to K, the gas 
consumption is found to be 24 cu.ft. per hour per I.H.P. 

Example 5. Calculation and use of the Diagram, Fig. 254, giving for the non-compression 
gas cycles Brown, Lenoir, and Otto and Langen, the work per pound of gases and mean 
effective pressure. Three points will be calculated: A, for Brown; B, for Lenoir; and C, 
for Otto and Langen, all for the initial conditions of Example 4, and 800 B.T.U. supplied. 

From Eq. (1029), 



l-^-222 T alog e (l+^r 



=778 X800 X(l -.713) -53.3 X492 log e 
From Eq. (1039), 
W^J\Q x -C p T b 



(- 



800 
239 X492 



= 124,400 ft.-lbs. 



=778(800 -.239X492) 
From Eq. (1058), 



>+,#-] 



('♦iSs)'"' '-']-».«»«- 



lbs. 



Wc=j\Qi-C v T b \og e (l+-^ 

= 778[800-.17x4921og/l+Y 7 ^^) =469,000 ft.-lbs.] 



992 ENGINEERING THERMODYNAMICS 

In the left-hand angle the point D representing by its abscissa the mean effective 
pressure for the Brown cycle is located by Eq. (1033). T aking a point for 100,000 ft.-lbs. 
work done, 

(m.e.p), =~ = 7 lXr) = ^^" — 675 = 9 - 75 lbs - P er sc L uare inch ' 

l ^ Tv 144/(l-^)g 144x778(1-713)^ 

The corresponding, value for the Lenoir cycle point E is located by Eq. (1043) : 

W W 

(m.e.p.) c = — = — 



UAVt 



W->] 



' = 19.68 lbs. per square inch. 



» 4txW [( 1+ jS§) ,B - 1 ] 

For the Otto and Langen the mean effective pressure for the same cyclic work of 
100,000 ft.-lbs. is represented by point F and given by Eq. (1062) : 



W 
(m.e.p.)/= — 



1447. 



(-<sr-j 



144X12.39 



100000 =1.855 lbs. per inch. 



247 V 248 ] 
+ .17X492/ J 



Illustrating the use of diagram, Fig. 254, the solution of the following problem is 
offered: Find the work per pound of working gases and the mean effective pressure for 
an Otto and Langen cycle receiving 500 B.T.U. per pound of gases. Starting at the 
500 B.T.U. point G, pass up to the cycle curve at H and then across to the point K 
on the work scale, reading 260,000 ft.-lbs. Passing horizontally across the point L and 
thence downward to point M the mean effective pressure is found to be 1.18 lbs. per 

square inch. 

Prob. 1. Find the temperature at which expansion ends for (a) the Brown and (b) 
the Lenoir cycles, receiving 300 B.T.U. per pound of gases, at an initial temperature 

of 160° F. ■ ■ 

Prob. 2. How much heat must be abstracted per pound of gases of Problem 1, and 

what is the corresponding thermal efficiency? 

Prob. 3. Compare the maximum and minimum pressures for the Brown, Lenoir, 

and Otto and Langen cycles receiving 700 B.T.U. per pound from an initial temperature 

of 200° F. 



HEAT AND WORK 993 

Prob. 4. What effect will high altitude where the atmospheric pressure is .7 the stand- 
ard have on the thermal efficiency, work per cycle, and mean effective pressure for these 
three cycles? 

Prob. 5. If a Lenoir cycle is to be carried out in a cylinder, with 500 B.T.U. per 
pound of working gases, and realizing 30 per cent of the cyclic efficiency, all heat 
losses taking place during the rise of pressure, what thermal efficiency may be 
expected and what mean effective pressure? 

Prob. 6. Compare the thermal efficiency of a perfectly executed Otto and Langen 
cycle for 400 B.T.U. supplied per pound of gases with another in which expansion 
proceeds only to half an atmosphere. 

Prob. 7. With the same data, Problem 6, make a comparison of mean effective 
pressures. 

Prob. 8. Compare efficiencies of two Lenoir cycles, one for perfect expansion and 
the other expanding to half the maximum pressure, in each case for 200 B.T.U. supplied. 

Prob. 9. With the same data of Problem 8 compare the mean effective pressures. 

Prob. 10. If a Lenoir engine used 40 cu.ft. of 600 B.T.U. gas per hour per I.H.P. 
requiring seven parts of air to burn it by weight, what fraction of its cyclic efficiency 
was realized? 

10. Stirling and Ericsson Cycles. Work, Efficiency and Derived Quantities 
for Isothermal Compression Gas Cycles, with and without Regenerators. 

Stirling Cycle IV 

There are two cases to be considered for this cycle, Fig. 255, first, that dealing 
with the heats involved as a whole, and second, that dealing with the heat de- 
rived from the source of supply, separately considered from the regenerator heat 
which is alternately added and abstracted and by hypothesis without loss. 
So far as the diagram points themselves are concerned, there is no difference 
between the two, and this is also the case for work and mean effective pressure. 

Let Qi =the whole heat supplied from B to C to D; 

Qi' = the heat supplied from the fire from C to D; 

Qi" = the heat supplied from regenerator from B to C; 

$2 = the whole heat abstracted from D to A to B: 

Q2 / = the heat abstracted by water jacket from A to B; 

Q2" = heat abstracted by regenerator from D to A ; 

p 
C = compression in atmospheres = ~ ; 

i£ = real thermal efficiency of whole cycle; 

E l = thermal efficiency of cycle referred to heat from source of supply 
(CtoD). 
Point B: 

T b = T a (a) 

P b = P a C (6) 

Vb=Va F b = 7? (c) 



(1075) 



994 



Point C: 



where 



Point D: 



ENGINEERING THERMODYNAMICS 



V c = V b = - 



(a) 



T c = T b +^ = T a (l+^Q=T a Z (6) 

T. 

to 



Pc = PbTFT = Pb% = PaCZ 
lb 



2 = 1+ 



Ql.'. 

Cpi a 



T d = T c =T a Z (a) 

F d = F« (6) 

P d = Pa^ = PaZ (C) 
■* a 






(1076) 






(1077) 







c 








I 


to" 


r 


3 


t 


05 


A 


0) 


\ 






3 




in 




1 




. H 




^^-- D 


B \ 


3. 



3000 



2500 



5? 2000 



1000 



500 





C £> 




£ X 




/ r 




7 7 




-Y £ 




7 A 




< t 


7 


r 


r 




2_ 




2 




/ ^ 




5^^ ^^ 




^ 









Volume Ln Cubic Feet 



10 12 V 



0.2 
Entropy 



Fig. 255.— Stirling Gas Cycle. Cycle IV. 



Q 2 = Q2' + Q 2 " = jPaFa loge |? + C.(!F<,- !F fl ) 



0,4 



= |7 7 a log e C+C„(^-7 7 a ). 



(1078) 



Oi = Qi / +Qi /, = TP^clo & ^+C f (!T < ,-!r ft ) 



R 



T d \og e C+C v (T d -Ta). 



(1079) 



HEAT AND WORK 995 

W = J(Q 1 -Q 2 )=R(T d -T a )\og e C. ...... (1080) 

E = 9 ^r = Y Z (1081) 

V1 jRT d \o&C+Cv(T d -T a ) 



E 



Ql ' jRT d log* C 



%r= l -Y d ^ 



This efficiency is therefore equal to the whole temperature range divided by 
the maximum temperature. 

r v =Va-V b =V a (l-l) ...... (1083) 

r P = P c -P a = P a (CZ-l) (1084) 

rv 144FJl-~ x 



B.T.U. perhr.perI.H.P. = ?P = P^ -. . (1086) 

Mi i d — l a 



As this cycle is executed on enclosed air and not by internal combustion 
of gas mixtures, the gas consumption has not the same significance as in 
the other cycles. If fuel is burned externally, the consumption per hour per H.P. 
will be measured only partly by the above value of heat consumption, which 
is only the heat that reaches the enclosed air, the rest of the generation escaping 
from the furnace, so that the other factor in fuel consumption will be a furnace 
efficiency which may be quite low and defined as the ratio of the heat appearing 
in the cycle along CD, Fig. 255, to the amount generated in the fire. While these 
engines had, as is indicated by Eq. (1082), a high cyclic efficiency, external com- 
bustion characteristics gave a higher fuel consumption than ought to corre- 
spond. This condition also applies to the following Ericsson cycle. 



996 



ENGINEERING THERMODYNAMICS 



Ericsson Cycle V 

This cycle when executed with regenerators alternately raising the tem- 
perature from B to C, Fig. 256, and lowering it from D to A, may, like the 
Stirling, have two efficiencies, one for the cycle regardless of the source of the 
heat, and the other considering only the heat received from the fire source. 



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The position of point B is the same as for the Stirling, and its coordinates 
need not be set down. 



Point C : 



where 



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. (1088) 



HEAT AND WORK 997 

Q2 = Q2 f + Q2" = - T PaV a log* ^ + C p ( T d ~ T a ) 
J V b 

= jT a \ogeC+C p (T d -Ta) .'. (1089) 



Qi = Qi+Qi" = -jT d loge C+C P (T d -T a ) . ..... (1090) 

W = J(Q 1 -Q 2 )=R(T d -T a )\og c C . . . . (1091) 

E= Qi^Q2 = J (1092) 



E f = q ^ = T ^ La = l-^ (1093) 



*). ...... (1094), 



r v =V d -V b =V a {Z 



r P = P b -P a = Pa(C-l). ....... (1095): 



Tf _R(T d -T a ) log e C 
144t> 



144 7 a Z 



-^-. ..... (1096) 

B.T.U.perhr.perI.H.P.=^=g^g. ...... . . . (1097) 



To show the important relations between the work done, or the efficiency 
of transformation, and the two prime variables, quantity of heat supplied 
per pound of working gases and the amount of compression, requires a series 
of curves, and four pairs would be necessary, some of which are omitted to save 
space as the practical value of these cycles is now small though their sci- 
entific value is great. The first pair, Fig. 257 and Fig. 258, for the Stirling 
and Ericsson respectively, show efficiencies and corresponding heat consump- 
tion in the working gas, and as supplied in the fire, for various furnace 
efficiencies, plotted to B.T.U. supplied per pound of working gases as the 
prime variable, three compressions, 5, 10 and 30 atmospheres each being 
represented by one curve. As heat supplied to the working gases may be 
derived from either fire alone, or from fire and regenerator together, a double 
heat-supplied scale is necessary and each curve must be referred to its appro- 
priate scale. These two sets show a very high possible efficiency when the fire 



ENGINEERING THERMODYNAMICS 

















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1002 ENGINEERING THERMODYNAMICS 

source of heat is considered alone, efficiencies rising to over 90 per cent, 
whereas half of this value approximately corresponds to charging all the heat 
supplied to gases. A regenerator of less than 100 per cent efficiency as here 
considered would yield efficiencies somewhere between these two limits. No 
difference in the efficiency of the two cycles is found when regenerator heat is not 
charged, but with it charged the Stirling has a higher efficiency than the 
Ericsson, the figures being 44 percent and 36 per cent for 1000 B.T.U. added 
with 10 atmospheres compression. 

The second pair of curves, Fig. 259 and Fig. 260, show efficiencies as functions 
of compression for various arbitrarily chosen quantities of heat added, as 
derived from fire alone, or from both fire and regenerator together. The third 
and fourth pairs necessary for a whole series of curves would give work per 
pound of gases and the corresponding mean effective pressures, first with heat 
supplied, and second with compression as the prime variables, but these are 
here omitted. 

As actually used in engines these cycles were operated with very low 
compression, about three atmospheres and with very small quantities of heat 
supplied, such as correspond to maximum temperatures not over 800° F., 
and as the curves show, rather lower but still fair efficiencies are obtainable, 
though mean effective pressures are hopelessly low compared to modern 
engines. These curves have been extended beyond the working ranges used 
in engines that have been built because they show possibilities that are truly 
remarkable though not yet realizable. It would require a somewhat severe 
stretch of imagination to picture a steam cycle offering any such efficiencies 
as are here shown — over 90 per cent — and while these prospects have as yet 
not been fruitful they certainly point a way for further investigation when 
high economy must be attained. 



Example 1. Calculation of Diagram, Fig. 255, the Stirling gas cycle, Cycle IV. The orig- 
inal condition assumed for point A is, P a = 1 atm., V a =12.39 cuit. (one pound air), T a = 

492° F. absolute, and in addition it is assumed that there is added to the cycle 1000 

p 
B.T.U. and that the compression, C-— = 7. 



Points 



P & =7 (assumed). 



V b = V a (jA =12.39 xf |] =1.77 cuit. 

T b = T a =492° F. absolute. 

(V \ ' 12 3S 

y) =.0686 log e ^ = .133. 



The whole heat added Qi = 1000 B.T.U. is divided between the lines BC and CD, 
only the former part raising temperatures from B to C, so to determine the temperature 



HEAT AND WORK 



1003 



at C, this part must be first found by subtracting from 1000 B.T.U. the heat added 
in the isothermal expansion, which is 

J Tcloge v c ' 



Q x = 1000 = C V {T C - T b ) + (C P - C v ) T c log c - d 

v c 

= . 17(7 7 c -492) +.0686!T C log e 7, B.T.U. 



Point C 



7 7 C =3568°F. absolute. 
V e = 7 6 = 12.39 cuit; 

P c =P*g =7x^=50.8 atm. 

<I)c - < I )6= C p log e -^ = .171oge 1 ^-=.337. 



Point D 



V d = V a = 12.39 cuit. 

T d = T c =3568° F. absolute. 

P d= P c Z- c == 5 ^ =7 .26atm. 

Vd 7 

V d 
<ba ~ <>c = (Cp - C„) log e — = . 133. 

V c 



Example 2. Calculation of Diagram, Fig. 256, the Ericsson gas cycle, Cycle V. The 
initial condition of point A, the heat added, and the compression, are taken as for the 
last example, as is also the condition at point B. The first determination necessary is 
therefore the temperature at point C, which is found as follows : 

Q^IOOO =C P (T C -T b ) +(C p -C v )Tc log e C 

= .239(T C -492) +.06867 7 c log e 7, B.T.U. 



Point C 



T c =3002° F. absolute. 

P c =Pb=7 atm. 

T 300? 

7.-7^ =1.77 X Tg f=10.8 cuit. 

T 3002 

4> c _ fo = C p log e ^ = .239 log e -^ = .432. 



Point D 



Pa =Pa = 1 atm. 

T d = T c =3002° F. absolute. 

V d = v£ =10.8x7 =75.6 cuit. 

- *e = (C p -Cv) loge C = . 565. 



1004 ENGINEERING THERMODYNAMICS 

Example 3. Calculation and use of Diagrams, Figs. 257, 258, 259, and 260, giving for 
the Stirling and the Ericsson cycles, efficiency and heat consumption for any compressions 
and for any amount of heat added. For any fixed cycle there may be two thermal 
efficiencies, so for the same data there will be two efficiency curves, the location of 
which is illustrated by the determination of one point for each. First, charging against 
the work done all the heat received by the cycle whether from regenerator, or fire, the 
efficiency will be as given b}^ the curve carrying point A, for 10 atm. compression, 
referred to the lower scale of heat supplied. From Eq. (1081), 



R 

a -a. 
E 



Ql -Q 2 7 ( ^- rj l0 ^ 
Qi Qi 



but T d is itself a function of Qi given by Eq. (1079), 

Qi=jT d \og e C+C v (T d -T a ) 
whence 

d C,+(C P -OlogcC 
Substituting this in the efficiency expression and solving, for C = 10 and Qi = 2000, 



» I Ql+C V Tg \ 

ha ~ ^Ac v + {C P -C v )\o ge C~ la ) l0geC 



JQx\C,+{Cv-C v )\o&C 
53.3 /2000 + .17x492 



778x2000\.17 + (.239-.17) log e 10 



i)lo. 



492 log e 10 =46.17 per cent. 



Charging against the work done, only such heat as is added directly from the fire 
during isothermal expansion the efficiency is given by the curve carrying the point B for 
10 atm. compression and 2000 B.T.U. total, the value of the efficiency is given by 
Eq. (1082), 



Qi' Qi-Qi'" 



hence 



E' = 



/',> <.,',<■ ; .-;;; ";':':,- ; -^\log e C 



-{T d -T a ) log e C ^\ C9+{ ct-C.)\o,,r 



fr*** Q ^^^^rc- T ) 



53.3/ 2000 +.17X492 

-492 ) log e 10 



_, 77 8\.17+.0691ogel0 

Eb = /^»»» ,~ — t^tt \~ = 91 . 65 per cent. 

2000-17^ 00+ - 17X492 -492 
2000 .171 17 + _ 0691o&]0 



HEAT AND WORK 1005 

This is plotted above the point on total heat supplied lower scale, of 2000 B.T.U. 
To locate a point on an efficiency curve referred to heat of fire alone, Q/ upper scale, 
the same Eq. (1082) is put in the following form, retaining the term Q x '\ 

Applying this to the point C, for which Q/=400 B.T.U. upper scale, and 10 atm. 
compression it becomes, 

53 3 
Ec ' = 1 ~778x400 X492 l0ge 10 =79 - 87 P er cent - 

: Passing into the left-hand angle the location of point D will illustrate the deter- 
mination of all others. For D a cyclic thermal efficiency of 30 per cent is assumed 
and a furnace efficiency of 40 per cent, whence from Eq. (1086), by introducing furnace 
efficiency, 

B.T.U. per hour per I.H.P. = — =^B. =21208 

^(cycle) X ^(furnace) .3X.4 

To illustrate the use of the diagram, Fig. 257, find the efficiency, cyclic and fuel 
heat consumption for a Stirling cycle, for 300 B.T.U. supplied from fire per pound of 
working gases, 30 atm. compression, and a furnace efficiency of 40 per cent. Starting 
at point E at the value 300 on the upper scale, pass vertically up to point F on the 
efficiency curve referred to fire heat, and horizontally to G, reading thermal efficiency 
of 62.8 per cent, and cyclic heat supplied 4050 B.T.U. per hour per I.H.P. Continuing 
across to point H on the 40 per cent furnace efficiency curve and down to fire heat scale 
at K, the fire heat supplied is found to be 10,200 B.T.U. per hour per I.H.P. 

A similar procedure applies to the curves, Fig. 258, for the Ericsson cycle, using, of 
course, the appropriate formulas for efficiency and needs no detailed explanation. The 
determination of| points on the diagrams, Fig. 259, for Stirling, and Fig. 260 for Ericsson, 
is made by the same equations as are used for Fig. 257 and Fig. 258, respectively. 

Prob. 1. Derive a formula for the efficiency of the Stirling cycle by the methods 
of this chapter, for any regenerator efficiency and show that it agrees with the different 
form found in Rankine's "Steam Engine." 

Prob. 2. Derive a similar formula for the Ericsson cycle. 

Prob. 3. An Ericsson hot-air engine with two atmospheres compression and a 
volume increase on constant pressure heating of 20 per cent burns 100 cu.ft. of 600 
B.T.U. per cubic foot fuel gas per hour per I.H.P. If the furnace efficiency is 30 
per cent and regenerator efficiency 100 per cent, what part of the cyclic efficiency is being 
realized? 

Prob. 4. If in Problem 3 the regenerator were only 70 per cent efficient,what would 
be the cyclic efficiency and per cent realization? 

Prob. 5. Compare the Stirling and Ericsson cycle efficiencies for equal compressions 
and equal temperature limits. 



1006 



ENGINEERING THERMODYNAMICS 



Prob. 6. What are the numerical values of the efficiencies of Problem 5? 

Prob. 7. For the data of Problem 3, find the amount of regenerator surface' neces- 
sary for the execution of 50 cycles per minute, if the rate of heat transfer were 3 B.T.U. 
per hour per square foot per degree difference. Use arithmetical mean differences. 

Prob. 8. Find the water jacket and furnace surface for any assumed reasonable 
values of water and fire temperature, corresponding to Problem 7, and for the same 
coefficient of heat transfer. 

11. Otto, Complete Expansion Otto, Atkinson, Brayton, Diesel, and Carnot 
Cycles. Work, Efficiency and Derived Quantities for Adiabatic Compres- 
sion Gas Cycles. 

Otto Cycle. Cycle VI 

As this and succeeding cycles include as a first phase, adiabatic com- 
pression to any arbitrarily fixed degree, the amount of this compression 
expressed by either the temperature, pressure or volume change which may 
characterize it, will constitute an additional independent variable or initial 
condition. 

It is most convenient to express this in terms of pressures; accordingly, let 

p 

C = compression in atmospheres = p-. Then the volume and pressure at point 

B, Fig. 261, representing the Otto cycle, are given by the ordinary adiabatic 
relation. 

Point B: 



Pb = PaC, 



1 V. 
1 •> 
PJ C~y 



V b =V a ' PX " 



(a) 
(b) 



'V \y -i 7-1 

Ay) =T a C— (c) 



(1098) 



Point C: 



Vc~- 


= Vb = ^ 

v b i , 

cv 










(a) 


Tc~- 


-•4:= 


^KSO- 


=nx= 


= T a C .1 


x, 


(6) 


Pc = 


T 

~ PbTFT ~ PbX = P a CX, 
■Lb 








(c) 



(1099) 



HEAT AND WORK 



1007 



Point D: 




V t =V a , 


(a) 


p d =pjpy '=p a cx{~y '=p a x, 


(6) 



T 

m rp J c rp y 

J- d — -L a rn — ■*■ a-^- ) 
1 b 



(c) 



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.2 .4 .6 .8 
Entropy 



Volume in Cu. Ft. 

Fig. 261.— Otto Gas Cycle. Cycle VI 



Q2 = C v (T d — Ta)=CvTa(X—l)=C v T a -^TTFT 



¥^ 



(a) 



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(1101) 



(1102) 



(1103) 



1008 



ENGINEERING THERMODYNAMICS 



rp = P c -P a = P a (CX-l), . 



(1104) 



(1105) 



(m.e.p.) 



_W_ 
144r F 



JQi 



1447 fl 



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B.T.U. per hr. per I.H.P. =2545 



T b 



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(1107) 



Chi.ft.gafl P erhr.perLH.P.= ^( S r^ r J, ....... (1108) 



Once more the efficiency of a cycle, this time that of the Otto, is given by an 
expression similar to that of the Carnot cycle, but in which the temperature 
range is not that for the whole cycle but only that for the adiabatic compression. 
Interpreted thus, the Otto cycle has the same efficiency as a Carnot, for the 
same amount of adiabatic compression. This is indicated graphically in Fig. 
261 by the dotted lines ABC'D'A representing the equal Carnot rectangle to 
same height AB, that of the temperature rise in adiabatic compression. 

Extended expansion, beyond the original volume as practiced by Atkinson, 
reaches its limit when the pressure falls to atmosphere if all expansion below 
atmosphere be excluded as impracticable. This cycle evidently will be more 
efficient than the Otto because work area is added without any additional 
heat supply, but the mean effective pressure will be much lessened as the 
volumes increase very fast in the lower ranges of pressures. Stopping the 
expansion at some pressure and volume greater than the initial, will produce 
a cycle having characteristics midway between the two limiting cases. This 
is really the Atkinson cycle. The complete expansion case will be called 
the Complete Expansion Otto and is of interest because of efforts to produce 
a perfect compound Otto gas engine for which it is a reference standard. 
These two modifications of the Otto will be considered separately as special 
cases of the Otto Cycle, Cycle VI, 



HEAT AND WORK 



1009 



Atkinson Cycle, Extended Expansion Otto Cycle. Cycle VI. 

All points up to and including point C are given by the same expressions 
as for the Otto, but points D and E, Fig. 262, are located at some volume 
greater than the volume at A, and as this is arbitrary, it adds one more 
independent variable, accordingly, 



Let 



n = 



Ve_ Volume after expansion 
V a Volume before compression' 





































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Volumes in Cu. Ft. Entropy 

Fig. 262.— Atkinson Gas Cycle. Cycle VI. 



Point D: 



Point E: 



V a n 



(a) 



P d =P c ^=P a Cx{^j = P a X^f (5) 


T^T a x(±y~\ ( C ) 


Pe = Pa (a) , 

7 a =7«n (t) 









(1109) 



(1110) 



1010 ENGINEERING THERMODYNAMICS 

Q2 = C v (T d -T e )+C p (T e -T a ) 

1 






= an Kn) 7 -l\+C p T a (n-l) (1111) 

Tr = J(Qi-Q 2 )=jJQi-C 8 !raw[z^y-ll-C 1 ,7 7 a (n-l)|. . (1112) 

^ 2=i -S([ x ® 7 -+^- c ^- 1 )} ( 1113 ) 



O1-O2 



^ = (Fe-F & ) = F c /n 



C^ 



(m.e.p.) = 



W 



W 



144r F i ..__ / 1 
lUV a (n r 

Cy 



r P =(P c -P a )=P a (CX-l), 



(1114) 



(1115) 



(1116) 



These expressions are not so simple as for the Otto cycle and will not be 
carried further; it is, however, clear that efficiency is greater, and now no 
longer independent of the amount of heat supplied, but on the contrary a 
function of it, and of the amount of extended expansion. 



































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Volume in Cu. Ft. Entropy 

Fig. 263.— Complete Expansion Otto Cycle. Special Case of Atkinson Cycle VI. 



HEAT AND WORK 



1011 



Complete Expansion Otto Cycle. Cycle VI. 
The only point to be located is that at D, Fig. 263. 
Point D: 



Pa = P a 



(a) 



Va 



i 



Vd = Vc © y = ~i {cx) * = VaX " {b) 

(c) 

Q2 = C P (T d -T a ) = C p T a (Xy -1). . 



T d = T a ^ = T a Xy 



W = J(Qi-Q 2 )=J[Qi-C P T a (Xy -1)1 






(1117) 



(1118) 
(1119) 
(1120) 



r v = (V d -V b ) = V a (Xy-— l 

Cy 



(m.e.p.) = 



W 
144r F 



W 



144 VJ X 



C y/ 



r P ={Pc-Pa)-Pa{CX-l) 



(1121) 



(1122) 



(1123) 



Brayton Cycle. Cycle VII. 

The first data to be determined for this cycle, Fig. 264, are for point C. 
Point C: 



P- Pb = P a C 



(a) 1 



Q 



7-1 



T c =T b [l+^rj=T a C y 7(6) 
Vc = V^=V b Y = V a Y&jy- (c) 



(1124) 



1012 

Point D: 



ENGINEERING THERMODYNAMICS 



Pa = Pa 



(a) 



T c 



■Lb 



V d =VaY 



(c) 



Q 2 = C P (T d - To) = C P T a (Y- 1) = C p T a 



> 






p 
































V 


7 


B 






C 


























































6 


































































5 


































































4 




































































































3 






































\ 






























2 




\ 


































\ 


\ 




At 


nos 


phei 


ic L 


ine 
















1 






A 
























D 







































10 30 30 40 50 60 70 V 
Volume in Cu. Ft. 



.2 .4 .6 .8 

Entropy 



Fig. 264.— Brayton Gas Cycle. Cycle VII. 
W = J{Qi-Q 2 )=JQi(i-t£) (a) 



JQ 



nwi 



E = \ 



7-1 

7 



-(ft) 



(c) 

(a) 
(6) 
(«) 



(1125) 



(1126) 



T 


























6000 




































c 
















r. Abs 


















































■a 


























h 4000 


























© 


























©3000 




































D 
























1 


















& 2000 








I 


















H 


p,y 


/ 


/ 




















1000 




v^ 




















C 


492 


.X 






















J, 




A 






















U 



1.2 <?> 



(1127) 



(1128) 



v = (V d -V b ) = V a \Y-(±yi . (1129) 



HEAT AND WORK 1013 

r P =(P b -Pa)=Pa(C-l) (1130) 

(m.e.p.)= Iii ^= L r^ (1131) 

; B " H u P perhr - per i= 2 -(^J eu»> 

Cu.ft. gas per hr.l ^2545/ T b \ (1133) 

. perl.H.P. i : H \T b -T a/ 



The efficiency and work are the same as for the Otto cycle, and the former 
is independent of everything but. the amount of compression, being the same 
as for the Carnot cycle with the same compression. This Bray ton cycle 
has the smallest pressure range for this efficiency, as the Otto has the smallest 
volume range, and the Carnot the smallest temperature range, the efficiency 
in each case being measured by the amount of compression. To make the 
comparison more clear the Carnot cycle rectangle for the same temperature 
rise in compression, that is, for the same height, is plotted in dotted lines to 
T$ coordinates ABC'D'A. 

Diesel or Incomplete Expansion Brayton Cycle. Cycle VII. 



A special case of the Brayton cycle is that of the Diesel which, executing 
all the processes in a single cylinder, can sweep through no more displacement 
volume on the working stroke than on the compression stroke, and therefore, 
cutting off the expansion at the initial volume. This is obviously a less 
efficient cycle as much expansion work is lost. It is represented in Fig. 265, 
ABC DA. 



Point D: 



V d = V a 



T d =T a (j^)=T a Y y 



VaY 



0) 



(a) 

= PaY y (b) 

(c) 



. (1134) 



Q 2 = C v {T,-T a )=C v T,{Yy-l) l 



(1135) 



1014 



ENGINEEEING THEEMODYNAMICS 

W=J(Qi-Q 2 )=J[Qi-C»T a (Yy-l)l 



(1136) 



Q1-Q2 1 C,T a (Yy-l) 



™ HI-H2 -. 



(1137) 



rF=(7.-75) = 7a[l-^) 7 ]=7«[l-(g 7 )] (H38) 



r P =(P b -P a )=P a (C-l). 



(1139) 















°_ G 


3T 


ry 


£ 


\ 




" t 


t > 


4 Sr T 


H \ 


ST 


">. 


I ^ 


X S 


J _u S U 


r s 


4 V 


V k ^S 


V ^ 


S^ "— =|D 


^ 




Atmospheric Lide ~~ , 

rrrrr rrn ' H-Hrmifm^ 



T 

6000 



1000 



2 1 



10 12 V 



~r ~r 








/- 


■ ±: t 


/ 


f 


% " 


-V 


Z 


/ 


r 


^ "^ ,0 h 


4- 2 / 


/ t 


/ z 




y / 


y / 


S* ~7 


s ~7 


B,/" 7 


/* 


S 




^^ 


«-*" 




1 


.1 .2 .3 .4 g> 

Entropy 



Volume in Cu. Ft. 

Fig. 265.— Diesel Gas Cycle. Special Case of Brayton Cycle. Cycle VII. 



(m.e.p.) = 



W = J[Qi-C v Tg(Yy-l)] 
144tv 



144 Va\l 



[>-©'] 



B.T.U. per hr. per LH.P.=— ^ 



(1140) 



(1141) 



Cu.ft. gas per hr. per I.H.P. = 



2545 
HE 



. (1142) 



HEAT AND WORK 



1015 



Carnot Cycle for Gases. Cycle VIII. 

After adiabatic compression from A to B, Fig. 266, the heat addition takes 
place iso thermally, thus locating the point C by the amount of this addition. 
When the pressure at B is small and the heat addition large, the point C may 
easily fall below atmosphere. This is of no importance to the cycle as a whole, 
but if expansion were to cease at the initial volume, then adiabatic expansion 
may not only be impossible but only a small amount of heat could be added 
to the gas isothermally and only a little work done. Moreover, the efficiency 
would be very low, likewise would the mean effective pressure be also low 



p 




























D 


Temp, in Degrees Fahr. Abs. 
























B 










































7 




























































































6 




























R 














c ■ 
















































a b 






































s' 


D' 




< 

a 




































y 


S 








i 4 




\ 


























r' 




* 










8 » 






\\ 






















^ 


„** 


* 










D 




a 3 






\\ 


f 




















A 
















2 




















































\ 




n' 


































1 


Atr 


tospl 


eric 


Line 


\] 












































A 



















































































1 




1 



Vol 


i 

una 


5 

e in 


2 
Cu 



Ft 


2 


5 


3 





V 











Enl 


• 
rop 


5 

y 






<P 



Fig. 266.— Carnot Gas Cycle. Cycle VIII. 



because adiabatics lie very close to isothermals, making the work area between 
them small for very considerable displacements. The original proposal for 
Diesel engines involved a graduated introduction of the oil fuel into highly 
compressed air to keep the heating line nearly of this isothermal form, but the 
small mean effective pressures obtained made the engines too big and there 
was substituted the modified Brayton cycle, just discussed, which follows 
by a faster oil injection and combustion, fast enough to keep the pressure 
from falling till the fuel supply is cut off. 

It, therefore, is of no importance to investigate this early Diesel proposal 
but the Carnot gas cycle will be analyzed not with any idea of its practical impor- 
tance, but rather to show the extremes of low mean effective pressures and 
large volumes that it involves in attaining the same thermal efficiency as the 
Otto and Brayton for equal compressions, they giving the same work with so 



1016 



ENGINEERING THERMODYNAMICS 



little volume range and such small pressure ranges respectively. As there 
is to be no practical application of this cycle, the several points will not be 
separately located, but the desired quantities set down by short-cut methods 






\ 



(1143) 



But 



77T *b 1 a -, 1 a 

E= ~-W- =1 -T b 



(1144) 



Therefore 



Also 



E = 



JQi 



W = JQ 



V a V b 



T b -T a 



ft.-lbs. 



'-it 



*-w'gg). 



V a 



•* d — r a — 

y a 



'-■(1) 



(1145) 



.(1146) 
(1147) 
(1148) 



r v =V d -V b =V 



log, 



.iiQi/1 



(m.e.p.) = 



W 



JQ 



I44 rt ; 



RT b \C 



(T b -T a \ 



i 

1 



»"i"^m-(¥] 



(1149) 



• . . (1150) 



These volumes are very large, and pressures, both final and mean, very low, 
so that a Carnot cycle for gases even if the apparatus for executing it were 
available, winch it is not would not be practicable, whereas for steam or other 
vapors it „ not very far from the attainable or actually used processes 



HEAT AND WORK 1017 

In the curves, Figs. 267 to 272, some of the important relations between 
the quantities involved in these cycles are more clearly indicated than by 
the equations they represent. 

Referring to Fig. 267, which gives thermal efficiency as a function of heat 
supplied per pound of working gases, a most wide variation is found for the 
different cycles. One, the Diesel, shows a decreasing efficiency with increase 
of heat supply, three, the Otto, Brayton, and Carnot, a constant efficiency, 
while the complete expansion Otto or perfect Atkinson increases in efficiency. 
This crossing of efficiency lines makes the relative efficiencies change places 
with the position chosen on the heat supply scale. Furthermore compression 
has the same effect, three different compressions, 5, 10 and 30 atmospheres 
being represented for each cycle, so any question of superiority of one cycle 
over another will in general depend on conditions. There are, however, certain 
clear relations, for example, the Diesel is always less and the perfect Atkinson 
always more efficient for the same compression than the constant value for the 
three, Otto, Brayton, and Carnot, as is shown by the three curves starting from 
the origin for a given compression, one rising (perfect Atkinson) , one horizontal 
(Otto, Brayton, Carnot), and one falling (Diesel), the whole set rising with 
rise of compression. The Diesel, it is interesting to note, is quite impossible 
beyond a given heat supply where adiabatic expansion ceases, due to constant- 
pressure heat supply for whole stroke. For 1000 B.T.U. supply per lb. of work- 
ing gases and 30 atmospheres compression the efficiencies are 71.8 per cent for 
the complete expansion Otto, 62.2 per cent for Otto, Brayton, and Carnot, and 
45.7 per cent for the Diesel. Equal heat supply and 10 atmospheres com- 
pression lowers these efficiencies to 63.4 per cent, 48.2 per cent, for the first two 
and is an impossible condition for the Diesel. Reduction of heat supply lowers 
the first, leaves the second unchanged, and raises the last, for example with 
30 atmospheres compression and 200 B.T.U. added per lb. of gases the 
efficiencies become 66 per cent, 62.2 per cent, and 58 per cent. 

The precise effect of compression alone is shown more clearly in Fig. 268, 
where all cycles are shown rising in efficiency, the curves of all being nearly 
parallel each exhibiting at first a rapid rise, which later becomes more gradual. 
The relative positions for a given heat supply are the same as indicated by the 
last set of curves. 

Mean effective pressure and work per pound working gases as functions 
of the two prime variables are shown in Fig. 269, with B.T.U. per pound of 
gases and in Fig. 270 with compression. The most striking thing brought out 
is the enormous superiority of the Otto and the hopeless smallness of the 
Carnot values. As the Otto and Diesel the two gas cycles of practical import- 
ance are frequent subjects of computation, the two curves of Fig. 271 for the 
Otto and Fig. 272 for the Diesel are given to separate large scales from which 
the mean effective pressure may be accurately read off, for 200, 500, and 1000 
B.T.U. added. Engineers engaged in this work are advised to reproduce 
these charts on a large scale, adding more heat supply curves to meet working 
conditions, as they are great labor savers and may be used in estimating actual 



1018 



ENGINEERING THERMODYNAMICS 




HEAT AND WORK 



1019 



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1020 



ENGINEERING THERMODYNAMICS 




HEAT AND WORK 



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HEAT AND WORK 



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1024 ENGINEERING THERMODYNAMICS 






horse-power after the determination by tests of performance or diagram factors 
so called. 

Comparison with the curves of the non-compression cycles shows how 
very much less effective the latter are, and why no engine embodying these 
non-compression cj^cles is ever likely to successfully compete with engines 
executing compression cycles, and also why in the practical construction of gas 
engines designers are ever on the alert to discover means by which the allow- 
able working compression may be raised. In fact, it is quite possible that 
one engine mechanism that permits of higher compression than another, even 
when the former is working on a less efficient cycle than the latter, may give 
a better actual efficiency. This has for some years been demonstrated by 
the less efficient Diesel cycle compared with the more efficient Otto, the engines 
using the former compressing only pure air to about 30 atmospheres without 
difficulty, while the latter when compressing explosive mixtures, are limited 
to compressions between four and twelve atmospheres by the temperature 
of ignition of the mixtures, the low limit for pure kerosene-air mixtures and 
the high for blast furnace gas. These compressions have resulted in better 
actual efficiencies in general for Diesel than for Otto engines, reversing the 
order of cj^clic efficiencies. 

Another point that must not be forgotten is, that in the execution of all 
these cycles both steam and gas, realization always falls short of promise, but 
not to the same degree in the different cases; thus, in round numbers the efficiency 
actually realizable in steam engines may easily exceed 70 per cent of the most 
closely representative Rankine cycle for the same pressures while few Otto cycle 
gas engines have so far been able to realize more than 60 per cent of the 
Otto cycle efficiency. 

Example 1. Calculation of Diagram, Fig. 261, representing the Otto gas cycle, Cycle VI. 
The initial condition represented by point A is assumed to be defined by P a =l atm., 
Fa = 12.39 cu.ft. (1 lb. air), and 7^=492° F. absolute, and it is also assumed that the 

compression is — = 7, with 1000 B.T.U. added after compression: 



Point B 



7 atm. (assumed). 



Ay)' = 12.39 (y) ^ n 



(g 



.286 j 



492 ( — ) =859° F. absolute. 



Point C 



T c = T b +^= 859 +-~ = 6741 ° F. absolute. 

T 6741 

P,= P^=7x--=54.9a t m. 

V c = V b = 3.09 cu.ft. 

-$a=C v log e y - .17 log e -gjjg- = .35. 



HEAT AND WORK 



1025 



Point D 



Pd=F c (yJ = P C ^=54.9 



G-) 



—•©- 



492X7.85=3861° F. absolute. 



Example 2. Calculation of Diagram, Fig. 262, representing the Atkinson cycle, Cycle 
VI or extended but incomplete expansion Otto cycle. All data are the same as for Example 
1, up to and including the point C. 



Point D 



Vd = 17 cu.ft. (assumed) 



T d = T c (^ ' =6741 



m - 



3408° F. absolute. 



Point E 



P e =P a = l atm. 
Ve = V d = 17 cu.ft. 

T e = T d ^ =3408 X^ =675° F. absolute. 
Pd 5.05 

- <!>« = C v log, ^ = .239 log, || = .076. 



Example 3. Calculation of Diagram, Fig. 263, representing the complete expansion 
Otto, the limiting case of the Atkinson cycle. All data of the two preceding examples 
are identical up to and including point C. 



Point D 



Pd=Pa = l atm. 

Vd = Vc(y) =3.09(54.9) 713 =53.7 cu.ft. 

T d = T a ~ =492 X^: =2132° F. absolute. 

— 4>c — <&c — 4>a = -35. 



Example 4. Calculation of Diagram, Fig. 264, representing the Brayton gas cycle, Cycle 
VII. All data are the same as in the preceding examples through compression and 
including point B: 



Point C 



Pc=Pb= 7 atm. 

T c = T» +77 =859 +^j- =5 o43° F. absolute, 

T 5043 

F c = 7 6 - c =3.09 X-^rz- = 18.14 cu.ft. 

T 5043 

> c - <ba = C P log, - c = .239 log, — = .423. 



1026 ENGINEERING THERMODYNAMICS 

Pd=Pa = l atm. 

Point D \ V « = V °Tr 18 - 14 (lf ) =727 CUit - 

Tt = T a I? = 492 ( ^- ) = 2890° F. absolute. 



Example 5. Calculation of Diagram, Fig. 265, representing the Diesel gas cycle, a 
special case of the Brayton cycle, Cycle VII. All data of Example 4 apply up to, and 
including point C. 



V d = V a = 12.39 cuit. 



Point D 



P.-ft(fi)"-»(S)"-^-L 



492X6.72=3328° F. absolute. 



Example 6. Calculation of Diagram, Fig. 266, representing the Carnot gas cycle, Cycle 
VIII. The data of examples preceding apply through the compression and include 
point C. 



T c = T b =859° F. absolute. 
V c = 8 cu.f t . (assumed) . 



Point C 



Point D 



Point D' 



p P b V b 7X3.09 _ + 
P c =-r— = — ^ — =2.70 atm. 



8 



V< 



fc - <k = (C p - C v ) log* —^ = .0686 log. 



T d =T a =m. 



(— )-■ 

\3.09/ 



= .0653. 



/TA 3 - 48 /4Q2\ 3 - 48 

p <=%!) - 2 - 7 W) =- 388atm - 

T . P a 7 a 12.39 Q1QQ 

F ^"p7 = "^8 =3L93cuit ' 



7 rf ' = 7« = 12.39 cuit. 



P<z'=P c U^. =2.7 



7/ 

P/ 



12.39 



= 1.465 atm. 



^' = ^-^=4.92x1.465=721° F. absolute. 

"a 



Example 7. Calculation and use of Diagrams, Figs. 267 and 268, ^'w'n^ for the adi- 
abatic compression cycles, Otto, Brayton, complete expansion Otto and Diesel, the 
efficiency, heat, and fuel gas consumption for any compression and heat supplied per pound 
of working gases. The efficiency and its derived quantities are given in Fig. 267 as 



HEAT AND WORK 1027 

functions of heat supplied and the calculation of these curves will be illustrated by 
the location of one point on each, A for the Diesel; B for Bray ton, Otto, and Carnot, 
all having the same efficiency for the same compression, and C for the complete expan- 
sion Otto. These points are all for 800 B.T.U. supplied per pound of working gases, 
after 10 atm. adiabatic compression. 

Point A is located by the formula for efficiency of the Diesel cycle given by Eq. 
(1137) # . 



jr, , t v T a 

E.-l— 7 



(-^«) r -'l 



=1 



W p TaC v 
.17X492 



800 



/ 800 \ 1A 1 

V239x492x(10)- 28 « + V "M =23 ' 68 Pef Cent 



The limiting case for the efficiency of this cycle is that receiving enough heat to 
make V c = V a (see Example 5). Whence 

Qi (limit) =C p T a C^~ \Cv-l) = .239x492xl0- 286 (10- 713 -l) =947.3 B.T.U. 

and this is indicated on the curve carrying the point A by a crossing line and appropriate 
lettering. 

The point B is located from Eq. (11036) giving, 

7-1 

P a \ y 



Eb = 1 ~\p b ) =1 -^TEi =1 -(^286=48.2 per cent. 

C y 

By means of Eq. (1120) the efficiency represented by point C is fixed, giving, 



= 1- 



CvTaC r-' J 

.239x4921"/ 800 V 713 ] 

L\l7x492xl0- 28 ^ + 7 _1 J =62 per eent - 



800 



The location of the lines in the left-hand angle needs no explanation in view of 
the examples for preceding gas cycles. 

Illustrating the use of the curve the solution of the following problem is traced 
graphically on the diagram. Required the thermal efficiency, cyclic heat, and fuel 
consumption for the Diesel cycle, supplied with an oil yielding 1500 B T u' per cubic 
foot in its vapor, the cycle receiving 600 B.T.U. per pound of working gases after 
10 atm. compression. From the 600 point E on the heat-supplied scale pass up to the 
10 atm. compression Diesel curve F, and horizontally across to the efficiency scale G 



1028 



ENGINEERING THERMODYNAMICS 



reading 28.6 per cent and 8900 B.T.U. per hour per I.H.P. Continuing across to the 
fuel calorific power curve of 1500 B.T.U. per cubic foot H, and thence down to K, the 
fuel consumption is found to be 6 cu.ft. 

The second set of efficiency curves, Fig. 268, is obtained and used in exactly the 
same way as is the first, the only difference between the two being the scales, so this 
series requires no explanation. 

Example 8. Calculation and use of Diagrams, Figs. 269, 270, 271, and 272, giving 
for the adiabatic compression cycles the work per pound of working gases and mean effective 
pressures. Explanations of the first, Fig. 269, will apply to all the others as they differ 
only in scales to facilitate the use of numerical problems. 



From Eq. (11026), 

W=JQi(l—^i 

C v 



For 5 atm. compression and 1000 B.T.U. supplied, 
W a =778 X1000 X (l -Jse) =287160 ft.-lbs. 



From Eq. (1136), 



W=J 



Qi-CvT, 



■[ 



0i 



^i+M - 



']}■ 



For 30 atm. compression and 1000 B.T.U. supplied, 



W b =77S 1000 -.17X492 



1000 



,239X492X(30) 



.286 



+ 1 



)"-]!■ 



: 355000 ft.-lbs. 



Q 



W+iV-i 



From Eq. (1119), 

W=J\Qi-C v T a 

I " V 7 ', 

For 5 atm. compression and 1000 B.T.U. supplied 

f r / 1000 

Wc=77S \ 1000-239x492^ 17x492><5 . 



+ 1 



)'"-'] 



=447300 ft.-lbs. 



Mean effective pressures are derived from these cyclic work values, for the location 
of points and curves of the left-hand angle, by the following equations for one point 
of each curve A', A", A'", B r , and C. 



HEAT AND WOEK 
From Eq. (1150) for the Carnot cycle, 

W 



1029 



W 
p- e ' p - )a ' = l447 






287160 



144xl2.39[log.-. y x * 2x5 , 86 ) - (-) J-X.0000121b.persq.ta. 



From Eq. (1131) for the Brayton cycle, 
W W 



(m.e.p.)a» = 



144r, 




144X12.39 



h 



1000 



239X492X5- 



+1 



- — ji2r\ =26.61 lbs. per sq.in. 



-c-n 



From Eq. (1106) for the Otto cycle 
W W 



(m.e.p.)a'" = 



144r, 



1447. l-g) 



287160 



144X12.39 



[-en 



= 235.81 lbs. per sq.in. 



From Eq. (1140) for the Diesel cycle, 
W W 



(m.e.p.)y = 



144r, 



144F, 



Mfl 



355000 



144 X 12.39[l-(i)' 713 ] 



= 218.79 lbs. per sq.in. 



1030 ENGINEERING THERMODYNAMICS 

It may be noted for equal works and compressions by the Otto and Diesel cycles 
the mean effective pressures are equal because the volume range is the same for both 

From Eq. (1122) for the complete expansion Otto cycle, 
W W 



(W)c = 144r 



v 1447, 



4(- s i3")--®'] 




144x12.39 



447300 

58.3 lbs. per square inch 






( m +1 w _Y!V 713 1 

V.17x492 X 5- 286+ / U/ J 



No explanation of the use of the diagram is needed nor is it necessary to give the 
detailed steps for, or illustrate the use of, the other work and mean effective pressure 
diagrams, Figs. 270, 271, and 272, for all are similarly derived and all serve to solve 
graphically, work and mean effective pressure problems for these cycles without calcu- 
lation and with sufficient accuracy for general use. 

Prob. 1. An Otto cycle engine is supplied with producer gas having 135 B.T.U. 
per cubic foot and the compression is 12 atm., heat per pound of working gases 700 
B.T.U. Find the cyclic efficiency, cyclic heat consumption, gas consumption, and 
mean effective pressure. 

Prob. 2. For the same data as in Problem 1, solve for the Bray ton cycle. 

Prob. 3. For the same data as in Problem 1, solve for the complete expansion Otto, 
and from the answers state the value of perfect compounding of Otto gas engines. 

Prob. 4. A Diesel engine is operated with 25 atm. compression, on oil having 
19,000 B.T.U. per pound, and the cut-off is such as to add 500 B.T.U. per pound of 
working gases. Find its thermal efficiency, weight of oil per hour per I.H.P., and mean 
effective pressure. 

Prob. 5. If 50 per cent of the cyclic efficiency were realized in the Otto cycle engine 
of Problem 1, all losses taking place during explosion, find the new answers for the same 
data. 

Prob. 6. If the Diesel engine of Problem 4 realizes 60 per cent of its cyclic efficiency 
and all losses occur during combustion, find the corresponding thermal efficiency, mean 
effective pressure, and oil consumption. 

Prob. 7. Otto and Diesel cycle engines are in commercial competition and the best 
of each have nearly equal thermal efficiency. If this be taken as 35 per cent actual and 
if each realizes 50 per cent of the cyclic efficiency, all losses taking place during com- 
bustion, and finally if the Otto cycle receives 800 B.T.U. per pound of working gases, 
what are the characteristics of a Diesel engine cycle of equal performance to an Otto 
with 15 atm. compression? 

Prob. 8. Solve Problem 7 for an Otto engine with 20 atm. compression. 

Prob. 9. Find what cut-off volume corresponds to various quantities of heat per 
pound of working gases for the Diesel cycle at various compressions from 15 to 30 atm. 
Plot a curve of relations. 

Prob. 10. Compare the terminal pressures and temperatures for all adiabatic com- 
pression cycles receiving 750 B.T.U. per pound of working gases after 15 atm. com- 
pression. 



HEAT AND WOEK 1031 

12. Comparison of Steam and Gas Cycles taking the Rankine as Standard 
for Steam, and the Otto and Diesel as Standard for Gas. Relations of 
Otto and Diesel to Rankine Cycles. Conditions for Equal Efficiency. Trans- 
formation of heat into work is a practical and economic possibility by the 
use of either a vapor or a gas as the working medium, and, of course, 
mixtures of vapors and gases as well. Such transformation is effected 
by a series of thermal processes, the efficiency of which is given by 
certain equations derived from analysis of the corresponding thermal 
cycles, which also give the corresponding work of the cycle, mean 
pressure, and other important quantities. Comparison of these equations for 
the different cycles establishes principles of practice that are invaluable as 
working guides in all sorts of ways. For example, the relative value of gases 
; and vapors as the working media fixes corresponding relations between 
the steam and gas power systems as such, independent of interferences and 
thermal losses imposed by the mechanisms of the actual machinery. This 
is the most valuable contribution of such thermal analysis of cycles because 
such conclusions could not have been obtained in any other way. For steam 
alone the analysis shows how fast efficiency increases with the rise of initial 
pressure and temperature and with the lowering of back pressure, and shows 
the gains to be expected by high boiler pressures or high superheats over low, 
and by most perfect condensing equipment over a poor equipment, and these 
gains are to be balanced against the additional cost of equipment involved, 
in arriving at a good engineering judgment of how far it is worth while to go 
in any given direction. The situation is the same with gas power efficiency, 
for the rate of increase of efficiency is fast at first with increase of compression, 
and slower as compression rises. Higher cylinder compression means more 
costly construction of engine, so that a means is available for comparing the 
increased cost with the gain to be derived. However, in this case there is another 
limit to the compression as has been pointed out and that is the ignition tempera- 
ture of the explosive mixture when such is the substance being compressed. 
Similarly, there is another variable in the whole gas power problem and that 
is the cycle itself, for it has been shown that for the same compressions the 
maximum pressure which controls the weight and strength to be given to the 
working parts is not fundamentally related to the conditions for high efficiency 
but depends rather on the cycle itself, being highest for cycles of the Otto 
group and lowest for those of the Bray ton group. 

The most striking general result of the cyclic analysis is the fact that in 
the abstract, neither steam nor gas as a working medium is always superior 
to the other as a transformer; both may yield high efficiencies if the proper 
and special conditions prescribed by the cycle are fulfilled. When, however, 
the possibility of their easy and cheap fulfillment are examined, then there 
is noticeable a superiority of gas cycles over steam so far as efficiency is con- 
cerned. On the other hand, judging by mean effective pressure standards 
the differences change again. Although in general the conditions for most 
efficient working are coincident with low mean effective pressures, which means 



1032 ENGINEERING THERMODYNAMICS 

that large cylinders are generally needed to fulfill the conditions of high 
efficiency, this handicap has been removed from the steam system by the substi- 
tution of the turbine for the cylinder machine but still remains with the gas 
system. However, the mean effective pressure for gas systems is not so much a 
function of the efficiency conditions as it is of the cycle. Nothing could be more 
convincing in this respect than the comparison of the Otto and Carnot cycle 
mean effective pressures, those for the Otto extremely high, higher several 
times over what is possible with efficient steam cycles, while those for the 
Carnot are vanishingly small. It is sometimes believed that the engine size 
or mean effective pressure handicap is to be removed in the gas power system 
by the same turbine method as has succeeded with steam. While this may 
some day work out, the difficulties of a practical sort are very great, and more 
important still, the handicap is nowhere near so great as it was for steam and, 
therefore, the necessity is less, provided the best gas cycle be chosen. This 
is beyond all question the Otto cycle, because of the high mean pressures and 
the corresponding high efficiency and from this cycle no departure is permissible 
without proof of its effectiveness in some practical respect with equally strong 
proof that a substitute is available, and one that gives a fair return for what 
is lost by comparison with the Otto. One such is the Diesel, in which the 
limitation of compression by ignition temperatures is removed by compressing 
pure air alone and to which fuel is subsequently added, the air compression 
being carried far enough to itself ignite the fuel on admission and burn it at 
constant pressure. Though a comparatively inefficient cycle for equal com- 
pressions, this may, and does in practice, yield with its 30 atmospheres com- 
pression better over-all results in efficiency than the Otto for the lower 
compressions of about 12 atmospheres, to which the latter has been confined 
by self-ignition limits even with the least sensitive fuel, in the ratio of about 6 
to 5. This is not, however, the case with the corresponding maximum and 
mean effective pressures, because Diesel cycle engines are approximately twice 
as costly as Otto cycle engines, due to very high maximum without corre- 
spondingly high mean effective pressure. Furthermore, Diesel cycle engines 
are confined to oil fuel to-day as it is still regarded as impracticable to so highly 
and separately compress gas fuel for injection into the air at the end of the 
compression, and gas fuel-air mixture compression is retained in Otto cycle 
engines in spite of the ignition temperature limitation the practice imposes. 

Another illustration of reasonable departure from the standard Otto would 
be the substitution of a cycle represented by a mechanism that would give 
speed regulation and reversibility, equal in perfection to that obtainable with 
steam, and which is impossible with Otto cycle engines, or to cite another 
case, the substitution of a cycle and mechanism that would permit of safe 
and direct internal combustion of solid fuel, also now considered impossible. 
Either of these things would offer advantages enough to offset cyclic deficiency 
measured by lower thermal efficiency or smaller mean effective pressures as 
compared with the Otto, if not carried too far. 

It appears then that while there are innumerable reasons for not taking 



HEAT AND WOEK 1033 

the teachings of the cyclic analyses too seriously, this is not to be regarded 
as proof of a lack of value. As a matter of fact cyclic analysis, such as has 
just been briefly developed and whether worked out graphically or algebraically 
is quite invaluable, but is not to be used or applied without a more or less 
cultivated engineering judgment. 

To close the comparison, it is desirable that the efficiency and work of 
the cycles, together with the quantities derived from them for the standard 
steam and gas cycles be brought together so as to show more clearly the con- 
ditions to be met in each case for the attainment of equally good results, 
especially with respect to efficiency, as there are practical ways of meeting 
the low mean effective pressure handicap, but none for overcoming inherently 
low efficiencies. The standard cycle for steam is the Rankine and those for 
gas the Otto and Diesel, so that the comparison need be made only for these; 
any other steam cycle should and may easily be compared with the Rankine, 
and the same is true for the gas cycle with respect to the Otto and Diesel. 
Thus, a given gas cycle is best judged by saying its efficiency is more or less 
than that of the Otto if it uses constant volume heating, or its mean effective 
pressure is as great, for either the same amount of heat supplied or for equal 
compressions, or perhaps for equal maximum pressures. If the gas cycle uses 
constant-pressure heating it may be compared in performance with the Diesel. 
This allows of judging all gas cycles by a gas cycle standard and all steam cycles 
by a steam cycle standard, the comparison of a steam cycle with a gas cycle 
being made through the relative position of the Rankine with the Otto and Diesel 
standards. This may at first glance look like a cumbersome and roundabout 
method, but it really is not. 

The comparison may be conveniently made by a diagram of Fig. 273, on 
which efficiency is plotted vertically, the curves for Otto and Diesel gas cycles 
to the left, and the Rankine steam cycle to the right, thus making a comparison 
of conditions easy through the common efficiency scale. Compressions for 
gas cycles are carried to 40 atmospheres so as to exceed somewhat the working 
maximum of the Diesel engine, but steam pressures are carried to 500 lbs. 
though 250 lbs. may be regarded as the corresponding used limit, because 
the efficiencies of the steam cycles would not otherwise rise to the same values 
those for gas. For a similar reason wet steam is not considered, back 
pressures are kept at the lowest practicable limit of half a pound absolute 
corresponding to 1.03 ins. Hg. and a temperature of 80° F.; superheats are 
carried in each case to an amount that brings the maximum temperature of the 
steam up to 600° F., beyond which dissociation, pipe oxidation, and 
expansion are excessive, though 100° less is a better practicable limit. The 
diagrams show that it requires some 450 lbs. per square inch initial pressure 
of steam to reach 35 per cent cyclic steam efficiency, a pressure beyond working 
practice, whereas less than five atmospheres compression, about the least 
ever used, gives an equal efficiency to the Otto cycle gas engine standard. For 
the same 35 per cent efficiency it appears that the Diesel requires 7§, llf, and 
18| atmospheres compression when there is added to the cycle 200, 500 and 1000 



1034 



ENGINEERING THERMODYNAMICS 


















































































































































































































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-II 



g. 273. — Comparison of Otto and Diesel Gas Cycles, with Rankine Steam Cycle, for 0.505 pound per square inch 
Absolute Back Pressure and Various Initial Pressures and Steam Temperatures Up to 600° F. 



HEAT AND WORK 1035 

B.T.U. respectively per pound of working gases. The most striking thing brought 
out by the comparison is the sustained rapid rise of efficiency for all gas cycles 
with compression, while the steam cyclic efficiency rises very slowly beyond 
200 lbs. initial pressure and by almost a straight line, superheat helping some 
but not much in proportion. 

This chart shows clearly the inherently higher efficiency possibility for 
gas systems over steam and, as realization has not been in the same proportion, 
it indicates that the gas engine mechanism as the means of executing the gas 
cycles must receive very great improvement before it can equal the effectiveness 
of steam mechanism. It is therefore of great practical importance that gas 
engine mechanism losses be most carefully studied more than they yet 
have been, for once brought to equality with steam, the cyclic promise may be 
realized and the fuel consumption of the world for power purposes cut in 
half. 

Example. Under what operating conditions may a steam engine realizing 60 per 

cent of the Rankine cycle efficiency be worked, to equal in thermal efficiency an Otto 

cycle gas engine realizing 40 per cent of the cyclic efficiency on 10 atm. compression. 

Reference to the diagram, Fig. 273, in the left-hand angle shows that the Otto cycle 

efficiency is 48.6 per cent. If 40 per cent of this is realized the actual efficiency is 

19.44 per cent, and if this is equal to that of the steam engine its cyclic efficiency must 

19.44 
be — '■ — =32.4 per cent. On the right-hand part of the diagram this would be realized 

for \ lb. back pressure and absolute initial pressures from 190 to 220 lbs. per square 
inch approximate with superheats from 200° to zero. 

More exact determinations may be made by separate reference to the charts at the end 
of the sections dealing with these cycles. 

Prob. 1. A steam engine operating on 200 lbs. initial pressure and 1 lb. back 
pressure realizes 55 per cent of the Rankine cycle. Under what operating conditions 
may an Otto cycle engine realizing 50 per cent of its cyclic efficiency equal the per- 
formance? 

Prob. 2. Solve Problem 1 for a Diesel cycle realizing the same fraction of its cyclic 
performance. 

Prob. 3. What would be the mean effective pressures for the three cases, the steam, 
and the O to and Diesel gas cycles? 

Prob. 4. Plot to T$> coordinates the pure cycles and estimate the probable diagram, 
accounting for the losses given, and check by areas the efficiencies reported. 

Prob. 5. Plot to PV coordinates the pure cycles and estimate the position of the 
probable diagram, accounting for the losses given, and check by areas the mean effective 
pressures reported. 

13. Gas Cycle Performance as Affected by Variability of the Specific 
Heats of Gases, Applied to the Otto Cycle. All results so far obtained for 
gas cycles have been derived on the assumption that the specific heats of 
gases are constant, but it is known that at high temperatures such as characterize 
the Otto, for example, and some others, that specific heat varies, so it is necessary 
for a rigorous treatment to examine the effect of this variability on the con- 
clusion of cyclic performance. 



1036 ENGINEERING THERMODYNAMICS 

When specific heats vary, it has been assumed in accordance with thermo- 
dynamic prediction, that they are functions of the temperature only, and 
this is true if the gases are perfect. Real gases depart from this law and the 
most notable example is superheated steam, for which the specific heat is 
fairly well established and now known to be a function of both pressure and 
temperature. It is, therefore, quite likely that practically all gases are similar 
in this respect, though no one is yet able to say just what is the true relation, 
as has several times been pointed out. The usual procedure is to assume 
a linear relation between specific heats and scale temperature above 32° F., 
expressing the specific heat at, or the mean specific heat to a given tempera- 
ture, as equal to the value at 32°, with a constant multiple of the temperature 
excess over 32°, added. While this is too simple a relation probably, to cor- 
rectly state the facts, it is yet too complicated to enable the derivation of cyclic 
performance equations that shall be simple enough to be useful in engineering 
practice, as will appear from what follows. Nevertheless, it is highly desirable 
to go through with the derivation, for the light it will throw on the nature 
of the effect of variability of specific heat, however inadequate the numerical 
results may prove. 

Careful studies of the indicator cards of Otto cycle gas engines by Clerk, 
Burstall and others has led to some conclusions as to specific heats that may 
be used for the purpose in hand, though always with doubt as to accuracy. 
Without in any way implying inferior merit in the work of others the data 
of Clerk will be adopted and his values are given below: 

C — 9fi^j_ 0000^1/ f * n * erms °f Centigrade scale temperatures; 

C. = .194+.0000283r*-32°l i. 

n — o«k i_ nnnnooQ ooo m terms of Fahrenheit scale temperatures; 

C, = . 1801 + . 0000283 T \ . , , _, , _ ., , . . . 

r — 9^1 1 -u nnnn983 T ln ^ erms °* Fahrenheit absolute temperatures. 



In general, then, 



R = J(C p -C v )=J(a'-«); 



«+$T 



Referring now to the cycle, Fig. 274, the first phase of which is adiabatic 
compression, it will appear that the first effect of the above variability of 



HEAT AND WORK 



1037 



specific heat relations is to cause the line AB to lie below the corresponding 
curve for constant specific heats A B', which latter is shown dotted. This is 
proved algebraically by starting with the general equation for a small increment 
of heat which for adiabatic changes is put equal to zero. 



dH = C4T+jPdV 



= j^(PdV+VdP) +jPdV = 0, 



60 



O 

S 40 



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Volume.in Cu. Ft. 

Fig. 274. — Comparisons of Otto Cycles for Constant and for Variable Specific Heat 



0.5 



whence substituting the new values for C v and R this becomes 

(<x+$T)(PdV+VdP) + (a'-a)PdV = 0, 
<*'PdV+aVdP+$T(PdV+VdP) = 0, 



dV dP T 
a y +o?£+$py(PdV+VdP) = 0, 



a '^+a^+pdr = 0. 



1038 ENGINEERING THERMODYNAMICS 

Integrating 

a , log7+alogP+^J T = Xi, 

-\ogV+\ogP+$-T=K 2> 
a a 

PV«e« =K. 





PaV a ' Y e" Ta = K = P b V b ' Y e« Tb (1151) 



or 



This is the new adiabatic relation between pressures and volumes, when 
specific heats are variable, but it also involves the temperature. By the 
general gas law, the other relations can be found as follows for T and V: 

T b _P b V b _ V b VSe" a _ (V a \y- 1 ^T a -T b ) 
Similarly for T and P 



y_-l f 



f r (^y~£^ (11 54) 

Finally for P and V 



PaVaV^^K, 

or 



g-^)^- 7 --™ (1155) 



By means of these three new adiabatic equations the compression line is com- 
pletely determined and can be plotted, though the solution is not by any 
means easy. The next phase is that of heat addition from B or B f , to C or C", 
and as the specific heat increases with temperature rise, the increase in tem- 
perature will be less for the variable than for the constant value of the specific 



HEAT AND WORK 



1039 



heat. The relation between temperature rise {T c —T h ) and the quantity of 
heat added is given by Eq. (1156): 



/ C v dT= I 



(«+$T)dT = <x(T c -T b )+^(T c 2 -T l ?) 



(Te-T b )[*+*/LTc+T>)], (1156) 



To express the final temperature T c in terms of the heat added, Qi, requires 
the solution of the second degree equation above and the solution is given 
by Eq. (1157): 



r.-^[Q 1+ ( a +|n)ft] + (|) a -|. 1 . . (1157) 



Of course, as before 



Pc = P b ~ 



By means of these equations the point C of the cycle is located and there 
remains only the fixing of D with reference to C and A. The same adiabatic 
relations as applied to the compression line A B also apply to the expansion 
line CD and in addition, V c = V b and F d = F . 



-(T d -T c ) 



(a) 



<T d -T c ) 



k-9P 

MrP » 



(1158) 



It is possible to find an expression for the pressure, volume, and temperature 
of points C and D, at the beginning and end of expansion, not merely as has 
been done in terms of equivalent values at the last previous point, but in terms 
of initial conditions at A , of the compression and of the heat added, but this 



1040 



ENGINEERING THERMODYNAMICS 



derivation results in very complex equations and is omitted to save space. 
The efficiency of the cycle is given by 



Qi Qi 



(r c -n)[a+|(r c +T e )J-(T d -r a )^+|(^+r a )] 



(T d -T a ) 



=1— 



(r c -r 6 )^+|(r c +n)] 



*+*(T d +T a ) 



a+|(zvw 



(a) 



(« 



(c) 



(1159) 



(5Te-!r 6 ) 
When the specific heat is constant, @ = 0, and this expression becomes 

Ta \Y a ~ l ) , r a 



77T/ -i J- d la 



»GH ~* 



which is what was previously found for the constant specific heat Otto cycle, 
but it must be remembered that for equal compressions and heat additions, 
the temperatures T b , T c and T& for the variable specific heat are not the same 
as those for the, constant value TV, TV and TV- 

Substitution of the various temperatures in terms of prime variables in 
the equation for efficiency gives an expression that cannot be solved at all 
and so is not worth deriving. Graphic methods are more satisfactory, as 
the plotting of the cycle shows the relations at a glance, and the degree of 
accuracy in evaluating work and heat area to get mean effective pressure 
and thermal efficiency is entirely a question of the sort of cross-section paper 
and the scale used. 

However, working on the temperatures as prime variables Wimperis 
secures a solution by a series of algebraic approximations that is sufficiently 
interesting to warrant quoting. Assuming equal temperatures for the points 
B, C, and D, for the two cases of constant and variable specific heat which, 
of course, cannot be for equal compressions and heats added, and calling 
the efficiency for the former case W , he finds 



1-^[(1-E')T C +Tc 



(1160) 



The efficiency by this equation is plotted in Fig. 275 for two different values 
of the naximum temperature, T c = 1000° C. = 1832° F., and T c = 1600° C. 



HEAT AND WORK 



1041 



= 2912° F., when T a = 400° C. absolute. To this chart a line is added giving 
Clerk's own calculations of the efficiencies as given in Table CXXVIII compared 
with those for constant specific heat for the same temperatures. These relations 

































































































































































































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Fig. 275.— Relation of Efficiency of Otto Cycle with Constant Specific Heat, to Efficiency 
of Same Cycle with Variable Specific Heat. 



are clearly of limited value since equal temperatures cannot result from equal 
compressions and heat additions, nor can any other temperatures for the 
cycle be equal if the maxima are. 



1042 



ENGINEERING THERMODYNAMICS 



Table CXXVIII 

CLERK'S COMPARISON OF OTTO CYCLE THERMAL EFFICIENCIES FOR 
CONSTANT AND VARIABLE SPECIFIC HEATS. L EQUAL LIMITING 
TEMPERATURES 





Thermal Efficiency. 




Compression. 










/Volume befores. 
^ Volume after / 


Variable Specific Heat. 


Constant 






Specific Heat. 




Max. temp. =2912° F. 


Max. temp. =1832° F. 




2 


.197 


.200 


.242 


3 


.286 


.293 


.356 


4 


.354 


.356 


.426 


5 


.384 


.394 


.473 


7 


.439 


.443 


.541 



Prob. 1. A gas engine operates on an Otto cycle, and the ratios of volumes before 
and after compression is 5. What is the cyclic, efficiency, assuming constant specific 
heats and what according to Wimperis' approximation? 

Prob. 2. If in a gas engine the temperature of compression is 500° F. absolute and 
1200 B.T.U. are added per pound of gas, what will be the temperature rise if the 
specific heat is considered constant, and what if it follows the values as given by Clerk? 
What will be the difference in pressure rise due to the two values of specific heat? 



14. Actual Performance of Otto and Diesel Gas Engines, and its Relation 
to * the Cyclic. Diagram Factors for Mean Effective Pressure and Thermal 
Efficiency. Effect of Load on Efficiency. Heat Balance of Gas Engines 
Alone and with Gas Producers. All modern gas engines are operated on the 
internal-combustion method of adding heat to the mass of working gas both 
explosively and non-explosively, a fact that is responsible for one group 
of limitations in the results obtainable. There are two direct consequences 
of importance, first, the charge must be renewed each cycle, fresh air and 
fuel coming into the working chamber and hot products of combustion leaving 
alternately, with the other phases between, and second the obtainable tem- 
perature rise with its corresponding pressure increase in Otto cycle engines, 
or volume increase in Diesel cycle engines, is fixed by fuel-air mixture prop- 
erties. Furthermore, as all these operations take place in cylinders having 
pistons that require lubrication and which must, therefore, be kept cool enough 
to prevent destruction of the lubricant, a heat exchange takes place between 
gases and walls alternately in each direction, that causes corresponding pressure, 
volume, and temperature changes in the gas, not contemplated by the cycle 
nor subject to computation or formulation. Besides these influences there 
are others chargeable to mechanical construction or adjustments that are 
responsible for further departures in the pressure, volume, and temperature 
changes from the ideal cyclic ones, as for example, too early an exhaust cur- 



HEAT AND WORK 1043 

tails expansion; too late a closure of admission valve similarly delays the 
beginning of compression and reduces the total amount. In Otto cycle engines 
where mixtures are exploded the combustion line may not be vertical and 
heating fail to be truly at constant volume, while in Diesel cycle engines where 
the fuel burns as fast as it enters, inaccurately graduated injection may cause 
the combustion to depart from the truly constant pressure sort, too rapid a 
feed causing pressure to rise; too slow a feed permitting a fall of pressure. 

It is not surprising, therefore, to find that the mean effective pressure, 
thermal efficiency, or other characteristic performances differ in real gas engines 
from their cyclic equivalents, but this does not in any way reduce the value 
of cyclic analysis, in fact it accents the necessity for it. There are two sorts 
of contributions of practical value in the cyclic analysis taken in conjunction 
with the performance of real engines. The first is of the fundamental scientific 
order, giving numerical value to the extent of the possible improvement of 
real engines of any one class and, pointing out just where the losses occur with 
the amount of each indicates where any improvement must be made. The 
second is of more direct practical value to builders and users of engines as it 
is concerned with the prediction of results, for a given cylinder and fuel just 
what horse-power and fuel consumption may be expected, or what cylinder 
size is required for a given output. In these latter cases the cyclic analysis is 
responsible for the form of equations for mean effective pressure and thermal 
efficiency, actual tests furnishing numerical values for the constants of 
proportionality. 

As cyclic performance is to be the basis of all computations on the approach 
to perfection of performance in real engines and of their probable power and 
efficiency, it is necessary to select the standard cyclic equations as a 
first step. It has been shown that comparatively simple performance equations 
are derivable for the Otto and Diesel cycles if the specific heats of gases are 
assumed to be constant, whereas it is known that they are not constant. But 
with any proposed law of variation in specific heats it has also been shown 
that the equations for cyclic performance are very difficult of any solution 
and impossible to exactly solve. This makes it difficult to decide on a course 
of procedure for practical computations in the first group of comparisons that 
are concerned with the approach to perfection of real engines. Strictly speaking, 
they should be compared with cyclic performance as computed on the basis 
of variability of specific heats, but in view first, of the uncertainty of the law 
of variation, and second, of the complexity of the cyclic equations, this is 
not yet a feasible thing as a matter of regular engineering routine, however 
desirable it may be in exceptional cases. 

It is, therefore, generally agreed that the cyclic results based on constancy 
of specific heats shall be used as a basis of comparison of real engine performance, 
and in the simple everyday predictions of probable power and efficiency this is 
quite as good as any other, because a single constant factor can include all cyclic 
departures as well as the losses in the engine itself. The next step in estab- 
lishing the cyclic standard, is to fix the physical properties of the working 



1044 ENGINEERING THERMODYNAMICS 

gases, for it must be remembered that these working gases include some fuel, 
some air, and some products of combustion left in the cylinder from a previous 
explosion. Every change in fuel, or in proportions of fuel, to air, to burnt 
gases, involves a different specific heat for the working mixture and even for 
a constant mixture, the expansion stroke, being made only with burnt gases, 
will have different specific heats than the compression stroke where the mixture 
is yet unburnt. Some authorities attempt to evaluate the specific heats for 
these various mixtures, but this practice is defensible only in those exceptional 
cases where approach to perfection is under study, and here variability of specific 
heat with temperature and pressure must be taken into consideration as well 
as variability due to chemical composition, a procedure so far impossible. 
This has led to the assumption that for the cycle tl\e working gases have the 
physical properties of air, or are air, have specific heats known to apply to 
air at 32° F., and that these do not vary. 

On this basis the thermal efficiency of real engines is compared with a so- 
called air-card standard which is the cycle most nearly corresponding, and 
with air as the gas. Compared to this air-card standard, actual performance 
is found to be surprisingly constant, regardless of the fuel, the compression, 
or the size of engine, the ratio ranging from .40 to .60 for Otto cycle 
engines and this is known as the diagram factor. In some cases the reason 
for an increase or decrease in this number can be found, but in general this 
is not possible. For example, it is generally higher in large than in small 
engines, higher when the cylinder and piston heat absorbing surface per cubic 
foot of working gases is small, but other influences may entirely counteract 
these and the precise effect of each separate influence cannot be evaluated. 

According to the above, Otto cycle engines yield efficiencies referred to 
indicated horse-power that are in round numbers somewhere about half the 
corresponding air-card efficiencies, so that if absolutely perfect they apparently 
would be only twice as good as they now are, or would consume only 
half the fuel they do. A little more precisely, Wimperis has made some com- 
parisons by means of his formula based on Clerk's linear variation of specific 
heat with temperature, and finds that when the air card with constant specific 
heat yields efficiencies of 54 per cent for a compression from 7 to 1 volumes, 
the air card with variable specific heats yields 44 per cent and 46 per cent, 
for a maximum "temperature of 1600° C. = 2912° F. and 1000° C. = 1832° F., 
respectively. If for these cases the diagram factor based on constant specific 
heats were .50, then based on variable specific heats, it would be .615 and .59 
respectively, showing the performance to be really nearer to perfection than 
at first appeared. 

A similar computation made by Meyer on test data of a Deutz engine, 
8.67 X 13 ins. with a volumetric compression ratio of 8.9, delivering about 10 H.P., 
showed air-card efficiencies of 39.75 per cent and 40.30 per cent, based on con- 
stant specific heat, and 29.69 per cent and 31.78 per cent, when based on his 
values of variable specific heat to which correspond the diagram factors of 
.665 and .655 for the former, and .89 and .83 for the latter. From the stand- 



HEAT AND WORK 1045 

point of constant specific heat the performance could be half as good again, 
but if the variability values are right the improvement possible is only about 
one-ninth better than was obtained, to be quite perfect. 

For these Otto cycle engines some little assistance in estimating the value 
of the diagram factor, is afforded by the determinations and the specific 
instances in the following Table CXXIX. Calculating from Meyer's 
test for a four-cycle engine 7.8X11.8 ins., 250 R.P.M., with four different 
compressions from 40 to 80 lbs. per square inch above atmosphere, Clerk 
finds the diagram factor practically constant at .58 even when the air-card 
efficiencies range from 33 per cent to 44 per cent. He also finds from 
Burstall's tests for the English Engineering Society Gas Engine Research 
Committee, on a four-cycle engine, 6X12 ins., 200 R.P.M., with compres- 
sions ranging from 30 to 105 lbs. per square inch above atmosphere, values 
of diagram factors from .38 to .59 while air-card efficiencies range between 
33 per cent and 47 per cent, though not regularly. He points out, however, 
a generally higher diagram factor for low maximum temperatures, the values 
ranging from .50 to .59, when the maximum temperature is about 1100° C. 
= 2012° F., whereas the factor lies between .40 and .50 for maximum tempera- 
tures of about 1750° C. = 3182° F. Clerk also reports values of the diagram 
factors of .6 and .66 respectively for a 10-in. National and a 26-in. Crossley 
engine and while the difference favors the larger engine it is not enough to warrant 
any conclusion. 

For some small gasoline engines, Callendar reports an increase of diagram 
factor with engine cylinder diameter as follows, .44 for 2.36 ins., .61 for 5.5 ins., 
.65 for 9.0 ins., .69 for 14.0 ins., but these figures do not apply generally as 
absolutely contradictory data are available, probably because of different values 
for other influences than cylinder size alone. Moreover, any values in excess 
of .6 for diagram factors for Otto engines must be regarded, to say the least, 
as abnormal. 

The load that a constant-speed engine carries will, of course, change its 
performance because the governing mechanism when in operation always 
introduces losses greater at one time than at another, and the hand-control 
mechanism does the same for variable-speed engines. Such variations as these 
are considered here as incidental rather than fundamental data, and a test series 
by Hopkinson on a four-cylinder Daimler automobile engine, 3.56X5.11 ins., 
will serve to illustrate the point. At speeds of 400, 600, 1000, 1100, and 1225 
R.P.M., the actual thermal efficiencies were 19.5 per cent, 21.5 per cent, 24.2 
per cent, 24.6 per cent, and 22.3 per cent, referred to I.H.P. As the com- 



\3.85/ 



pression ratio by volumes was 3.85, the cyclic efficiency is, 1 — lr"rr) =41.8 

\3.85/ 

per cent, so that at the successively higher speeds the diagram factors apparently 

were .48, .53, .58, .59, and .54. Really these are not the diagram factors, 

because at low speeds the engine was throttled and the cycle changed by the 

lowered suction line. 



1046 ENGINEERING THERMODYNAMICS 

Table CXXIX 
DIAGRAM FACTORS FOR OTTO CYCLE GAS ENGINES 



Engine. 



Size in Inches. 



Bore. Stroke. 



Test 
Authority. 



Compression. 



Vol. before 



Vol. after 



Pre ss, after 
Press. before 



Efficiencies, Per Cent 



Actual. 



Air Card 
Standard 



Dia- 



Four cycle. 



Four cycle. 



7.8 



40 H.P. four cycle 



Cockerill 

Delamarre 

Cockerill 

Letombe 

Winterthur 

Cie. Berlin Anhalt 

Benz 

Soest 

Deutz 

Tangye 

Fetu 

Schmitz 

Otto-Deutz 

Niel 

Winterthur 

Schmitz 

Winterthur 

Benier 

Tangye 

Dudbridge 

Tangye 

National 

Guldner 

Catteau 

Tangye 

Four cycle 



51 . 18 
33.465 
22.64 
23.622 
23.622 
20.47 
16.92 
16.73 
15.75 
14.173 
14.5 
13.78 
13.78 
13 

13.78 
12.2 
.85 



.8 



1 

1 

1 

1 

1] 

10 

10 

10 

9.85 

9.85 

9 

7 

6 

6 

6 

6 

6 



11 



12 
12 
12 
12 
12 
12 
12 
12 



55.07 

39.37 

37.4 

31.5 

31.5 

29.92 

27.56 

22 

22.83 

22.87 

22 

22 

21.26 

22.83 

19 

17.7 

18 

17.7 

17.3 

20 

18.6 

19 

19 

18 

15.75 

15.75 

18 

16 

12 

12 

12 

12 

12 



Meyer 



Burstall 



Hopkinson 



Hubert 

Witz 

Frangois 

Witz 

Allaire 

Witz 

Mathot 



6.37 



Witz 
Mathot 



Witz 
Mathot 



Witz 

Mathot 

Schrotter 

Witz 

Hirsch 

Burstall 



44 



78 



3.73-6.45 



3.03-8.13 
3.03-8.13 
3.03-8.13 
3.03-8.13 
3.03-8.13 
3.03-8.13 
3.03-8.13 
3.03-8.13 



9.18 

10.35 

5.8 

7.28 

8.03 

11.2 

8.17 

13.06 

7.35 

11.55 

4.83 

9.12 

9.12 

9.4 

11.58 

7.75 

11.3 

10.32 

4.39 

10.64 

4.83 

5.81 

6.8 

•5.88 

10.6 

10.6 

12.59 

10.2 



25 

24.4 

21.4 

18.8 

18.9 

21.2 

21.9 

23.1 

16.6 

18.7 

17.2 

18.1 
33.5-37.0 
depending 
upon load 

22.9 

25.0 

19.75 

24.3 

27.3 

25.6 

26.9 

23.8 

31.3 

30.4 

30.6 

18.0 

24.2 

38.8 

31.8 

31.6 

31.3 

25.2 

13.75 

29.8 

29.2 

27.4 

30.1 

21.2 

39.0 

33.9 

37.2 

25.8 

21.0 

18.0 

18.0 

17.6 

16.4 



44 
42 
37 
33 
33 
36 
43 
47 
33 
36 
43 
47 
52 



46.9 
48.7 
39.7' 
43.4 
45.0 
49.9 
45.2 
52.0 
43.6 
50.3 
36.2 
46.9 
46.9 
47.3 
50.4 
44.5 
50.1 
48.7 
34.3 
49.2 
36.4 
39.7 
42.4 
39.9 
49.1 
49.1 
51.5 
48.6 
42.8 
29.6 
42.8 
33.3 
32.7 



HEAT AND WORK 

Table CXXIX — Continued 

DIAGRAM FACTORS FOR OTTO CYCLE GAS ENGINES 



1047 







Size in 


Inches. 


Test 
Authority. 


Compression. 


Efficiencies, 


Per Cent. 


Dia- 


Eng : ne. 


Bore. 


Stroke. 


Vol. before Press, after 


Actual. 


Air Card 
Standard. 


Factor. 




Vol. after 


Press. before 




Four cycle 


G 


12 


< t 


2.04 




16.2 


34.6 


.468 






6 


12 


11 


2.17 




15.6 


26.2 


.595 






6 


12 


< i 


2.04 




15.5 


24.6 


.63 






6 


12 


l c 


2.04 




15.4 


24.6 


.626 






G 


12 


" 


2.78 




15.2 


33.3 


.456 






G 


12 


1 1 


2.7 




15.0 


32.7 


.459 






6 


12 


" 


2.44 




14.4 


29.6 


.486 






6 


12 


tl 


4.0 




13.6 


42.8 


.318 






6 


12 


{t 


4.0 




13.4 


42.8 


.313 






6 


12 


11 


1.75 




12.6 


19.5 


.646 






6 


12 


i ( 


2.7 




11.7 


32.7 


.358 






6 


12 


" 


2.22 




19.4 


26.9 


.721 






6 


12 


" 


2.94 




20.0 


35.0 


.572 






6 


12 


( ( 


4.0 




■ 22.7 


42.8 


.53 






H 


13 


Meyer 


3.75 




32.7 


41.2 


.794 






$h 


13 


" 


3.6 




26 . 8 


40.3 


.665 






8* 


13 


t i 


2.84 




20.2 


35.2 


.574 



Compression pressure ratio has been calculated assuming an initial pressure of 14.7 lbs 
per square inch. 



Examination of a very large number of tests showed that for various fuels 
used in typical Otto engines, the diagram factors lie somewhere within the 
following limits, for natural gas, coal gas, carburetted water, and producer 
gas, .45 and .55, for blast furnace gas, gasolene and kerosene vapors pre- 
viously made and taken in during suction, from .40 to .50, but when injected 
during compression, the mixture non-homogeneous, the factors may fall quite 
low or be very high, depending on circumstances difficult, if not impossible, 
to predict. These figures are for ordinary commercial engines, not for special 
engines operated under finest test conditions where the factor slightly exceeds .70. 

One most interesting and valuable series of tests for diagram factor is 
that conducted by Strong for the U.S.G. Survey on a 10-H.P. Nash and a 
15-H.P. Otto engine, using gasolene and alcohol, the results of which are shown 
in Fig. 276. In this series the compression was varied from 85 to 215 lbs. 
per square inch above atmosphere and both fuels were tested in each engine 
with most unusual accuracy of method. At 6 atmospheres absolute com- 
pression, the diagram factors were .65 for gasolene and .716 for alcohol, at 8 
atmospheres, .66 and .724, respectively, this being the maximum compression 
for gasolene, but at compressions of 11 and 15 atm. absolute for alcohol, the 
factors were .732 and .74. Strong finds that the efficiency is given by Eq. 
(1161). 



1048 



ENGINEERING THEEMODYNAMICS 



E (for gasolene) = 1 



(a) 



('^) 19 ( fc ) 



Th e'ore, the diagram factor is as given by Eqs. (1162) and (1163). 



(1161) 






F (for gasolene) 




5 10 15 

Compression Pressure in. Atmospheres Abs. 

Fig. 276. — Variation of Otto Cycle Gas Engine Diagram Factors with Compression, in Small 

Gasolene and Alcohol Engines. 

Diesel cycle engines have diagram factors at about their rated load, which 
corresponds to fuel admission for about 10 per cent of the stroke, of very 
closely, .50, though it must be admitted that there are insufficient data available 
to make this quite as well established as the factors for Otto engines. 



HEAT AND WORK 1049 

Prediction of thermal efficiency for a given engine is, through the use of 
established diagram factors reduced to a simple operation of finding the cyclic 
or air-card efficiency for the allowable compression pressure, and multiplying 
it by the factor. From this the heat consumption per hour and gas con- 
sumption follow directly. 

Let E = thermal efficiency of air card; 

Ei = thermal efficiency of engine referred to I.H.P. ; 
E B = thermal efficiency of engine referred to B.H.P, 

Em = mechanical efficiency = T ' ' ' ; 

i.ri.r. 

F = diagram factor ; 

i7 = B.T.U. per cubic foot gas, standard conditions of 29.92 ins. Hg and 
32° F. 

Then 

Ej = EF (1164) 

Eb=EjEm = EFE m (1165) 

B.T.U.perhr.perI.H.P.=^=ff. ...... (1166) 

B.T.U.perhr.perB.H.P. = ?g=^ = ^. . . (1167) 

Cu.ft. gas per hr. per I.H.P. (Std.) = ^ = ^i| (1168) 

2545 2545 2545 

Cu.ft. gas per hr. per B.H.P. (Std.) = # ^ = E 2 E M H = EFE M H ' ' ^ 1169 ^ 

That these equations may be numerically evaluated requires some data 
on mechanical efficiency and usual compressions in engines, which are 
given in the following Tables CXXX, and CXXXI, the former showing 
mechanical efficiency as a function of type of engine structure, and the latter 
allowable compressions as functions of the fuel class, from 2 atm. to 35 atm. 
The allowable compressions are always limited by the temperature of ignition 
of the mixture, when mixtures are under compression and by the effectiveness 
of the internal wall cooling. When only air is compressed, the compression 
may be anything but is generally limited by the cost consideration, high 



1050 



ENGINEERING THERMODYNAMICS 



compressions producing large working forces and requiring heavy and more 
expensive engine parts. Mechanical efficiencies include all losses, mechanical 
friction, fluid friction, the work of charging and exhausting, and in addition the 
precompression or pump work in two-cycle engines. 



Table CXXX 
MECHANICAL EFFICIENCIES OF GAS ENGINES =Em- 



B.H.P. 



I.H.P. (motor cylinders) 





Mechanical Efficiency. 


Type of Engine. 


Four Cycle. 


Two Cycle. 


Small high-speed automobile multi-cylinder, single acting 


.74 

.85 
.87 
.84 
.82 
.80 
.90 
.86 
.84 
.83 
.81 
.77 




" single-cylinder boat engine, single acting 


68 


' { or medium single-cylinder stationarv, single acting 


7 


' ' " ' * two cylinder stationarj^ single acting 




" " ' * three-cylinder stationary, single acting 




11 " " four-cylinder stationary, single acting 

Large single-cylinder stationary, single acting 


70 


• ' ' two-cylinder stationary, single acting 


to 


' ' four-cylinder stationarv, single acting 


80 


Double-acting single cylinder 


75 


' ' li tandem, two cylinder 


73 


" " " twin four cylinder 


69 







Table CXXXI 

ALLOWABLE COMPRESSION OF GAS ENGINES IN STANDARD ATMOSPHERES 

ABSOLUTE 



Fuel. 



Compression in Standard Atmospheres (Abs.) for Water- 
cooled Engines. 



Gasoline in suction mixture 

Kerosene in suction mixture 

Alcohol in suction mixture 

Oil injection in hot bulb before compression 

Oil injection after compression 

Carburetted water gas 

Coal gas 

Natural gas 

Producer gas 

Blast-furnace gas 

Coke oven gas 



6 to 8 

7 with water injection 

3 with hot mixture without water 

9 to 13 
4 

18 in Franchetti-Otto cycle engine 
35 in Diesel 
6 to 8 
6 to 9 

8 to 11 
8 to 11 

10 to 13 
8 to 10 



Predictions of gas-engine horse-power are best made by establishing values 
for the mean effective pressure of the whole cycle referred to one stroke, with 
which the horse-power of any gas engine, two cycle or four, and of any number 



HEAT AND WORK 



1051 



01 cylinders can be found, by multiplying this mean effective pressure by the piston 
area in square inches, the stroke in feet, the total number of cycles per minute in 
all working chambers, and dividing the product by 33,000. The prediction of 
mean effective pressure of gas-engine cycles is, therefore, a matter of con- 
siderable practical importance not only as directly fixing the horse-power, but 
as an indication of good working of the whole mechanism, since any loss in 
weight of charge by suction resistances, or heating on entrance, as well as 
any failure, to compress as much as the fuel will stand, or to realize the full 









































p 

m 








c 

I 










































































\ 






































\ 






































\ 






























20 








\ 


\ 




































\ 






i 


90 
[eatS 


B.T.I 
ippre 


J. per 
ssion- 


Cu.Ft 
Tactoi 


•--.45 






















\ 






































\ 


































































10 
















































































B 


\ 






































\ 


\ 


















D 
























































Attn 


ospher 


c Line 




















A 




















i 















5 10 

Volume in Cubic Feet 



V 15 



Fig. 277. — Otto Cycle Standard Reference Diagram without Suction or Exhaust Lines. 



temperature, pressure or volume increase on combustion, or to completely 
expand and expel burnt gases with least resistance, all these effects will be 
shown in reduced mean effective pressures. Unlike the case for steam engines, 
the mean effective pressure of gas engines is intimately associated with thermal 
losses and closely associated with thermal efficiency, so a purely thermal 
analysis of mean effective pressure of gas engines is necessary. 



1052 



ENGINEERING THERMODYNAMICS 



Referring to the ideal diagram of the cycle, Fig. 277, without any losses, 
except the heat suppression durring combustion, 

Let ( V a - V b ) = displacement = volume of fresh charge per cycle ; 

V b = clearance volume = volume of neutral gases added per cycle; 
W = foot-pounds work per cycle; 
E = air-card thermal efficiency; 
Ei = thermal efficiency referred to I.H.P.; 
Qi = heat supplied per cycle; 
(m.e.p.)'=mean effective pressure, pounds per square inch for com- 
pression and expansion strokes referred to one stroke; 
# = B.T.U. per cubic foot gas (std.), low value. 



FQi (1170) 



(1171) 



Then 
whence 

But 

Oi = 
V a - Vb Cu.f t fresh mixture taken in per cycle Displacement per cycle 

Therefore, if the working charge during suction be always at 32° F. and one 
standard atmosphere pressure, then 



7-1 

W = 144(m.e.p.) \V a - V b ) = 77SEFQi = 7881 - (j^ 

<».e.,r= 5 .4(pgv.) HIT .■ • • 

Heat liberated per cycle Heat per c ycle 



T , * = B.T.U. per cu.ft. fresh mixture (std.) = -—r 
V a —Vh a-p-i 



(1172) 



where a = cubic feet of air per cubic foot of gas for best mixture. 

Substituting Eq. (1172) in Eq. (1171), the mean effective pressure is given 

byEq. (1173). 

rz i 

/p \ y 



(m.e.p.)' = 5.4i^ 



1- 



I\ 



-»>M>-m 



(a) 



(1173) 



It appears from this equation that the same diagram factor F applies to 
mean effective pressure as to thermal efficiency. For convenience in cal- 
culation the values of the bracket are set down in Table CXXXII and 

TT 

and these are to be multiplied by — , values for which have already been 

calculated in the previous chapter for various typical gases, and by the diagram 
factor {F) applied to efficiencies previously discussed, 



. 






HEAT AND WORK 




1053 






Table CXXXII 






MEAN EFFECTIVE PRESSURE FACTORS FOR OTTO CYCLE ENGINES 


Px 
Pa 




[*-(£)*] 


«[i-©fl 


Px 
Pa 


(£)♦ 


h-on 


"[i-©fl 


Atmos 








Atmos 










1 


.0000 


.000 




.2082 


.4002 


2.519 


1.0 


1.000 






9 


9.318398-10 


9.008591-10 


0.401209 




.878 


.0508 


.274 




.1931 


.4821 


2.604 


1.2 


9.43442-10 


8.705522-10 


9.438140-10 


10 


9.285714-10 


9 . 083092-10 


.415710 




.786 


.0917 


.495 




.1804 


.4900 


2.680 


1.4 


9 . 895623-10 


8.962180-10 


9.094798-10 


11 


9.256148-10 


9.095450-10 


.428074 




.715 


.1256 


.079 




.1695 


.5083 


2.746 


1.6 


9.854271-10 


9.098990-10 


9.831008-10 


12 


9.229156-10 


9.700154-10 


.438172 




. 050 


.1540 


.835 




.1601 


.5195 


2.807 


1.8 


9.817662-10 


9.189181-10 


9.921799-10 


13 


9.204328-10 


9.715552-10 


.44S170 




.610 


.1797 


.971 




.1518 


.5295 


2.861 


2.0 


9.784979-10 


9.254451-10 


9.987009-10 


14 


9.181337-10 


9.723891-10 


.456509 




.569 


.2017 


1.090 




.1445 


.5378 


2.911 


2.2 


9.755412-10 


9.304706-10 


.037324 


15 


9.159935-10 


9.731355-10 


.463973 




.535 


.2213 


1.190 




. 1380 


.5471 


2.956 


2.4 


9.728421-10 


9.344981-10 


.077599 


16 


9.139914-10 


9.738099-10 


.470717 




.505 


.2389 


1.291 




.1322 


.5549 


2.998 


2.6 


9.703591-10 


9.378234-10 


.110842 


17 


9.121108-10 


9.744231-10 


.476849 




.479 


.2549 


1.377 




.1269 


.5021 


3.037 


2.8 


9.680601-10 


9.406300-10 


.138918 


18 


9.103376-10 


9.749837-10 


.482455 




.456 


. 2694 


1.450 




.1221 


.5088 


3.073 


3.0 


9.659199-10 


9.430398-10 


.163016 


19 


9.086604-10 


9.754990-10 


.487608 




.436 


.2827 


1.528 




.1177 


.5751 


3.107 


3.2 


9.639179-10 


9.451403-10 


. 184021 


20 


9.070693-10 


9.799751-10 


.492369 




.417 


.2951 


1.594 




.1136 


.5810 


3.139 


3.4 


9.620372-10 


9.469925-10 


. 202543 


21 


9.055556-10 


9.704109-10 


.496787 




401 


.3065 


1.656 




.1099 


.5805 


3.169 


3.6 


9.602641-10 


9.486402-10 


.219020 


22 


9.041126-10 


9.768283-10 


.500901 




.385 


.3171 


1.713 




.1005 


.5915 


3.197 


3.8 


9.585869-10 


9.501223-10 


.233841 


23 


9.027337-10 


9.772131-10 


.504749 




.372 


.3271 


1.767 




.1033 


.5967 


3.224 


4.0 


9.569957-10 


9.514615-10 


.247233 


24 


9.014135-10 


9.775741-10 


. 508359 




.359 


.3364 


1.817 




.1003 


.6014 


3.249 


4.2 


9.554822-10 


9.526804-10 


.259422 


25 


9.001471-10 


9.779127-10 


.511745 




.347 


.3451 


1.865 




.0970 


.0058 


3.273 


4.4 


9.540391-10 


9.537983-10 


.270601 


26 


8.989305-10 


9.782339-10 


.514947 




.336 


.3534 


1.909 




.0950 


.0100 


3.296 


4.6 


9.526601-10 


9.548255-10 


. 280873 


27 


8.977597-10 


9 . 785344-10 


.517962 




.327 


.3612 


1.952 




.0925 


.6141 


3.318 


4.8 


9.513399-10 


9.557760-10 


.290378 


28 


8.900310-10 


9.788204-10 


. 520822 




.3168 


.3086 


1.991 




.0902 


.6179 


3.338 


5.0 


9.500736-10 


9.560579-10 


.299197 


29 


8.955430-10 


9.790918-10 


.523536 




.30S0 


.3757 


2.030 




.0881 


.6215 


3.358 


5.2 


9.488569-10 


9.574794-10 


.307413 


30 


8.944914-10 


9.793504-10 


.526122 




.2998 


.3023 


2.066 




.0800 


.6251 


3.377 


5.4 


9.476861-10 


9.582401-10 


.315079 


31 


8.934741-10 


9.795963-10 


.528581 




.2921 


.3887 


2.100 




.0841 


.6285 


3.396 


5.6 


9.465580-10 


9.589048-10 


.322266 


32 


8.924893-10 


9.798305-10 


. 530923 




.2849 


.3948 


2.133 




.0823 


.6318 


3.413 


5.8 


9.454694-10 


9.590410-10 


. 329028 


33 


9.915347-10 


9.800546-10 


.533164 




.2781 


.4007 


2.165 




.0800 


.6349 


3.430 


6.0 


9.444178-10 


9.002770-10 


.335394 


34 


8.906086-10 


9.802691-10 


. 535309 




.2716 


.4003 


2.195 




.0789 


.6379 


3.446 


6.2 


9.434006-10 


9 . 008794-10 


.341414 


35 


8.897094-10 


9 . 804746-10 


. 537364 




.2656 


.4110 


2.224 




.0773 


.6408 


3.462 


6,4 


9.424157-10 


O. 014480-10 


.347104 


36 


8.888355-10 


9.807623-10 


.539341 




.2598 


.4108 


2.252 




.0758 


.6436 


3.477 


6.6 


9.414611-10 


9.619886-10 


.352504 


37 


8.879056-10 


9.808616-10 


.541234 




.2543 


.4217 


2.278 




.0744 


.6463 


3.492 


6.8 


9.405351-10 


9.625025-10 


.357643 


38 


8.871583-10 


9.810434-10 


. 543052 




.2491 


.4265 


2.304 




.0730 


.6489 


3.506 


7.0 


9.396359-10 


9.629909-10 


.362527 


39 


8.863525-10 


9.812192-10 


.544810 




.2441 


.4311 


2.329 




.0717 


.6514 


3 . 520 


7.2 


9.387620-10 


9.634558-10 


.367176 


40 


8.855671-10 


9.813881-10 


.546499 




.2394 


.4355 


2.353 




.0705 


.6539 


3.533 


7.4 


9.379120-10 


9.639008-10 


.371626 


41 


8.848011-10 


9.815511-10 


.548129 




.2349 


.4398 


2.376 




.0693 


.6563 


3.546 


7.6 


9.370847-10 


9.643265-10 


.375883 


42 


8.840733-10 


9.817082-10 


. 549700 




.2306 


.4439 


2.399 




.0681 


.6586 


3.558 


7.8 


9.362789-10 


9.647334-10 


.379952 


43 


8.833147-10 


9.818609-10 


.551227 




.2264 


.4480 


2.440 




.0670 


.6608 


3.570 


8.0 


9.354936-10 


9.651239-10 


.383857 


44 


8.826105-10 


9.820076-10 


. 552694 




.2225 


.4518 


2.446 




.0659 






8.2 


9.247276-10 


9 . 654984-10 


.387602 


45 


8.819134-10 








.2187 


.4556 


2.462 




.0649 






8.4 


9.339801-10 


9.658584-10 


.391202 


46 


8.812316-10 








.2150 


.4592 


2.481 




.0639 






8.6 


9.332501-10 


9 . 662040-10 


.394658 


47 


8.805644-10 








.2115 


.4628 


2.500 




.0630 






8.8 


9.325369-10 


9.665384-10 


.398002 


48 
49 


8.799114-10 

.0020 

8.792717-10 







1054 



ENGINEEKING THEEMODYNAMICS 



The actual indicator card is subject to losses in suction and exhausting 
which make it take on the form of Fig. 278, shown to a large scale incomplete 
and to a small scale complete. The exhaust stroke ends at H, at a pressure 
greater than atmosphere, so that no new charge at atmospheric pressure can enter 
till an expansion, actually or in effect, has lowered the pressure to that at point 
/. Here some volume of atmospheric fresh mixture begins to enter in effect, 
and this continues till point A has been reached, where compression begins at 
a pressure less than atmosphere. This resisted suction is characterized by a 





















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15 



Volume in Cubic Feet 
Fig. 278. — Otto Cycle Gas Engine Diagram with Suction and Exhaust Losses. 

negative mean suction pressure to be subtracted from the positive mean 
pressure of the upper loop. There is, however, another suction effect and that 
is a volume increase of the original fresh cool charge, due to mixture with hot 
burnt gases in the clearance and to its passage over hot metal surfaces, reducing 
the weight of the charge, and the real volumetric efficiency. Various tests 
have shown that the mixture suffers a rise of temperature during suction, 
ranging from 125° F. to 200° F., and mean suction resistances depending on 
speed and valve openings from less than 1 lb. square inch to over 5 lbs. These 
data together lead Gtildner to assign values of real volumetric efficiency 
according to the following Table CXXXIII with values of terminal suction 



HEAT AND WOEK 



1055 



pressure, p a , from which the mean suction resistance is estimated at two- 
thirds of the terminal. 

A similar situation exists for the exhaust stroke, where a mean resistance 
is encountered ranging from half a pound to three or four pounds per square 
inch, depending on the design and valve setting, and this is added to the 
suction resistance. The sum is to be subtracted from the mean pressure of 
the compressing and expanding area A BCD A of Fig. 277, to get the mean 
effective pressure of the whole, referred to one stroke, neglecting the small 
curved triangular area JKL in comparison with the rest. 

Table CXXXIII 

GULDNER'S VALUES OF OTTO ENGINE REAL VOLUMETRIC EFFICIENCY 
WITH ESTIMATED MEAN SUCTION RESISTANCES 



Type of Engine. 



Slow-speed mechanical valve 

Slow-speed automatic valve 

High-speed mechanical valve 

High-speed automatic valve 

Very high -speed air cooled automatic valve 



True Volumetric 
Efficiency EV. 



.88 to 
.80 to 
.78 to 
.65 to 
.50 to 



93 

.87 
.85 
.75 
.65 



Terminal Suction 
Pressure, Lbs. 
per Sq. In. Abs. 



12.8 to 13.7 
12.5 to 13.3 
11.8 to 12.5 
11.5 to 12.2 
8.8 to 11.0 



Estimated Mean 
Suction Pres- 
sure, Lbs. per 
Sq. In. • 



1.2 to .7 
4.5 to .9 
1.9 to 1.5 
2.1 to 1.3 
3.9 to 2.5 



If the gas and air mixture exist at a temperature external to the cylinder, 
other than 32° F. and at a pressure other than one standard atmosphere, then 
the heat per cubic foot of mixture measured external to the cylinder, will not 

TJ 

have the value previously assigned, but a correction can be applied. 

Let T m = absolute temperature F of the mixture external to the cylinder; 
" hm = absolute pressure of the mixture external to the cylinder, inches Hg 



Then 

| B.T.U. per cu.ft. of fresh 1 
\ mixture externally measured J 






. (1174) 



It has been shown that the mean pressure of the compression and expansion 
strokes referred to one stroke is given by an expression containing the term 



Q, 



Heat per cycle 



V a — Vb Displacement per cycle" 



1056 ENGINEERING THERMODYNAMICS 

But 
(Displacement per cycle) XE V = cu.it. externally measured mixture, per cycle 
or 

Qi I Heat per cycle 

= UivX 



Va—V b \Cu.ft. of externally measured mixture per cycle 

= ^yX(Heat per cu.ft. of mixture externally measured) 



= ME 



AM T t (n75) 



Substituting Eq. (1175) in Eq. (1171) giving the mean effective pressure 
for a compression and expansion stroke, and subtracting the mean resistance 
on suction p s and on exhaust p e there results for Otto cycle gas engines Eqs. 
(1176) and (1177): 

(m.e.p.)=5.4X.06^(^- 1 )g[l-(g)- 29 ]- Ps - P< . . (1176) 

=-™ F H^M l -(W 9 }-"<- ■ ■ ■ (n77) 

For Diesel cycle engines the following Eq. (1178) will serve: 

(m.e.p.)=M y X(m.e.p. of the cycle)— p s — p e . . . . (1178) 

As is the case with steam engines, the efficiency and fuel consumption 
of gas engines are not the same at all loads, even when expressed in terms 
of indicated horse-power and the variations are due to the kind of control 
used. There is, however, an interesting relation that should be noted, based 
on the fact that the thermal efficiency referred to I.H.P. would be constant 
if, (a) there were no suction and exhaust resistances ; (b) the ignition were 
always adjusted to give vertical explosion lines; (c) all the fuel used were 
burned at the right time. There are three typical methods of control, each 
yielding a characteristic fuel or heat-consumption load curve, the quality 
control mainly used in oil engines, the hit and miss in cheap engines that need 
not regulate closely, and the suction throttle, with sometimes a fourth, suction 
cut-off. The first two have always the same compression characteristics and 
valve resistances, while with the throttle resistance varies all the time with 
load so it would be expected that two characteristic curves would result. For 
the former, fuel per hour or B.T.U. supplied per hour, gives a straight line 
when plotted to load, except where, at overload, excess and unburnt fuel 
causes a change of curvature. This is equivalent in steam engines and tur- 
bines to the straight Willans line and the curve which characterizes the 
throttle type throughout is expressed as a second degree equation similar to 



HEAT AND WORK 



1057 



that found for cut-off- governed steam engines. Several efficiency curves of 
each engine type are shown in Fig. 279 on actual and in Fig. 280 on a fractional- 
load basis to bring all sizes in accord. From these fuel consumption or heat- 
supplied curves could be found having equations of the first degree if straight 
or of the second degree otherwise, but this is omitted here to save space. 

Any examination of the question of gas-engine efficiency would be incomplete 
without at least a brief review of the ultimate disposition of thle heat 
supplied, that does not become converted into work. The heat balance of gas 
engines and complete plants is the term applied to a tabular or graphic state- 
ment of the disposition of the heat supplied in fuel, in the form of gas or oil 
for engines, or in the form of coal or coke for gas producer and engine plants 
together. These balances are, of course, different for different sizes and styles 
of engines, for different fuels and engine loads, but more widely different for 
complete plants of many units operating on a fluctuating load. In a recent 
paper by Andrus and Porter some data were presented showing for average 
plant working conditions in England that the overall efficiency of complete 
gas plants averages about 14 per cent from useful power to heat of coal, and 
over periods ranging from one month to a year. The minimum reported 
was 12.4 per cent and the maximum 15.05 per cent. They also compared 
similar data for steam plants which showed average of 6.83 per cent overall 
efficiency for periods of one to twelve months, but here the lowest was only 
3.70 per cent and the highest 9.30 per cent, showing in round numbers about 
twice as -good a return for gas systems as for steam. Comparing engines alone 
it will be found that the proportion is different, charging heat in steam or 
in gas as heat supplied, the performances are closer on the average, though 
the best gas or oil engine performance remains about twice as good as the 
best steam engine or turbine performance, in efficiency. 

One interesting overall estimate of heat balance by Coster for marine 
plants is as follows, Table CXXXIV, 

Table CXXXIV 
COMPARATIVE HEAT BALANCES OF PRODUCER AND ENGINE PLANTS 





Gas Producer 
Plant. 


Steam Plants. 




Turbine. 


Reciprocating 
Engines. 


B H.P 


25,0 
1.4 

30.4 

20.0 
3.2 

20.0 


13.0 

1.0 

56.0 

30.0 


10.8 
1 2 


Friction loss 


Exhaust loss 


58 


Gas-engine jacket loss 




Gas-engine radiation, etc 




Producer or boiler loss 


30.0 






100.00 


100.0 


100.0 



1058 



ENGINEERING THERMODYNAMICS 



























































References 

1. Westinghouse-3 Cyl.Vert.-Natural Gas -Mixture Throttled 

B.H.P. 100 R.P.M. 270- Robertson, 1899 

2. Nurnberg -Blast Furnace Gas - Gas Throttled 

B.H.P. 1200 R.P.M. 106 -Riedler, 1905 

3. Diesel -Kerosene- Cut-off Governor- B.P.H.70-R. P.M. 158 -Meyer, 1903 

4. Banki -Gasoline with Water Injection- Hit and Miss Governor 

B.H.P. 25 - R.P.M. 210- Jonas and Tahorshy, 1900 

5. Charon- Illuminating Gas - B.H.P. 50 - Allaire, 189t 

6. CockerhiM Blast Furnace Gas- B.H.P. 600 - Hubert. 1900 

7. Westinghouse - Horiz. double acting - Producer Gas - 

B.H.P. 500 - R.P.M. 150- Alden and Bibbins, 1907 

8. Hornsby-Akroyd - Kerosene - B.H.P. 10 - R.P.M. 260 

9. DeLa Vergne F.H. - Solar Fuel Oil - B.H.P. 85 

R.P.M. 180 - Towl, 1911 

10. Piece - Arrow - Gasoline - B..H..P. 48 - R. P. M. - variable Chase, 1912 

11. Franklin- Air - Cooled- Gasoline - B.H.P. 20 -R.P.M.- variable 

Evans and Lay, 1909 












































































































































































































































































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HEAT AND WORK 



1059 





















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'd'H'a 03 paaaajay; Queo aaj) ^ouatoHjg; itjraaeqi, 



1060 



ENGINEERING THERMODYNAMICS 



The numbers give the per cent of the heat of coal converted into work, 
or otherwise disposed. Two balances, one by Bibbins and one by Mathot, of 
a gas producer plant complete are given below. Table CXXXV. 

j ' ! I j Table CXXXV 

HEAT BALANCES OF GAS PRODUCER PLANTS (Per Cent of Coal Heat) 



Authority. 


B.H.P. 


Friction and 
Pumps. 


Exhaust and 
k Radiation. 


Jacket. 


Producer Loss. 


Bibbins 


18.36 
21.00 


3.37 
4.00 


28.81 
23.00 


25.22 
27.00 


26.22 


Mathot 


25.00 







In the following Table CXXXVI the heat distribution for several different 
engines alone are compared to show the range of variations but without any 
idea of explanation of differences. 

Table CXXXVI 
HEAT BALANCES OF GAS AND OIL ENGINES (Per Cent of Gas or Oil Heat) 



Engine and Authority. 



I.H.P. 



B.H.P. 



Friction. 



Exhaust. 



Jacket. 



Radiation 
and Un- 
accounted 
for. 



Donkiii 

Beck engine, Kennedy 

Griffin engine, Kennedy ........ 

Atkinson engine, Kennedy .... 

Otto Crossley engine, Kennedy. 



R.P.M. 
187 
247 

187 
247 



a/Q 

7.11, Slaby 
7.35, Slabv 
7.43, Slaby 
7.40, Slaby 



Comp. Ratio. 

2:67 

2,67 

4.32 

4.32 

General, Mathot 

Westinghouse, Bibbins 

300 H.P. engine at 197 HP., Eberlv 
294 H.P., Eberly 
335 H.P. , Eberly 

6 H.P. engine, LC.E 

24 H.P. engine, LC.E ,. 

Deutz 2 H.P., Wimplinger 

Giildner 20 HP., Schroter 

Walrath 75 H.P, Geer and Yane- 

lain 

300 H.P, Goldsmith and Hart- 
wig . 

Hornsby, Robinson 

De la Vergne F. H, Towl 

Pierce- Arrow, Chase 



22.32 

19.4 

21.1 

25.5 

22.1 



18.0 

18.1 

24.4 

23.7 

33.0 

29.48 

43.5 

45.8 

41.5 

31.8 

33.3 

21.5 

42.7 

27.1 

24.4 

21 

40.14 



28.0 
24.9 
33.5 
32.2 
30.9 
26.7 
28.3 
16.1 



21.3 

17.1 

-18 

27.52 

18 



5.0 

4.58 
10.0] 
13.6 | 
10.6J 

5.1 

5 

5.4 



5.8 

7.3 
3 

12.62 



43.29 

42.9 

39.8 

37.9 

35.5 



30.8 

36.3 

21.8 

26.8 

31. H 

36. 

24. 

23. 

24. 

41. 

37 

25 

24 



23.4 

50.6 

29 

20.03 



32.96 

33.0 

35.2 

27.0 

43.2 

51.2 

45.6 

53.8 

49.5 

36.0 

34.22 

34.3 

31.8 

33.8 

27 . 1 

29.6 

50.4 

33.2 



1.43 

4.7 

3.9 

9.6 

.8 excess 



\t 



1.9 excess 
1.5 " 
.1 " 



3.1 



49.5 

25.0 
50 

26.50 
29.4 



13.33 



* Including radiation, f Inclrdirg pumps. J Including external radiation. 



HEAT AND WOKK 1061 

There is no doubt that one of the prime factors in the failure of the per- 
formance of real gas engines, to agree with that of the corresponding cycle, 
is the combined effect of leakage and heat exchanges between walls and 
working gases, and it is certain that all jacket loss is derivable from heat trans- 
mitted from gases to walls. These exchanges have been much studied, but 
it must be confessed with as yet little conclusive result. Therefore, these 
results will be omitted but the most promising mode of attack is given below, 
with, however, the warning that it is quite useless if any leakage is taking 
place, and if the physical constants of the gases are not established. Just as 
investigators have been misled by air-compressor lines into believing leakage 
to be evidence of cylinder-wall cooling, so have gas engine investigators been 
finding heat losses from pressure volume measurements, which are nothing 
more than leaks. One additional cause of difficulty with the gas engine and 
that which makes the T equivalent of its PV product of, to say the least, 
doubtful value is the non-homogeneity of the charge, both chemically and 
physically, a fact often forgotten. 

If, however, all these precautions are observed, then the gains and losses 
of heat by the gas may be found by studying the PV lines in accordance with 
the following algebraic relations: 

dH = C v dT+PdV. 

Therefore, 

dV~ Lv dV^- 



But 

dV RV ' ' dV 



T =j(p+v d 



Whence, if dH is in foot-pounds, 



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'P-NT^l-! 



T-1V ' dV 



=M yP+v ^)- ••••••• $»> 



Accordingly, if on any part of the PV diagram or indicator card the pressure 
P, and volume V, be known, and the change of pressure dP to the change of 



1062 ENGINEERING THERMODYNAMICS 

volume dV measured a short distance on each side of the point, the heat change 
for that particular volume change is given by the above equation. It must be 
noted, however, that this is true for constant specific heats only, but even 
so, it is of some value and changes in sign indicate whether heat is being 
received (+), or lost ( — ), by the gas at that time. 

Prob. 1. An engine operates on the Otto cycle and the pressures after and before 
compression are 85 lbs. and 14 lbs. per square inch gage. The diagram factor being 
taken as .5 and the mechanical efficiency as .8, what is likely to be the hourly consump- 
tion per B.H.P., of gas having a heating value of 600 B.T.U. 

Prob. 2. For a large 4-cycle single-cylinder stationary engine operating on natural 
gas, find the probable consumption of gas per B.H.P. per hour, taking the necessary 
information from Tables CXXIX and CXXX and assuming 900 B.T.U. as the heat- 
ing value of the gas. 

Prob. 3. An Otto cycle engine works with a ratio of pressures of 5 and uses producer 
gas No. 6 of the producer gas table of Chapter V. What will be the horse-power if 
this engine is a single-acting 2-cycle one, running at 150 R.P.M. and having a cylinder 
20X30 ins.? 

Prob. 4. Estimate the horse-power and fuel consumption for a gas engine of any type 
and size on any fuel. 

15. Actual Performance of Piston Steam Engines and Steam Turbines 
at Their Best Load and its Relation to the Cyclic. Effect on Efficiency of 
Initial Pressure, Vacuum, Superheat, Jacketing, and Reheating. Heat 
Balances of Steam Power Plants. Steam plants including boilers, engines 
and auxiliaries, are very much more complex both structurally and with 
reference to thermal changes than are gas plants, so it is not surprising that 
analysis of thermal losses and heat distribution is more difficult. There are 
hundreds of thousands of these plants in existence bearing little or no similarity 
in detail in the stationary class, though the representatives of the locomo- 
tives and marine groups do fairly well agree with their respective type forms 
though less so to-day than a few years ago. It would be a very lengthy and not 
very profitable procedure to study analytically all existing test data of steam 
plants and of the separate component units, so the treatment of the subject 
will be mainly confined to methods of determining the influence of controlling 
factors with just enough data to serve as illustrations. 

The overall heat balance will, of course, be most variable, but the com- 
parison of the balance of a large street railway steam central station, and 
a locomotive is given in the following Table CXXX VII, together with 
Gebhardt's estimates of a small non-condensing and a simple condensing 
stationary plant. 

Those losses that occur in boilers have already been examined and plant 
auxiliary requirements are items that need no separate detailed thermodynamic 
analysis, but the performance of the engine or turbine does. It is not so much with 
the actual efficiency of a given steam engine that this analysis is concerned but 
rather with the best obtainable efficiency under whatever conditions prevailed, 



HEAT AND WOEK 



1063 



Table CXXXVII 
STEAM PLANT HEAT BALANCES 







Stationary Plant. 








Simple Non- 
Condensing, 
Gebhardt. 


Refined 
Condensing, 
Gebhardt. 


Large Central 
Station Con- 
densing, Stott. 


Locomotive, 
Goss. 


4 

Suppliec 
Boiler 


1 in coal 


1.00 


.50 


1.00 


.180 
.050 
.070 


1.00 


.227 
.024 
.080 


1.00 




f Sensible heat of stack. . . 

1 Unburned fuel loss 

1 Radiation, etc 


.19 
.17 
.07 


Piping 

and 

auxil 


Used for boiler feed 

Used by condenser pumps 
Piping and auxiliary 
losses . . 


:::: 




.162 


.010 
.024 

.021 


.099 


.014 
.016 

.023 






iaries 


Returned by feed heater 
and economizers 








~ Discharged by engine 
exhaust 




.022 




.624 
.183 
.037 
.146 




.601 
.114 
.008 
.106 






Engine • 


Converted into I.H.P. . . 

Engine friction 

. Converted into B.H.P. . 




.054 


Totals (I.H.P. excluded) 


1.00 




1.162 


1.162 


1.099 


1.099 


1.00 





with a view of making clear a policy or the value of imposing one or another 
condition of service. It has already been shown that the steam consumption 
per H.P. hour varies with load according to a characteristic curve always 
having a minimum point, and since efficiency is directly related, so must^very 
steam engine have a best efficiency at some load. This best efficiency is 
worthy of study and as was done with the gas engines, the method to be 
used is that of comparison with the corresponding cyclic efficiency, using 
for this purpose the Rankine cycle. This comparison will show, first, 
the approach to perfection of the actual machine, but also it will demon- 
strate, secondly, the specific value of high initial pressures, low terminal 
pressures, superheat or initial wetness. Exactly similar methods of comparison 
will demonstrate the value of reheat in compound engines and throw some 
light on the desirability or undesirability of using jackets on steam cylinders 
of piston engines, and on the limiting conditions of staging and vane speed 
of turbines. 

The first step in this analysis is to set down some tests of steam engines, 
giving their best load thermal efficiency with the initial and final conditions of 
the steam, and by comparison with the Rankine cycle establish some cyclic 
efficiency factors. A number of these, for both piston and turbine engines 
are given in Table CXXXVIII. 



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1068 ENGINEERING THERMODYNAMICS 

Examination of this table brings out clearly two important characteristics 
of steam engines and turbines, first, that the efficiency factors for both classes 
are substantially the same and approach 80 per cent as a high limit, which 
is better by about 15 per cent than that found for gas engines; second, that 
the factor may be very low, less than 50 per cent. This is due to the fact 
that piston engines leak, suffer heat exchanges, and expansion is limited by 
the valve adjustment and cylinder ratios, and in turbines whereas expansion 
is always complete, the vane and steam jet speed relations may be improper 
or friction losses occur in the flow path. Where such turbine losses exist 
they differ not so much in range from the different losses in piston engines 
even at best load. It is most interesting to note that so far as capabilities 
of conversion of steam heat into work are concerned, as measured by this factor 
there is little to choose between these two classes of machines, though piston 
engines appear to be able to make a little better use of the steam than tur- 
bines, when constructed to do so, and allowed to in service. 

In all cases where cylinder leakage, heat exchange, and expansion losses, 
or turbine velocity and flow friction losses, are not intentionally suffered to meet 
load or first-cost conditions, the factor for steam engines and turbines is con- 
sistently and materially higher than in gas engines. This indicates that there 
is much more room for improvement of efficiency in gas than in steam engines, 
though it is not yet clear whether the present modes of operation forbid its 
realization or not. The doubt is, however, all on the side of the gas engine 
as the nature of the losses in steam machines are pretty well understood and 
the value of the various factors that control them have been experimentally 
determined many times. 

It takes no elaborate thermal study to see that the higher the initial 
steam pressure and the lower the terminal pressure the better should be the 
efficiency and it remains to see whether it is really so. As experimental data 
to illustrate this point the test results of Goss on locomotives will be used. 
Several series of runs are available, some with superheated steam and some 
with saturated, each with boiler pressures from 100 to 250 lbs. per square inch 
gage, some of which are shown graphically in Fig. 281, on which lines are located 
to indicate the Rankine cycle efficiency and water rate at the left. These 
results are replotted in the center, to a scale of efficiency factors, as functions 
of boiler pressure, but it must be remembered that in no case is expansion 
complete in a locomotive and the degree must be less for the higher pressures. 
The corresponding T<& diagrams are also given at the right. Examination 
of these curves indicates clearly that no great improvement in efficiency or 
water rate follows the pressure rise, and what changes occur are of the order 
of first decreasing steam consumption and later increasing it, the minimum 
in one case being at 200 lbs., per sq. inch gage and in the other a little less with 
the exception of the saturated steam which consistently improves in performance. 
Of course, the Rankine cycle regularly improves so it appears that as pressures 
rise the performance departs more and more from that of the Rankine cycle, 
the factor being highest at 140 lbs., 64 per cent for superheated, and at 100 



HEAT AND WORK 



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1070 



ENGINEERING THERMODYNAMICS 



lbs., 55 per cent for saturated steam, proving an increase in losses due to 
leakage, initial condensation, and curtailed expansion. To really prove the 
value of high initial pressure would require test data of several engines each 
designed for a successively higher pressure and operated under its own best 
conditions, but no such data are available. The real limit to high pressures is 
not thermal, but one of expense, gains due to excess pressures over 175 lbs. for 



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Fig. 282. 



-Effect on Steam Engine Performance of Raising the Initial Steam Pressure 5 
and Lowering the Back Pressure by 5 lbs. per sq.in. 



stationary plants and 200 lbs. per sq.in. gage for locomotives are obtained at 
too great cost generally to warrant it. 

Lowering the back pressure should have the same effect as raising the 
initial pressure, in improving efficiency, though to a very much greater degree 
as appears from both the pressure volume and temperature entropy diagrams, 
Fig. 282. These are drawn for adiabatic expansion from 175 lbs. per square inch 
initial pressure absolute with dry saturated steam, to one atmosphere, ABCD 



HEAT AND WORE: 



1071 



and for a 5 lb. per square inch rise of initial pressure there is added the area 
BCEF, which is small compared to that obtained from a 5 lb. per square inch 
reduction of back pressure, given by area ADGH. Whether a real engine is 
capable of taking advantage of this or not depends on its structure and 
i adjustment. For example, if the cylinder volume is represented by AD' 
on the PV diagram, the gain due to reduced back pressure will be AD'G'H 
instead of ADGH, since expansion must stop at D", so that piston engines 
that ordinarily do not completely expand the steam, can hardly be expected 
to utilize all the available gain. The situation is better for turbines where 
the expansion is always complete, but even here the full advantage is not 
realizable, unless the vane speeds and steam-jet velocities are correctly related. 
If so related for the higher back pressure, a reduction of it will result in higher 
jet speeds and unless vane speeds and R.P.M. correspondingly increase, the 
steam will leave the vanes with some residual velocity and some kinetic energy 
not utilized. 

All experimental data prove that both piston engines and turbines do gain 
in efficiency with lowered back pressure and very much more than with 
corresponding increase of initial pressure, the amounts, of course, depending 
on the pressures themselves, and it is also experimentally established that 
turbines are benefited more than piston engines. To illustrate the effect 
of back pressure on a turbine, the test data of a 1250 K.W. Westinghouse 
turbine is given below in Table CXXXIX and Fig. 283, and the corresponding 
Rankine cycle efficiencies and cyclic efficiency ratio calculated and added. 

Table CXXXIX 

STEAM TURBINE EFFICIENCY WITH VARYING BACK PRESSURE WITH STEAM 
APPROXIMATELY AT CONSTANT INITIAL PRESSURE DRY SATURATED 



Initial pressure pounds per square inch absolute 

Back pressure inches Hg absolute 

Pounds of steam per B.H.P. hour (best) 

Pounds of steam per I.H.P. hour for 94% mech. efficiency 

Actual thermal efficiency referred to I.H.P. per cent 

Rankine cycle efficiency per cent 

/ Actual efficiency \ 

\Rankine cycle efficiency/ 



161 
5 
14.39 
13.53 
17.22 
25.30 

68.05 



162 
4 
14.10 
13.25 
17.42 
26.18 

66.56 



161 
3 

13.59 

12.78 
17.92 
27.08 

66.16 



161.6 

2 
13.49 
12.68 
17.84 
28.64 

62.29 



The curves of actual and Rankine cycle efficiency are not parallel, so the 
efficiency factor varies from 62 per cent for 2 ins. Hg to 68 per cent for 5 ins. 
Hg, indicating that not so much use is made of the high velocities of the lower 
back pressure as of the lesser jet velocities of the higher back pressure. 

There are no similar tests available for reciprocating engines but the 
performance of a large one, 28X54X48 ins., non-condensing and with 
a 26-inch vacuum, will serve to bring out its ability to utilize as much of the 
available gain as the turbine, in this particular case, more. This engine 
required 19.4 and 13.65 lbs. of steam per hour per I.H.P. for non-condensing 



1072 



ENGINEERING THERMODYNAMICS 



































































































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HEAT AND WORK 1073 

and condensing operation with a back pressure of one atmosphere and 4 ins. Hg 
absolute respectively, and an initial pressure of 165 lbs. per square inch 
absolute. These water rates for initially dry steam correspond to actual 
thermal efficiencies of 12.86 per cent, and 16.83 per cent, for which the 
equivalent Rankine cycle efficiencies are 17.3 per cent, and 26.29 per cent, 
so that the ratio 

/Actual thermal efficiency\ 

\Rankine cycle efficiency/ 

is equal to .743 for one atmosphere back pressure, and .644 for 4 ins. Hg 
absolute back pressure. 

The turbine is not only able to utilize a fairly large fraction of the available 
gain due to reduction of back pressure, but is very much cheaper to build than 
piston engines, especially for very low back pressures which would require 
in piston engines abnormally big low-pressure cylinders. Thus the practice 
has been established of using turbines for the low ranges of pressure in preference 
to piston engines, and of combining reciprocating engines and turbines to 
work together, the latter taking the exhaust steam from the former at what- 
ever release pressure it is discharged. This practice has been found especially 
valuable in adding capacity to existing overloaded reciprocating engines, and the 
most notable example of it is the installation at the 59th Street power station of 
the Interborough Ry. in New York City. Here the original installation con- 
sisted of 15,000 K.W. twin horizontal- vertical compound engines 42X86X60" 
ins. Need for more power led to an investigation of the relative advantages 
of installing some new high-pressure turbines, or of adding low-pressure turbines 
to existing reciprocating engines, and the latter was adopted on a basis of 
manufacturers' guarantees for both, as it offered 8 per cent better efficiency 
than the former, and a 7500 K.W. low-pressure turbine was added to one 
piston engine unit for trial. Most unusually complete tests were conducted 
which proved beyond question the value of the combination, an average 
thermal efficiency between loads of 6500 K.W. and 15,500 K.W., of 20.6 per 
cent was obtained. The engine alone showed a best thermal efficiency of 
10.3 per cent on 177.7 lbs. per square inch initial pressure gage, with 8.9° F. 
superheat, and a back pressure of a little less than 2 ins. Hg absolute, for which 
the Rankine cycle efficiency is 15.6 per cent, so that the engine realized at 
best 66.2 per cent of the Rankine cycle. When operated together, however, 
the rise of back pressure on the engine enabled it to realize a larger per cent 
of the Rankine cycle, the figure rising to 79 per cent, while the best that was 
obtained for the low-pressure turbine was about 64 per cent, though not at 
the same load. Operated together, the overall performance was the best 
on record, the thermal efficiency being 21.8 per cent, which corresponds to 
about 78 per cent of the Rankine possibilities, a truly remarkable performance 
probably not exceeded in service by any single unit of whatever type. 
At this time the initial pressure was 199.1 lbs. per square inch absolute for the 



1074 



ENGINEERING THERMODYNAMICS 



engine, and 10.35 lbs. per square inch absolute for the turbine with a final 
back pressure of .46 lb. per square inch absolute, the initial steam being 
about 1 per cent wet. 

The value of a piston engine for steam in the lower pressure ranges has 
been experimentally established by Carpenter and Sawdon on a Shuman 
24X24 ins. simple engine with the following results, Table CXL. Allowing for 
experimental inaccuracies it is reasonable to assume that the efficiency ratio 
is constant at about 50 per cent, which is considerably lower than is easily 
obtainable in low-pressure turbines. 

Table CXL 
USE OF LOW-PRESSURE STEAM IN PISTON ENGINES 



1 



6 



Initial pressure pounds per square inch abs 

Initial steam quality (per cent dry) 

Back pressure, inches Hg absolute 

R.P.M 

B.H.P 

Pounds dry steam per hour per B.H.P. . . 

Actual thermal efficiency (per cent) 

P . /Actual thermal efficiency \ 
\ Rankine cycle efficiency / 



8.12 
89.4 
1.92 
126.6 
11.3 
36.8 
6.40 
.524 



8.15 
97.7 
1.90 
143.8 
12.1 
43.0 
5.51 
.438 



15.56 
90.4 
1.31 
129.5 
12.6 
35.7 
7.57 
.479 



15.67 
95.6 
1.98 
143.7 
18.5 
31.0 
7.43 
.467 



15.8 
96.9 
1.90 
133 
22.2 
28.8 
8.22 
.509 



15.8 
97.2 

1.83 
126.9 
21.0 
29.4 
7.97 
.499 



Superheat is another important factor in steam-engine efficiency, as exam- 
ination of the T$ diagram will show, though curiously enough and uniquely the 
actual benefits sometimes exceed the expected. In Fig. 284 is shown a series of 
Rankine cycles for 190 lbs. per square inch absolute initial, and 2 ins. Hg absolute 
back pressure, for every 50° of superheat up to 250°, and the value of the super- 
heat is shown at the bottom as computed in two different ways. The curve 
(1) shows the efficiency of the whole cycle ABCNA to ABCHIKA, and indicates 
a slow and regular improvement with increase of superheat from 29J per cent 
efficiency for zero superheat to 31 per cent efficiency for 270° superheat. The 
other curve (2) gives the efficiency of the superheat part of the cycle alone as 
represented by the areas CDMNC to CHINC, and this also increases with the 
amount of superheat and faster than the whole Rankine cycle of which it is a 
part. According to this, a gain in thermal efficiency with increase of super- 
heat is to be expected very nearly proportional to the superheat as line (1) 
is nearly straight, but as maintenance costs increase rapidly beyond a moderate 
superheat practice has fixed on values about 100°, occasionally going as high as 
150°, as the economic limit. 

These expected gains must now be compared with actual ones and as 
data some of the tests of locomotives will be used as representing the piston 
engine, and compared with Hodgkinson's turbine tests. The locomotive 
tests showed an improvement in efficiency, depending on the boiler pres- 



HEAT AND WORK 



1075 



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Superheat (Deg. Fahr.) 



250 c 



Fig. 284. — Effect of Superheat on Efficiency of Transforming Heat of Steam into Work. 



1076 



ENGINEERING THERMODYNAMICS 



sure, of 2 per cent for 120 lbs. gage pressure, 2.3 per cent for 180 lbs. gage, 
and only 1.3 per cent for 240 lbs. as would be expected, though another factor 
of variability of amount of superheat is introduced by the peculiarities of the 
locomotive structure. All gains are based on the performance with saturated 
steam as a standard of comparison. The superheat varies in these machines 
with the boiler pressure and rate of boiler firing and in these tests ran over 
and under 150°. The net results are given in Fig. 285, and compared with 
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150 200 250 

Steam Pressure, Lbs. per Sq. In. Gage 



100 150 200 250 

Steam Pressure, Lbs. per Sq. In. Gage 



Fig. 285.— Effect of Superheat on Locomotive Engine Efficiency. 



superheat are much higher, and also shown by the greater distance between 
the actual and the Rankine curves for saturated and superheated conditions. 
This is a most interesting observation, usually explained on the theory of 
lessened leakage with superheated steam, and less wall heat absorption due 
to the insulating action of a layer of gaseous steam which is always more 
effective than a film of water, as has already been established. 

Somewhat better data are available for showing the improvement in 
efficiency due to superheat with turbines, as tests are available at constant 
pressures with superheat as the only variable and Hodgkinson's results for a 



HEAT AND WORK 1077 

1250 K.W. Westinghouse turbine will illustrate this. With an initial pressure 
of 162 lbs. per square inch absolute and a back pressure of 2 ins. Hg absolute 
♦ the water rate per I.H.P. hour on an assumed mechanical efficiency of 94 
; per cent was for dry saturated steam, 12.7 lbs., 12.4 for 75° superheat, and for 
100° superheat 12.05 lbs. These data correspond to actual thermal efficiencies 
of 17.80 per cent, 17.55 per cent, and 17.86 per cent and the corresponding 
Rankine cycle efficiencies are 28.68 per cent, 28.90 per cent, and 29.01 per cent, 
so that the cyclic efficiency ratios are 0.6206, 0.6073 and 0.6156. 

The use of superheated steam in small engines according to some tests 
analyzed by Tolz showed improvement in efficiency with rise of initial tempera- 
ture according to the curves of Fig. 286 for four engines, though the initial 
steam temperature instead of the superheat itself is given as the prime variable. 
Tests on the use of superheated steam in pumping engines has led to the 
estimates by Foster of a saving over saturated steam, depending on the 
efficiency of the unit previously, as follows: 6, 10, 20, 40 per cent, when the duty 
is 150, 100, 50, and 10 million foot-pounds useful pump work per million heat 
units supplied. These figures are higher than those of the Jacobus tests of 
a New York pumping unit with and without superheated steam, which showed 
a gain of 18.5 per cent with 207° superheat and 80 lbs. gage pressure when 
the duty was about 28 million, based on heat in the steam. 

The fact that changes in initial pressure, back pressure and superheat 
are associated in any one type of engine with a definite and regularly varying 
change in efficiency has led builders to adopt curves of correction per pound 
per square inch pressure or per degree superheat change, for their machines. 
Thus Emmet states with respect to vacuum, that near 27 ins. Hg the change 
in efficiency per inch is 6.6 per cent, near 28 ins., 7.8 per cent; near 29 ins., 
9.5 per cent; and according to Parsons each inch between 23 and 28 ins. affects 
efficiency 3 per cent in a 100-K.W., 4 per cent in a 500-K.W., and 5 per cent 
in a 1500-K.W. turbine. If the turbine is correctly designed for best economy 
the gain per inch of vacuum is almost exactly that for the Rankine cycle, but 
if not so designed, no prediction can be made. It is pretty well agreed that 
superheat corrections are properly fixed for prevailing pressures at about 
10 per cent per 100° F. superheat, up to about 100° F. superheat as a max- 
imum, but it does fall off in the higher ranges, being about 8 per cent per 100° 
between 100° and 150°. However, as superheat affects steam jet velocity, just as 
does high initial or low back pressure, it is clear that the realization must de- 
pend on design proportions. With respect to these corrections, therefore, this 
discussion must be considered merely as pointing out a method for their 
determination rather than fixing them in general, as this latter can be done only 
for particular speeds and dimensions. 

The next factor in steam engine efficiency is the cylinder jacket of piston 
engines, the analysis of which cannot be based on the Rankine cycle. It is 
expected that the jacket will impart heat to the working steam during expan- 
sion by condensing some live steam outside the cylinder wall. Exact analysis 
thermally, of this sort of expansion is quite impossible because the law of heat 



1073 



ENGINEEEING THERMODYNAMICS 



gain as expansion proceeds in the working steam depends on laws of transfer 
that are not established. Sometimes it is assumed that heat will be received 
at a rate to keep the working steam at constant quality and this will serve to 







400 500 600 

Initial Steam Temperature Deg\ Fan. 



Fig. 286. — Effect of High Initial Temperature of Superheat on the Thermal Efficiency of 

Steam Engines. 



show the sort of effect that the process of heat addition during expansion may 
be expected to produce. Such a case is illustrated in Fig. 287, for initially 
dry saturated steam at 160 lbs. per square inch absolute pressure, expanding 
to 2 lbs, absolute, ABC J) and compared with the corresponding Rankine cycle 



HEAT AND WORK 



1079 



ABCE, work being determined graphically by measuring T3> areas. Accord- 
ing to this the saturated expansion cycle has an efficiency of 24.11 per cent, 
which is less than the 26.07 per cent for adiabatic expansion, so from purely 
thermal grounds a loss rather than a gain is to be expected from jackets on 
this assumption of the cycle. However, this does not prove that in real engines 
jackets may not be efficiency aids, since in the case of superheat more gain 
may be realized than expected, but on the other hand with pressure range 
increases, less was realized so test data are necessary once more. A compound 
engine 9X16X14 ins. at 265 R.P.M. tested by Carpenter at 112 lbs. per square 



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Fig. 287.— Rankine Cycle Modified by Saturated Expansion Assumed to be Typical of Steam 

in Jacketed Cylinders. 



inch gage, at 22-in. vacuum showed a minimum steam consumption per I.H.P. 

hour of 18.6 lbs. with both cylinders jacketed, and 19.1 lbs. without jackets, on 

5 
initially dry steam, which is a reduction of -^— = 2.6 per cent, but at some 

loads there was an actual loss by the use of jackets. This is true for practically 
all cases that have been subject to test, and the conclusion is, that a gain of about 
2 per cent may be expected from jackets when the engine works always at 
the load at which the gain is realizable, which is possible for pumping engines, 



1080 



ENGINEERING THERMODYNAMICS 



for example, but when engines have to do work at variable loads the net effect 
is either a loss or so small as not to warrant the expense of obtaining it. 

Reheating the steam between the high- and low-pressure cylinders as a 
means of improving economy, is in much the same class as jacketing cylinders, 
in fact it is really a process of jacketing receivers of multiple-expansion 
engines. What may be expected thermodynamically from perfect reheating, 



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that is, up to the initial temperature, of steam initially dry saturated at 160 
lbs. per square inch absolute pressure expanding adiabatically to 2 lbs. absolute, 
is shown in Fig. 288 for the case of equal division of work without reheating. 
The work diagram without reheat is bounded by ABC DA, the heat of reheat 
which first dries and then superheats the steam is given by the area EFJGIE, 
the work derived from the heat of reheat by FJGHDF and the complete work 
diagram with reheat is ABCFJGHA. Evaluation of these areas shows the 



HEAT AND WORK 1081 

thermal efficiency without reheat to be 26.07 per cent, with reheat 25.03 

per cent, which is a slight loss. Tests made by the writer some years ago 

on a triple expansion Corliss engine of 200 H.P. with and without reheat 

in the receivers showed a net gain of about 1.5 per cent by reheating at one 

load and a loss or no measurable difference at other loads, which agrees with 

practical experience in the use of reheating receivers. These receivers are 

always troublesome to keep tight and well drained, so the tendency is, therefore, 

to use simple well-lagged receivers and apply heat in the form of superheat 

initially which does yield a gain in efficiency at all times. 

One interesting test on a large Mcintosh & Seymour engine, 29X60X56 ins. 

in the power station of the Boston Edison Co., by Cooke, shows the combined 

effect of both reheat and jackets, data being available with heaters both 

in, and out of service. At full load and with an initial pressure of 160 

lbs. gage and 95° of superheat, 26-in. vacuum, the indicated water rate was, 

for two runs, 10.85 and 11.18, or 11.01 lbs. mean, with both jackets and 

reheaters in use, and without them 11.55 lbs. Therefore, the gain by the use 

.54 
of both was — ■— = 4.7 per cent at full load. At half load the water rates were 
11.55 

.28 

10.33 with reheaters and jackets, and 10.61 without them, or a gain of = 2.64 

10.61 

per cent. While no data are available for other loads, it is quite probable that 
the gain would change to a loss. 

In nearly all cases it has been found that expected gains in efficiency due 
to increases of initial, decrease of back pressure, increase of superheat, use of 
jackets and receiver reheating, do not agree with that realized and the reason 
must be sought in interferences due to leakage and heat exchanges between 
walls and working steam. Analysis of these influences, often as it has been 
attempted and elaborately pursued, has as yet failed to give results worthy 
of reproduction, so these factors must as yet be classed with the unknown 
so far as any prediction is possible, just as is the case with the wall heat 
exchange and leakages in the cylinders of gas engines. One of the most 
interesting methods of attack yet offered, however, and sufficiently suggestive 
to warrant reference to it, is that recently used by Clayton, as it deals directly 
with the detection of missing water and leakage and their separate evaluation, 
though it has not yet been applied to enough cases to warrant any general 
conclusions on these troublesome quantities. By making very careful tests 
of piston engines for the determination of steam consumption and studying 
the correct indicator cards when the clearance was accurately known, he secured 
data for the plotting of the pressure volume expansion line and used for the pur- 
pose logarithmic cross-section paper on which any equation of the form of 
PV S = Constant, becomes a straight line if s is constant. As might be expected 
from the purely thermal investigation of the values of s for steam, it was found to 
be substantially constant over the length of one expansion line if not too long, but 
a variable with respect to the initial wetness or quality of the steam at cut-off. 
Knowing the quality of the steam supplied to the cylinder, any direct knowledge 



1082 ENGINEERING THERMODYNAMICS 

of the volume or from it the quality of the steam in the cylinder at cut-off, 
gives by differences the initial condensation if leakage is absent; this difference 
is itself the missing water per charge of steam- water mixture supplied. The 
slope of the expansion lines gives the value of s, and if s is a function of initial wet- 
ness only, it follows that the slope of the expansion line is a measure of missing 
water. Preliminary tests showed s to vary from .70 to 1.34, depending on en- 
gine type, size, speed, pressures, and ratio of expansion, but it always increased 
in any one engine with increase of cut-off and was always above 1.0 and as high 
as 1.34 for superheated steam, while for wet steam it was usually less than 1.0 
and as low as .7. For a 12X24 in. Corliss engine in particular, no value of s 
below 1.00 was found for values of cut-off quality above 80 per cent, and no 
value above 1.10 for cut-off quality less than 72 per cent, and in the region of 
s = .90 to s= 1.00, cut-off quality ranged from 50 to 70 per cent, the several values 
corresponding equally to saturated steam with long cut-offs and superheated 
steam with short cut-offs, indicating an independence of cut-off quality with 
respect to cut-off and a primary dependence of s on cut-off quality. The average 
of all results is given by Eq. (1180) and the maximum departure is 4.6 per 
cent from it. 

Cut-off quality = 1.258s -.614 (1180) 

Further tests showed these constants to vary with the pressures and 
speeds but no general equation was derived from which it could be evaluated. 
The conclusion is, therefore, that in any one engine or perhaps in all engines 
of one type, such an equation as the above holds and can be experimentally 
found, so that once established a single indicator card may serve to determine 
for missing water as well as the indicated water, provided leakage is absent or its 
quantity evaluated. Incidentally, the failure in real engines of the old assump- 
tion of logarithmic expansion is well established and renders the use of the expo- 
nential equations of Chapter III for horse-power and indicated water rate, more 
necessary than they have been regarded in the past. 

Prob. 1. Steam is being used in an engine at 100 lbs. per square inch initial pressure 
and 20 lbs. per square inch absolute back pressure. Show by PV and T$ diagrams the 
efficiency gain, if initial pressure be doubled while back pressure is held constant, and 
the gain if the initial is held constant and the back pressure lowered to 5 lbs. per square 
inch absolute. The engine cylinder volume may be taken as being 75 per cent of the 
volume of 1 lb. of steam at 20 lbs. 

Prob. 2. By means of a T§ diagram show the increase in efficiency due to super- 
heat of 100°, 200°, 300°, 400°, 500°, in a turbine using steam at an initial pressure 
of 150 lbs. per square inch absolute and running on a 3 in. vacuum. 

Prob. 3. An engine is running with an unjacketed cylinder on an initial pressure 
of 125 lbs. per square inch absolute and a back pressure of 5 lbs. Show by the T$ 
diagram the change in efficiency if there was a jacket imparting sufficient heat to cause 
the expansion line to be saturated. 

Prob. 4. The engine of Problem 3 is run under the same conditions except that 
there is initially 200° of superheat. Without the jacket adiabatic expansion occurs 



HEAT AND WORK 1083 

and with it the steam is constantly superheated 200°. What will be the difference 
in efficiencies with and without the jacket in this case? 

Prob. 5. A compound engine runs on an initial pressure of 185 lbs. gage and a back 
pressure of 5 lbs. absolute. The cut-off is so adjusted as to give a receiver pressure of 
30 lbs. absolute. Show by T$ diagram the efficiency with and without complete reheat 
for (a) no initial superheat; (6) 200° superheat and reheat to this temperature; and (c) 
200° superheat and reheat to saturation temperature of initial pressure. 

Prob. 6. Find the efficiency for the engine of the last problem, considering the low- 
pressure cylinder to be jacketed and low-pressure expansion to follow the saturation 
law. 

Prob. 7. Predict the thermal efficiency, heat consumption, and water rate for any 
piston steam engine or turbine under any working conditions, at best load. 

16. Flow of Hot Water, Steam, and Gases through Orifices and Nozzles. 
Velocity, Weight per Second, Kinetic Energy, and Force of Reaction of Jets. 
Nozzle Friction and Reheating. Relative Proportions of Series Nozzles 
for Turbines for Proper Division of Work of Expansion. Any expansive 
fluid such as steam, either superheated, dry saturated, wet, or even a 
liquid at the boiling-point, as well as all the gases, should on suffering a 
loss of pressure in passing through a nozzle or orifice of any kind, expand adia- 
batically and transform into work some of its heat content. The amount 
is given in Fig. 289 by the areas ABJ to ABDEFGA, depending on the quality, 
b for 1 lb. of steam or hot water. What actually occurs is an approximation 
to this, as close as interferences permit, and these interferences are of the 
friction and impact order. In the nozzle itself friction will have the effect of 
lowering velocity developed previously, and so converting kinetic energy 
back into heat and as this is a continuous process, the net effect is the 
same as if heat were added during expansion and the expansion line will 
take a position somewhat as shown in Fig. 290 instead of being vertical as it 
would be for an adiabatic change. The usual way of defining this line and 
the energy change responsible for it is to say that x per cent of the energy 
developed by a partial expansion, assumed for a small drop in pressure to 
be adiabatic is, after the expansion is completed, added to the fluid as heat 
at constant pressure. Thus in Fig. 290 let the initial condition be represented 
by B, C, D, or E, for the cases of water, wet, dry saturated, and superheated 
steam respectively. If expansion proceeds adiabatically from D for example, 
the work developed in the form of kinetic energy is represented by the area 
BDHA. Adding at constant pressure an amount of heat eqvivalent to 

/ x \ 

\i~n/ X( area BDHA), the final condition will be as represented by the posi- 
tion of point H' ', and a line joining a series of such points, is the representa- 
tion of expansion with reheat constantly going on. If all the work developed 
be converted back into heat the expansion line would follow the law of con- 
stant energy, and then the kinetic energy of the jet and the velocity would 
both be zero. Of course, this limiting case is never reached in a nozzle, but 
is reached when gases and vapors escape through so-called porous plugs or 



1084 



ENGINEERING THERMODYNAMICS 



diaphragms or when a jet once formed, loses velocity by impact, and for these 
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gases that are perfect, the final temperature would be the same as the initial, 
since the internal energy is a function of temperature only, but real gases 



HEAT AND WORK 



1085 



will always suffer a small temperature change usually to be neglected in 
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and Available Gain in Kinetic Energy. 



the adiabatic and the constant energy line, just where, depends on the amount 
of reheat. 

After the jet leaves the nozzle with the above velocity and corresponding 
kinetic energy, it may for a time retain all of it, lose all of it or lose part, depending 
on subsequent friction and impact. In the throttling steam calorimeter the 
jet is brought to rest by impact and all kinetic energy converted back into 



1086 ENGINEERING THERMODYNAMICS 

heat, so the final condition of the steam will be that corresponding to a heat 
content equal to what it had originally, but at the lower pressure, and like- 
wise the same as if it. had escaped through a porous plug emerging with no 
velocity, continuously expanding along the constant total heat line. Relations 
between initial and final qualities can be read off on T# or Mollier diagrams 
by following constant total heat lines. About the same thing happens when 
steam or air passes through reducing valves — some velocity is developed and 
subsequently all, or nearly all is lost by impact on walls or on the mass of fluid 
on the low-pressure side, and experience shows that the loss is practically 
complete so that perfect gases will not suffer any temperature change ultimately, 
and steam or hot liquids will have a final quality due to equal total heats. 
Hot liquids thus will always be partly converted into vapor as in boiler blow- 
offs and refrigerating expansion valves, while wet steam may be dried wholly, 
or in part, or more superheated than it was originally. 

In steam turbines the jet once formed is brought to moving vanes and 
there has its direction of motion changed so that it leaves the vanes with a 
velocity with respect to them, but ideally with no velocity with respect to the 
casing or nozzle. When this process is perfectly executed the whole kinetic 
energy of the jet is imparted to the moving vane wheel, but actually there is 
developed some vane friction which acts as additional reheat giving the 
steam a new quality higher than it had before it reached the vane, and which 
is to be measured by an added amount of energy equivalent to the vane reheat 
expressed as y per cent of the energy of expansion. This is very important 
in those turbines that have a series of successively expanding nozzles, as 
work to be developed in the second depends in part on the nozzle and vane 
reheats of the first. Such reheat is, therefore, not as much a loss in such multi- 
stage turbines as in those of one stage, but always constitutes one of the main 
energy losses in these machines, the other losses being leakage and windage 
friction of the rotor in the steam atmosphere, which latter adds more reheat. 
The whole reheat is commonly taken as 40 per cent, on the average. 

To calculate the velocity of a jet two factors are necessary: first, the work 
that would be done by adiabatic expansion, and second, the reheat factor. 
This reheat factor is for turbine nozzles about 10 per cent but is not established 
for other forms, like valve orifices. The determination of velocity without 
nozzle friction has already been explained; the PV formula of Chapter I is 
to be used for gases, but for steam the Mollier diagram is the most expeditious, 
as both the work done and velocity can be read off directly. Determinations 
of final quality of the fluid jet can also be made for the case of no reheat from 
the same diagram directly, and by a simple additional step the quality can 
be found graphically as illustrated in Fig. 291. This is a section of the Mollier 
diagram for steam, where AB represents adiabatic expansion, during or after 
which, 20 per cent of reheat occurs. To find the final quality lay off BC= .2 X AB, 
so that AC represents the effective work and BC the heat of reheat. A hori- 
zontal through C cutting the constant-pressure line through B locates D, the 
condition of final quality. It is interesting to note that if this heat is constantly 



HEAT AND WORK 



1087 



added, that is, if the reheat is a constant fraction in the nozzle, the quality 
at any pressure can be found by a similar construction used by Roe. With 
B as a center, an arc CE is drawn, and a tangent to it drawn through A. 
Similar tangent arcs drawn from other centers give the means of finding the 
quality at the pressure indicated by that center, by the intersection of hori- 

































































































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Fig. 291. — Graphic Method of Locating Nozzle Steam Expansion Line and Final Quality 
with Given Per Cent Friction Reheat. 

zontals tangent to the tops of the arcs, with constant-pressure lines through 
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in the nozzle. If it receives vane or other reheat beyond the nozzle the jet 
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1088 



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1090 



ENGINEERING THERMODYNAMICS 



range of pressures over which the flow does not change with pressure change, 
from the other region where it does. In the first Chapter it was shown by 
PV analysis that this critical pressure was a function of the exponent y given 
by Eq. (1181). 



/ Final pressure \ 



At the critical point, T ... T I 

\Initial pressure/ 



(M" 



(1181) 



This is easily fixed for gases not suffering wide temperature changes because 
for them y is constant, but for vapors the variability of y makes it not so 
easy to fix. However, for the conditions in turbine nozzles the ratio is usually 
taken at the round number of .58 though it depends on the quality, the pressures 
and temperatures of the steam. From the tables of Section 4 of this Chapter, 
values of f for all sorts of steam conditions determined by the T$ method, 
can be used in the above equation. Careful investigation of this critical 
phenomenon seems to indicate that at some point in every orifice the 
pressure falls to this critical value of itself, and that the weight of flow is fixed by 
the pressure at that point regardless of how much lower the pressure may be 
beyond. Furthermore, it appears that at this critical point of the orifice 
or throat, each fluid acquires a fixed velocity peculiar to itself about the same, 
according to Emden, as the velocity of sound in the medium, and for steam 
this is between 1400 and 1500 ft. per second for such conditions as are met 
in turbine nozzles. Of course, beyond this critical point, further expansion 
may take place, increasing the velocity above the critical value but not the 
weight, provided the orifice or nozzle is so shaped as to keep the flow axial and 
prevent sidewise dissipation, in which case the nozzle is an expanding nozzle. 

To make this condition of affairs more clear the curves of Fig. 292 and 
293 for steam and for air, have been calculated from Eq. (25), Chapter I, using 
y = 1.4 for air, and by the Mollier diagram for steam. To this diagram are 
added some curves of experimental flow laws stated 'symbolically below: 



Lbs. dry saturated steam 
per sq.in. area per hour. 



Lbs. superheated steam per 
sq.in. area per hr. 






= J 



Wp 97 (Grashof) (a) 

51.43p (Napier) (6) 

49.6p (Harter) (c) 
3.6p (16.3-96 log p) (Rateau)(d) 

60p 97 [1 + .00065 X (degrees 

superheat)] (Moyer) (a) 

49. Qp to 45. Op for superheat 
0° to 185° F. at 160 lbs. sq. 

in. (Harter) (6) 



(1182) 



(1183) 



r Lbs. air per 
\ sq.in. area 
I per hour 



3816 



4 



P2(P1-P2) 



for less than 2 atm. to atm. 
(Fliegener) (a) 



Vi 



1900—^= for more than 2 atm. to atm. 



vr, 



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(6) 



(1184) 



HEAT AND WORK 



1091 



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1092 



ENGINEERING THERMODYNAMICS 



The superheated steam Napier coefficients and corresponding experimental 
steam weights are shown in Fig. 294 reproduced from Harter's report. 

For steam, the best and most used relation is that of Grashof, which for 
wet steam, such as always is found entering low-pressure turbines, becomes 



(Lbs. wet steam per square inch area per hr.) 



60 v' 



97 



Vdryness fraction 



(1185) 



By means of these equations the size of nozzle throat or narrowest point 
actually or, in effect, to pass a given weight of steam or air can be calculated, 
or inversely, the weight that will pass a given area. The form of the nozzle 
however affects the result by making the actual minimum area differ from that 
which is effective, because of stream contraction. The effective area may be 
only 60 per cent of the actual throat area if sharp corners exist. The product of 
weight per hour and foot-pounds per pound of steam gives the kinetic energy of 
the jet in foot-pounds per hour, which has, of course, a horse-power equivalent. 
This jet horse-power will not be realized, however, unless another condition is ful- 
filled and that is a proper shape for the nozzle beyond the throat and a proper 
ratio of maximum terminal nozzle area, to that of the throat, to enable the full 
expansion work to be realized beyond the nozzle throat. The ratio of the largest 
area of the nozzle that at the mouth, to the least, that at the throat, is the ex- 
pansion ratio of the nozzle and for it, two usable values not in accord, are given 



/Mouth area\ 
\ of nozzle / 
/Throat area\ 
\ of nozzle / 



.1550 






1 881 



when ^< 25 

V2 



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(Moyer) (b) 



175P 

XP2 



.7] when ^> 25 (Moyer) 



(c) 



(1186) 



by Eq. (1186) ; where pi and P2 are the initial and final working pressures for the 
nozzle. Between throat and mouth all sorts of curves and angles are used, with 
however, not so much difference in results as might be expected; about the 
commonest practice is to use an angle of 20° and a straight-line element. Too 
sudden an expansion due to too wide an angle will cause the steam to bounce 
from side to side and set up waves that result in increased frictional reheat. 
Too long a nozzle also brings about increased friction; too large a mouth results 
in overexpansion, with subsequent compression and more waves; too small 
a mouth incomplete expansion, all of which is well illustrated by one of 
Kneass' pictures, Fig. 295. As practically all nozzles must be used under 



HEAT AND WORK 



1093 



variable conditions of pressure and quality it is necessary in their design to 
avoid hair-splitting and use some judgment in estimating which to select for 
a mean or standard condition. 

The reaction force on the nozzle due to the exit of the jet is equal to the 
weight in pounds per second discharged -^-32.2, into the velocity in feet per 
second, as in Eq. (1187). 



(Reaction in lbs.) 



— — [(lbs. steam per sec.) X (jet velocity ft. per sec.)] (1187) 




Fig. 295. — Kneass Steam Nozzle Flow Lines. 



A series of steam turbine nozzles of different forms were tested for efficiency 
of conversion of heat into kinetic energy by means of measured weights and 
reaction forces, by Selby and Kemble and the best results were obtained with 
a straight taper of 14° 31', rounded entrance, having a searching tube in the 
center and a throat diameter of .3949 in. gross, .0734 in. net, a length of 3^V 
ins. and a mouth diameter of 1.156 ins. gross, 1.0005 ins. net. With initial 
absolute pressures of 100, 115, 130, and 145 lbs. per square inch, back pressures 



1094 



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1096 ENGINEERING THERMODYNAMICS 






of 1.116, 1.288, 1.46, and 1.632 lbs. per square inch absolute, the flow 
was 389, 446, 502, and 558 lbs. per hour and the ratios of the work 
equivalent to the kinetic energy computed from flow weights and reactions, 
to the Rankine cycle were, 98.3, 97.9, 97.5, and 97.1 per cent respectively. 
These figures are given to show how nearly perfect such nozzles may be made, 
and how simple is the means of experimentally proving their efficiency. 
Similar experiments made in the laboratory of the writer on both compressed 
air and steam proved that air could give quite as high efficiency as the steam, 
in the nozzle expansion, which indirectly proves also the perfect feasibility 
of carrying out one important step in the development of a gas or mixed gas 
and steam turbine. 

When several nozzles are in series and the same steam weights pass through 
all, their absolute and relative sizes can be fixed by the previous relations 
if the pressures in the intermediate chambers are fixed, because each may be 
treated independently of the other except for initial quality of approach 
to each, which must be determined. Thus, assume expansion to take 
place in four pressure stages or four nozzles or sets of nozzles, each with 
x per cent uniform frictional reheat, and with y per cent vane reheat between 
nozzles, then will the whole expansion process be according to the line 
ABCDEFGHI of Fig. 296 to both T<$> and H$ coordinates for equal division 
of pressures from 200 lbs. to 1 lb. per square inch absolute, and with originally 
dry saturated steam. This line is determined by drawing the four constant 
pressure zones, each 50 lbs. per square inch, across the two diagrams, dropping 
a vertical AM on the H$ chart, giving the Rankine cycle work for the first 

stage pressure range, laying off B'M= \T^j XAM, the nozzle reheat loss, and 

B'B" = (tt";) XAM. the vane reheat loss, referred to the same Rankine cycle 

basis, so that point B locates the condition of the steam at exit from first 
nozzle, and C its condition at the entrance to the next. The work of the first 

stage is, 1 — f — — J X (Rankine cycle work) for the two pressures, and the 



first-stage efficiency is, 1 — f j 



A similar construction applies to the 



other stages, and where graphically carried out will show unequal work and 
stage efficiencies even with equal per cents of reheat, when the stages are 
equally divided as to pressure range. It is more usual to divide the work 
per stage equally and if the stage efficiencies were 100 per cent this would 
be easily carried out by dividing the vertical AN representing the whole 
expansion work on the H$ diagram into four equal parts and through each 
drawing the constant pressure lines as in Fig. 297. With any nozzle and 
vane reheat losses, however, this will not give equal stage work, but is a good 
starting point for a graphical trial and error method of finding that pressure 



HEAT AND WORK 1097 

division that will give it, after which the nozzle sizes can be found to maintain 
that pressure relation. 

Prob. 1. Steam at 100 lbs. per square inch absolute containing 5 per cent moisture 
expands in a nozzle to 10 lbs. per square inch absolute with 20 per cent nozzle reheat. 
What is the final quality, velocity, and horse-power per square inch of nozzle and what 
fraction is each, of what would result without nozzle friction? 

Prob. 2. How would the results of the previous problem differ by changing final 
pressure to 1 lb., initial pressure to 150 and 200 lbs., and initial quali y to 50 per cent, 
100 per cent or 100° superheat, in any combinations? 

Prob. 3. Air at 150 lbs. per square inch absolute and 1000° F. expands into a standard 
atmosphere, where its temperature in the jet is 300° F. If the reheating had been uni- 
form, plot the T$? expansion curve, find s and the reheat factor. 

Prob. 4. What would be the nozzle reaction per square inch of orifice for Problems 
1 and 3 and what dimensions should expanding nozzles have? 

Prob. 5. Water from a 150-lb. gage boiler is blown off. What per cent will evap- 
orate? How much will the evaporation change with initial pressure, plot a curve? 

Prob. 6. Air flows from a tank where the pressure is 150 lbs. per square inch absolute, 
to a standard atmosphere through a f-in. diameter throat nozzle. How many cubic 
feet of free air must be supplied per hour to maintain the tank pressure and what com- 
pressor horse-power will be required if two stage? 

Prob. 7. A steam turbine has four f-in. diameter nozzles in parallel and the back 
pressure is 2 lbs. per square inch absolute. How much steam per hour and what jet 
velocities result for, (a) 150 lbs. per square inch absolute and 100° superheat initially, (b) 
same pressure and dry saturated steam, (c) 100 lbs. per square inch absolute and dry satu- 
rated steam, (d) same pressure as in (c) and 75 per cent quality. Use Mollier diagram. 

Prob. 8. Compare results of Problem 7 with those computed by the formulas of 
Grashof, Napier, Rateau, Harter, Moyer. 

Prob. 9. For same initial pressure and temperature and same final pressures com- 
pare the preceding results on steam flow with air flow, using rational formula and 
Fliegener's empiric form. 

Prob. 10. If four nozzles of a steam turbine are in series, and the first ^-in. diameter 
is supplied with dry saturated steam at 180 lbs. per square inch absolute, what must 
be the sizes of the other three 

(a) to equally divide the pressure range with no nozzle and no vane reheat 

(b) " " " " " 20% vane reheat 

(c) ." . c ' " 20% nozzle and no vane reheat 

(d) " " " " 20% 

(e) to equally divide the temperature range with no ' ' 
(/) " " " " no 

(g) " " " " 20% 

(ft) " " " " 20% 

The back pressure is 2 lbs. per square inch absolute. 

17. Flow of Expansive Fluids under Small Pressure Drops through Orifices, 
Valves, and Venturi Tubes. Relation between Loss of Pressure and Flow. 
Velocity Heads and Quantity of Flow by Pitot Tubes. When gases and vapors 
flow with a small difference between initial and final pressures, the flow charac- 



20% 


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( i 


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( i 



1098 ENGINEERING THERMODYNAMICS 

teristics have a special engineering importance great enough to warrant more 
precise determinations of equations and constants than those applying to flow 
in general or with large pressure drop. The loss of pressure through the 
valves of air compressors, steam and air engines and gas engines, and its 
relation to velocity and weight of flow per square inch of opening, are 
examples of one class of application. Others of a similar sort include the flow 
of gas through burner nozzles, air from ventilating duct openings and through 
fan passages; and various meter orifices including the Venturi and Pitot tubes 
involve in their use, similar relations. 

As a starting-point for the analysis of all these cases of flow, the flow through 
an orifice, assumed to be adiabatic, must be adopted. Accordingly, calling 
Pi the initial and P 2 the final pressure in pounds per square foot, Eq. (25), of 
Chapter I, gives the flow when these differ not too much, 



(Lbs. per sec. per sq.ft.) =| = V2g^^[(^) 7 -(fj * ]. . . (1188) 



Applying this to air passing through a round orifice of d ins. diameter, this be- 
comes for air at 60° F., Eq. (1189). 






flibs. air per sec. through! __ nsifi 7 2 JPi\ 1 1 j 

1 circlejof d ins. dia. at 60° F. J " ' U * lbd \ Vi IR P 1A25 R p 1712 \ 



r>/rr 

Substituting Vi = -5— > this becomes 



. . (1189) 



s. air per sec. through 1 _ nnfUQ 7 2p / 1 1 . 

ircle of d ins. dia. at 60° F. J " ' UUU4yrt ^\H P 1A2b R P 1J12 a) 

= .0707d^^ 5 -^ 7T2 (b) 



(1190) 



With orifice friction and poor orifice forms the actual area may not be 
effective, so there must be introduced coefficients representing these influences 
as experimentally established. It is found, however, that the actual flow 
values agree fairly well with a simpler and more practical formula for very 
small pressure drops. Such a formula can be developed on the theory that when 
the drop in pressure is small the work derived by expansion may be neglected, 
and the fluid may be considered as non-expansive, and of a density correspond- 
ing to the mean pressure on the two sides and to the original temperature, 
supposed to remain constant. For this case 



HEAT AND WORK 
Let h w = the difference in pressure in inches of water ; 
" H A = the head in feet of air corresponding to hw', 

P2+P1 



" P = mean pressure, lbs. sq.ft. = 



2 ' 



1099 



F A = cu.ft. per lb. air = 53.34—; 



11 A=sq.ft. area of orifice; 

" u = velocity in feet per second ; 

11 w = pounds air per second. 



Then since 5.2 pounds per square foot is equal to one inch of water column, 



H A = V A (P 2 -Pi)=5.2V A h w , and —t^ = u, 



so that 



Pounds air per sec. per 
sq.ft. of orifice 



w _ _w_ _ \Z 2gH A 

a~Va~ ~v7 

V2gX5.2h w V A 



-4 
-4 



64.4X5.2^ 



64.4X5.2^.P 



53.34r 



= 2.506 



-Jh w 



(a) 
(b) 
(c) 
(d) 

(«) 



(1191) 



For a round orifice of d ins. diameter this becomes 1 



Lbs. air per sec. through circle! 
of d ins. diam. J 



•7854d 2 v/0 / P 

144 X 2.506^/^ (a) 



= M36Qd\hw 



4 



T 



(b) 



(1192) 



1100 



ENGINEERING THERMODYNAMICS 



If the discharge is from a low-pressure chamber into a standard atmosphere, 
this becomes very nearly 



| Lbs. air per sec. to atmos. through 
I circle of d ins. diam. 



.6283d 2 



; \h w 



(1193) 



The best determinations of the constants for either the approximate formula, 
Eq. (1192), or the exact one, Eq. (1190), are those made by Durley, whose 
calculation on the relation of the results of the two with a 100 per cent coeffi- 
cient are given in Fig. 298 for pressures up to 100 inches of water for air at 60° 



90 

80 

70 

u 
o 

£60 
























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0.2 0.4 0,6 0.8 1.0 1.2 1.4 1.6 1.6 2.00 2.20 2.10 2.60 
.Pounds of.Airper.Second 

Fig. 298. —Calculated Ideal Frictionless Discharge of Air to Atmosphere from Circular 
Orifice 3 inches diameter. Initial Temperature 60° ¥., Barometric Pressure 14.7 pounds 
per square inch. 



F. discharging to atmosphere, 30 ins. barometer, and his experimental coefficients 
for various orifices in Table CXLI on the opposite page. His results show that 
the coefficient for small orifices increases with the head but more and more slowly 
and is constant for a 2-in. diameter orifice, after which it decreases, and more, the 
larger the orifice. Increase of diameter involves smaller coefficients, but temper- 
ture has no effect between 40° F. and 100° F. as proved, and this is probably so 
also for even wider ranges. The orifice must be not more than gV the area of the 
approach chamber for these coefficients to apply. They are all for a plate .057 
in. thick and for sharp edges; changing the thickness or rounding the edges will 



HEAT AND WOEK 



1101 



Table CXLI 

COEFFICIENT OF DISCHARGE C FOR VARIOUS WATER HEADS AND 
DIAMETERS OF ORIFICE (Durley) 



Diameter of 
Orifice. Inches. 


1-inch Water. 


2-inch Water. 


3-inch Water. 


4-inch Water. 


5-inch Water. 


_5_ 
16 


.603 


.606 


.610 


.613 


.616 


1 

a 


.602 


.605 


.608 


.610 


.613 


1 


.601 


.603 


.605 


.606 


.607 


14 


.601 


.601 


.602 


.603 


.603 


2 


.600 


.600 


.600 


.600 


.600 


2h 


.599 


.599 


.599 


.598 


.598 


3 


.599 


.598 


.597 


.596 


.596 


3§ 


.599 


.597 


.596 


.595 


.594 


4 


.598 


.597 


.595 


.594 


.593 


4-i 


.598 


.596 


.596 


.593 


.592 



Weight of air discharged per second = , 6283 Cd 



- 2 \h\ 



h JL lbs. 



(Barometer at 30 inches.) 



change the coefficient, and Moss reports some tests with rounded approach 
orifices in which the coefficient was as high as (7 = 0.942. 

When the thickness of the plate is great in proportion to the diameter of 
the orifice, the latter becomes a short tube for which differences enter, having 
new experimental values not as numerous as they should be, as this is the case 
for gas burner nozzles. It appears, however, that the old results of Weisbach 
on air for both small orifices and short tubes are pretty good and these are 
reproduced in Table CXLII below. 



Table CXLII 

VALUES OF C FOR AIR FLOW (Weisbach) 

Orifice of diameter = .394 ins. 

R P 1.05 1.09 1.43 1.65 1.89 2.15 

C 555 .589 .692 .724 .754 .788 

Orifice of diameter = .843 ins. 
Rp .... 1.05 1.09 1.36 1.67 2.01 

C 558 .573 .634 .678 .723 

Short tube, diameter = .394 ins. and length = 1.181. ins. 
R P .. .. 1.05 1.10 1.30 

C 730 .771 .830 

Short tube, diameter = .557 and length = 1.673 ins. 

R P 1.41 1.69 

C 813 .822 

Short tube, diameter = .394 ins. and length = .630 ins. rounded entrance 

R P 1.24 1.38 1.59 1.85 2.14 

C 979 .986 .965 .971 .978 



1102 ENGINEERING THERMODYNAMICS 

His general conclusions on the coefficients for these small orifices give them 
values between .97 and .99 for orifices with rounded entrance, .56 to .79 for 
sharp edge in thin plates, .81 to .84 short straight tubes, .92 to .93 for short 
tubes with rounded entrance and .90 to .99 for converging orifices. 

When the fluid is not air, the flow is to be determined by the general law 
of proportionality as a function of density. The velocities due to pressure 
drop for different gases and vapors in these cases are to each other inversely 
as the square root of the density; thus, for any gas of specific volume, V G , or 

density, Tr , its velocity is given in terms of that of air by, 
Vq 



Ug=u a —7==u ax \^ (1194) 

/ 1 \ V A 



wV 
Putting this in terms of weights by the relation,— p = u, 



Wg=Wa Yo\v: =Wa ^v g (1195) 



so that flow weights are to each other directly as the square roots of the densities. 
One very common and important case of orifice flow with small pressure 
drop is that through the valves of gas engines and compressors, valves 
of the poppet type with both flat and conical seats, the flow being in both 
directions, that is, toward the cylinder and from it, while the valve always 
seats away from the cylinder in the direction of the action of the superior 
pressure. The only series of experimental determinations of the coefficients 
for such flows is that made under the writer's direction in the laboratories 
of Columbia University on both mechanically operated and spring-closed 
valves, all, however, of small size. One valve had a flat seat 1.58 in. diameter, 
and the other two, conical 45° seats 1.5 and 2.0 ins. diameter respectively. These 
were supplied with measured air flowing in either direction and with the valves 
first fixed at known lifts, and later rising and falling with the piston known 
amounts and at measured rates. From the data obtained the relation between 
flow and pressure drop up to 30 ins. of water was obtained and compared with 
the flow-pressure drop relation for no losses for which, u = \/2gH A - The ratio 
of the measured velocity or volume to that computed for no loss, as above, 
gives a coeffic ent of efflux for the given pressure drop and the results for the 
more common con cal valve are reproduced graphically in Fig. 299. Double 
results are given for air measured at meter density and at atmospheric density, 
the difference being due to the pressure drop experienced when the flow was 



HEAT AND WORK 



1103 




xnuja jo iirapgpoo 



1104 



ENGINEERING THERMODYNAMICS 



steady and valve fixed, the valve being 2 in. diameter lifting .6 in. maximum. 
The corresponding coefficients are not easily obtained for intermittent flow, the 
more direct method being to compare momentary velocities actual and computed, 
at each point of the rise and fall of the valve for the measured pressure drop, 
and plot a curve of pressure drop to velocity as in Fig. 300. To give a direct 
comparison between the steady and the intermittent flow, the corresponding 
curves for steady flow are plotted in Fig. 301. It is impossible to give any 
single generalization of these results, as the coefficients varied from 0.2 to 1.0, 
depending on type of valve, its lift and the rate of flow as well as direction of 
flow, and the only thing that appears to be sure is, that such flow is very 



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5000 



10000 
Gas Velocity-feet per min. 



15000 



20000 



Fig. 300. — Intermittent Flow through Poppet Valves. Relation between Pressure Drop and 

and Velocity. 



complex indeed, and may involve very serious losses^ but on the other hand 
may be very highly efficient. 

Venturi tubes are usually made by a converging pipe of straight cone form 
and small angle, ending in a short small diameter cylinder, which discharges 
into a diverging cone of smaller angle, ending in the original diameter, and 
while designed originally for measuring the flow of water have been found! of 
great practical value in measuring gases, the writer having used them for tihis 
purpose in diameters from 1 to 16 ins. The larger sizes are especially 
useful, as gas meters of the volumetric displacement sort of equal capacity, are 
prohibitively costly. In such a tube the same weight of fluid is passing each 
point, and neglecting frictional losses and static head differences which apply 



HEAT AND WORK 



1105 



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5000 



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Velocity Ft. per Min. 



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Velocity Ft. per Min. 



15000 



20000 



Outer Scale Inches of Water. Inner Scale Coefficient. 

Fig. 301. — Steady Flow through Poppet Valves. Relation between Pressure Drop and 

Velocity. 



1106 ENGINEERING THERMODYNAMICS 

to tubes set vertically, but not to horizontal ones, the sum of the velocity head 
and pressure head is constant. Therefore calling Pi and P 2 the upstream and 
throat pressures lbs*, per square foot, and A\ and A 2 the corresponding areas in 
square feet, the equation of condition is, 

Pl +f = P2 +f • ••••••• (H96) 



If the fluid be of constant density as it passes from the large area to the small 
one, the velocities must vary inversely as areas, since equal Volumes pass each 
point in this case. 

^=4 2 (H97) 

u 2 Ai 



Therefore the throat velocity 112 will be given by Eq. (1198), for constant density 



u 2 =--=^===V2g(P 1 -P 2 ) (1198) 

, VAi 2 — A 2 2 



When gases and vapors suffer very small pressure drops Eq. (1198) may be 
applied to their flow in Venturi tubes, but when the difference between up- 
stream and throat pressures is so small as to warrant the use of this relation, 
it is too small for the accurate readings needed to determine the quantity flowing. 
Therefore, for such cases it is necessary to return to the fundamental equation 
for expansive fluids, equating the increase in kinetic energy to the work of the 
cycle as for turbine nozzles. 



iH9fl 



^V^-^TT'l' ' I 



In Eq. (1199), hi is the density of the fluid on the high pressure or upstream 
side in pounds per cubic foot. As equal weights must pass both the upstream 
and throat sections, the velocities, densities and areas must be related as in Eq. 
(1200): 



Therefore 



W = W\ =W 2 =AiUili =A2U2h2 = A2U2%l(-^Y (1200) 

j(^) 7 (1201) 



1 
A2I2 A 2 /P 2 

A\ 01 A\ 



HEAT AND WORK 1107 

Substituting Eq. (1201) in Eq. (1199) and solving for u 2 results in Eq. (1179) 



-^u, 



^2\ 2 /P 2 



) T - 



Introducing this value of u 2 in Eq. (1200) the weight of flow is found as in Eq. 
(1203): 



[ Lbs. of gas or vapor 1 
per second J 




-® 



7-1 

y 



i-mw- 



(1203) 



In the actual use of Venturi tubes the value of 



(?) 



should be between 



.95 and .995, which values correspond nearly to throat velocities from 300 to 
100 ft. per second for low-pressure gases near one atmosphere in pressure, but 
for higher gas pressures very much higher values may be used. For steam the 
Mollier diagram may be advantageously used in Eq. (1204) : 



Work of Rankine cycle between 



r , .z__ri w orK or nanKme cycle oetween 

77 1- ^ ] =788X Pi and P 2 , in heat units by 

1 Ol L V-t 1/ -I TV /T IT T 

- { Mollier diagram 



(1204) 



At the works of the Lackawanna Steel Co., Coleman made a check test on 
one of these meters, using throat velocities of 572 and 208 ft. per second with 
steam § per cent wet at 110 lbs. per square inch pressure, and weighing the feed 
water, found errors of only + .056 per cent and — .84 per cent, which is sub- 
stantially nothing. Checking similarly another meter on air in a 6-in. line, by 
measuring velocity heads across both upstream and throat diameters by 
Pitot tubes, the weights found by the former were .88289 lbs. per second and by 
the latter .88986, giving .9922 as the ratio of Venturi to Pitot indications. Other 
tests with the Pitot tube along one diameter only, gave results that averaged 
.9963, in a 10-in. Venturi tube and .9976, in a 6-in. tube. The Pitot tube, of 
course, gives the velocity head of the gas in inches of water directly at the point 
of measurement the center of a small area a, and the average gas velocity over 

the whole area is, = (u) (average) . In the same plant the writer checked a 

2a 



1108 ENGINEEKING THEEMODYNAMICS 






16-in. Venturi tube against the displacement of the gas pumps of a 1000 H.P 
Korting two-cycle gas engine, and against a regular proportional gas meter in 
the blast-furnace gas pipe, and the Venturi results agreed with the pump dis 
placement within 1 per cent whereas the proportional meter disagreed with all 
other data. 

The Pitot tube carries a double orifice at the end of a double tube that can 
be inserted in a pipe through stuffing boxes to allow the orifice to be located 
anywhere along the pipe diameter. One orifice points upstream and measures 
the velocity head by reducing to zero the velocity of the small portion of the 
stream that strikes it, to which is added the static gas pressure, while the other 
measures the gas pressure proper by an orifice tangent to the flow. The dlf 
ference in pressure between the two orifices is the velocity head only. When 
the fluid is of constant density with respect to pressure, the velocity is given by 
Eq. (1205) 

u = V2gH, (1205) 

where H is the head in feet of fluid corresponding to the destroyed velocity. 
Usually water or mercury is used in the Pitot tube giving velocity head in inches 
of water or mercury. These heads are related to gas heads according to Eq. 
(1206), where H w , and H M , are heads in feet of water and mercury: 



jr _tj v /lbs. per cu.ft. of water \ . . 
w \ lbs. per cu.ft. of gas / 
_ /lbs. per cu.ft. of mercury \ „ . 

M \ lbs. per cu.ft. of gas / 



. r (1206) 



When air is the fluid its density is to be determined from its pressure, tem- 
perature, and gas constant, i£ = 53.34, if it is dry, but when moist by the Smith 
sonian Eq. (1207), where p v is the partial pressure of the water vapor in the air. 

| Lbs per | ^ [- .080723 ] / barometer in ins. Hg-.378^ \ f 

cu.lt. ot -I ! + .0020389(^-32°) \\ 29.921 )• V*>0 

{ moist air J 



These relations are proper for such very small velocities, as correspond to 
no appreciable change of density of the gas for the pressures used, but to get 
measurable velocity heads, the density variation cannot as a rule be neglected, 
and in such cases the standard nozzle relation must be used as a starting-point. 



2 V^H8P] : (1208) 



,2_,,„2 

~2g 



HEAT AND WORK 1109 

If P 2 is the pressure measured at the impact orifice, u 2 =Q, and, U\=u. 

7-1 

Substituting Eq. (1209) in Eq. (1208) there results the velocity expression 
Eq. (1210), 



2g y-1 



RT! 



f) 7 -l] (1210) 



This is developed into a series by Taylor as follows: 

7-1 7-1 



Put pr = l+K, and expand, i^) y =(1+K) T into a series, which when 
multiplied by ~~ gives Eq. (1211). 

*-f ^m- K i a± w ?:L} } (-> 



Pi 



Substituting Eq. (1211) in Eq. (1210), putting #Ti = y, and solving for u, 
the velocity is found as in Eq. (1212) : 



u 



-^^F^WM^W^ ■ ^ 



Introducing the readings of P 2 and Pi, in inches of mercury h M 2 and h M i, and 
multiplying by 60, the velocity in feet per minute for air, is given by Eq. (1213) : 



Velocity of 
air in feel 
per minute 



air in feet = 4046.16 J (^j^ ¥1 )(l-.355K-{-.202K 2 -.137X 3 ) . (1213) 

nor tyii'tiii + o \ 1 / 



Calculation of velocity by this Eq. (1213) and by the equation neglecting den- 
sity changes, gives comparative results plotted in Fig. 302, reproduced from 
Kneeland. This shows that at velocities of 23,000 ft. per minute, the difference 
is 15 per cent, at 11,000 ft. per minute, about 6 per cent, which are appreciable, 
so that in using Pitot tubes for gases and vapors high velocities should be used 
and the formula Eq. (1213) will give the velocity for the observed pressures. 



1110 



ENGINEERING THERMODYNAMICS 







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HEAT AND WORK 1111 

Prob. 1. In a test on air flow sharp-edged orifices of sheet iron were used, to which 
Durley's coefficients apply. For the following data what was the volume and weight 
of air flowing? Air temperature =70° F. 

Orifice dia. in. Head in inches of water. 

1 1.4, 1.95, 2.8, 4.1, and 4.8 

2 1.25, 1.7, 2.25, " 4.8 

3 1.60, 1.95, 2.65, " 5.0 

Prob. 2. What size of orifice would pass 200 cu.ft. of air with 5-inch water pressure 
drop, if (a) the coefficient were 100 per cent, (b) if Durley's coefficient were used. 

Prob. 3. Air flows from a tank into the atmosphere at the rate of 75 cu.ft. per minute. 
The pressure in the tank is 25 lbs. per square inch absolute, and temperature 60° F. 
According to Weisbach how big should the hole be? 

Prob. 4. A gas burner is supplied at 2-inch water pressure and passes 5 cu.ft. of 
gas per hour, the density being 60 per cent that of air. What area of burner is needed? 
Check this by comparing with a standard gas tip. 

Prob. 5. A Venturi in an air pipe showed an upstream pressure of 17 lbs. per square 
inch absolute and a throat pressure 2\ ins. Hg less. The throat diameter is 6 ins., pipe 
diameter 20 ins. At 60° F., what quantity was flowing? 

Prob. 6. If the pipe diameter of Problem 5 had been 12 ins., what would be the differ- 
ence in pressure between throat and upstream in inches of water when the upstream 
velocity is 6000 ft. per minute? 

Prob. 7. Gas is being measured by the Pitot tube. The pressure is 50 lbs. per square 
inch gage and the tube reading is 2\ ins. Hg. If the gas density is .8 that of air, pipe 
4 ins. diameter, temperature 40° F., what is the flow rate, (a) by exact and, (b) by approxi- 
mate formula? 

Prob. 8. A steam Venturi tube gave the following readings: 

Dia. Pressure, lbs. Sq.in. Gage Quality. » 

Throat 2ins. 90 100% 

Upstream 4 ns..< 100 95% 

What weight of steam was flowing and what was the upstream velocity? 

18. Flow of Gases and Vapors in Pipes, Flues, Ducts, and Chimneys. 
Relation between Quantity of Flow and Loss of Pressure. Friction Resist- 
ances. Draught and Capacity of Chimneys. Pipe or conduit flow of gases 
an- vapors, is characterized by comparatively small losses in pressure com- 
pared to the mean pressure, that is by high values of the ratio of initial to final 
pressure, also by complete conversion into heat of all the kinetic energy lost by 
friction, and by appreciable gains or losses of heat by wall conduction and sur- 
face radiation, gains when the fluid temperature is lower than the surroundings 
as it is with refrigerating media, and losses otherwise as for steam. All relations 
between the rates of flow and the change of fluid conditions corresponding to 
it can be numerically evaluated, only when friction energy and heat conduction 
carj be evaluated, and these quantities depend so much on structural details 
as to defy algebraic analysis. Accordingly, while some thermal relations for 
pip*r flow may be set down, the really useful equations for engineering work 



1112 ENGINEERING THERMODYNAMICS 

are empiric, and will probably always so remain except as to form, which may 
change as more knowledge becomes available on gas and vapor friction, and 
on the transmission of heat to or from fluids through pipes and insulating 
coverings. 

Reduction of pressure with no conduction losses in a pipe line would result 
in the doing of work and the development of kinetic energy equal to that of the 
Rankine cycle for steam and its equivalent for gases. As, however, neglecting 
the change of density corresponding to the pressure drop, the velocity is constant 
in a uniform pipe, the above gain in kinetic energy is the frictional energy, 
which being converted into heat characterizes the process as one of constant 
total heat if in a non-conducting container. The velocity attained should, 
therefore, be that due to attaining the work of a cycle bounded by a constant 
total heat line between two constant pressure lines, corrected for actual thermal 
loss by conduction. Equations might be set down giving the velocity or flow 
per second in terms of initial and final pressures and conduction loss, but this 
would be of no assistance whatever, in fixing those pressures in terms of pipe 
dimensions, which is what is needed. This desired relation between pressures 
lost in a pipe of given dimensions and the corresponding energy was first estab- 
lished on somewhat questionable empiric grounds by Unwin, who, reasoning 
from experimentally established laws for water, put the loss of pressure in 



terms of head of fluid as- proportional to, (a) the velocity head, H=. ( 

(b) the length (L) directly; and (c) inversely as the mean hydraulic radius, 

/ area \ 

m = : : — ) . The name, coefficient of frictio 

Xpenmeter/ 

of proportionality and it is usually designated by £ 



/ area \ 

m = : : — ) . The name, coefficient of friction, is given to the constant 

Xpenmeter/ 



(Loss of fluid head due to friction) =U ~)-==tJH—. . (1214) 

\2g/m m 



With this as a basis and on certain reasonably fair assumptions, the relation 
between flow velocity and pressure drop can be derived. These assumptions 
are, that the fluid behaves like a perfect gas according to PV = wRT, that the 
work of friction is the product of friction head in feet of fluid and weight 
of fluid flowing, that negative work is done by the fall of pressure equal to 
the change of pressure into the volume, and finally, that the change in kinetic 
energy in passing from high to low pressure is equal to the sum of the pressure 
change work, and friction work, neglecting all other influences. 



Change in kinetic energy in 1 f Work done by 1 ( w ] of 
passing from high to low = by change of - friction 
pressure regions in pipes J 1 pressure J ^ 



HEAT AND WORK 1113 

These quantities are evaluated algebraically as follows: 

Let u = velocity, mean, in feet per second; 

Pi and P2 = initial and final pressures, pounds per square foot; 
pi and p2 = initial and final pressures, pounds per square inch; 
V = cubic feet per pound fluid at mean pressure; 
Um = velocity in feet per minute; 
w = pounds fluid per second; 
A = area pipe in square feet; 
m = mean hydraulic depth; 
L = length of pipe in feet ; 
D and d = diameter of pipe in feet and inches respectively. 

Then for an infinitely small pipe length dL and a correspondingly small 
velocity and pressure change du and dP 



\ Change in kinetic energy at ends 
of pipe length 



= 77-[(u-\-du) 2 — u 2 ]=— udu (a) 
wdH (b) t 



(1216) 



y2 2| dll 

because H = ^-, and dH = , and the square of a differential may be neglected 

in comparison with the differential itself. 

Also 
(Work done by change of pressure) = — wVdP= =- dP. . . . (1217) 

and 

(Work of fnction) = (Loss of head in feet) X (Pounds of fluid) 

= VK—w (1218) 



Therefore, substituting these values, Eqs. (1216), (1217), and (1218) in 
Eq. (1215), 

wdH+ ^ dP ^ H ^ w = 0; 
P m 

f+gdP+^ = 0. ..... (1219) 



But 



RT_RL 2 9RT _ 2gA 2 RT = 2gA 2 RT JlgA 2 P 
HP'u 2 p ~ /wV\ 2 p w 2 V 2 P w 2 R2T2 r> w2RT ' ' 



2? wr p 2 



1114 ENGINEEKING THEKMODYNAMICS 

Substituting Eq. (1220) in Eq. (1219) gives Eq. (1221), 

f +S™^=°- •••••• om 

Integrating between head limits Hi and H 2 , pressure limit- Pi and P 2 , 
and length limits and L, 






But 

hence 

Therefore, 



Hi ' w 2 RTi K \ x J ' * m 



H 2 Pi , Au AuPi 

and w = -==- = 



Hi P 2 ' Fi P2 7 ! ' 

jA^^jRTi 

w 2 RTi u 2 Pi 2 ' 



ioge S + ^ (P22 - Pi2)+ ^ =o - ■•■■• (1222) 



This gives for the velocity at the beginning of the L foot section 



„- «*™'-™ (1223) 



Pi 2 



(^S) 



For pipe lengths of usual size and the small allowable pressure drops, or 
small ( p^ ) , it is usually fair to drop ( log e -^ ) , in comparison with the quantity 
to which it is added, and doing this the velocity becomes as in Eq. (1224) 



\ gmRTi{Pi 2 -P 2 2 ) \ gm{Pi 2 -P 2 2 ) e . A , 100il , 

U = ^j UZP? = \J — ^p^— feet per second . (1224) 

When the pipe is of circular section 

area xD 2 _Z)_ d _&, 

m_ perimeter~4XxZ)~T _ 12X4~48 

p 

p = — — , and u m (feet per minute) = 60w (feet per second) ; 

u m =(Const.)X^ dRT{ £f~ P22) = (Const.) X ^ (? ^~ P22) ft. per minute . (1225) 



HEAT AND WOEK 1115 

The corresponding weight is given by, w m = Au m h = —r-u m % = .0054:5d 2 u m §. 



(Pounds of fluid per minute) = (Const.) X a/ r l 2 ( a ) 



= (Co„ S t.)x/-^f^ 



U>Pi 



(1226) 



In all ordinary cases of pipe flow the ratio of the absolute pressure at the 
end of a length to that at the beginning is sensibly equal to 1, at least to 
greater limits of accuracy than the coefficient of friction can be established 
or the premises of the above formula can be accepted as sound. When 

this is so the ratio *—- becomes 

Pi 



(Pi—P2)(pi+P 2) 
Pi 



(pi-P2)U+^j=2(pi-p 2 ), (approx.) 



which on substitution gives the second form of Eq. (1226) the following value 
Eq. (1227) : 



w m = (Constant) X yj d ° * {V * V ^ (1227) 

The first determination of the value of the coefficient of friction based on 
experiments of some accuracy is that made by Unwin on data obtained from 
tests of the compressed-air mains of Paris, all being less than one foot in diam- 
eter and his value is given by Eq. (1228), 



= .(Wl+^) (1228) 



Thi? indicates that the velocities vary as the square root of diameter divided 
by tile above fraction of diameter, whereas for liquids, velocities by some for 
mulas are given as functions of the square root of diameter, and by others gen- 
erally believed to be better, as increasing with diameter faster than the square 
root. No better values appearing for many years, engineers have adopted 
and used the above results with practical success, though it must be confessed 
without scientific accuracy, and experience has led to Various values of the 
con^ants, the most used of which is the Babcock value in Eq. (1229). Careful 



w 



= 87 m*=$ (1229) 



1116 ENGINEERING THERMODYNAMICS 

analysis of existing information by Martin in 1903 led to the adoption 
by him of a different constant a/7000 giving the following series of formulas, 
all equivalent, but put in convenient form for solution of any desired quantity. 






. pmp<^B>, rs ,A (mo) 



Kd 



„_w_ 7000rf 5 (pi-p 2 ) ._ ud? , - 

•'J Ll(u*$ =3 -° 56 ' * " ' ( } 



- 3 iS-T= 256 / J2 'F^= i^ir; ■ ■ 







^( 1+ t) ™( 1+ t) M 1+ t). 



(V2 Vi)— 700 oad 5 7000d 5 65360d ;•••'■• (1233) 



^( 1+ t) ™( 1+ t) u2L ( 1+Z ~f) 



Hg ~ 49 W ~ 49d 5 " 454d " B '* U ^ 4) 



/.0002w 2 L .0002 7 2 SL , , , 10Q _. 

d= Vte^^ = Vis^r (approx - ); (1235) 



r _ 7000(p 2-pi)Bd 5 _ 7000(p2-7>i)^ 5 n9 _. 



( 1+ ¥)^ 



where (?>2— Vi) =loss of pressure in pounds per square inch; 
Hg=Ioss of pressure in feet of fluid; 

V = cubic feet of fluid per minute; 

u = velocity in feet per second ; 

8 = density of fluid in pounds per cubic foot; 

w = weight, of fluid flow pounds per minute; 

d = diameter of pipe in inches; 

L = length of pipe in feet; 
27.7(p2 — pi) =loss of pressure in inches of water; 
2.04(p2 — Pj) =1oss of pressure in inches* mercury. 



HEAT AND WORK 



1117 



These formulas are applicable to all fluids, steam, both saturated and super- 
heated, compressed air, ammonia, carbon dioxide and to the combustible gases, 
but not with equal degrees of accuracy over the whole range of kinds of fluid, 
rates of flow and density. All sorts of special tests are available, giving data 
to check the formula for one case under test and a review of these tests 
would be of value here, did it lead to any suggestion for a better formula, but 
no such suggestion is available, so cases of individual departure from the best 
general law ever proposed will be omitted with the warning that engineers 
working with some one fluid over a definite range of conditions of service deter- 
mine for themselves the correct numerical constant to be used in the general 
equations to meet those conditions. 

One general investigation for steam recently conducted by Eberle under 
the auspices of the German Society of Engineers has led to an important 
and useful conclusion, inasmuch as it established that the coefficient of 
friction is independent of pipe diameter, steam pressures, quality and velocity 
for very considerable ranges. The velocities varied from 2000 to 15,000 
ft. per minute, pressures from three to ten atmospheres, with steam saturated 
as well as superheated to 180° F., and in pipes from 3 ins. to 6 ins. diameter ap- 
proximately, while the coefficients varied some 20 per cent. Gutermuth 
had used a constant coefficient of friction, which is 1.43 times mean of Eberle's, 
and while the latter's value is most accurate for high pressures, the Gutermuth 
value is recommended by Rietschel for low pressures such as are found in 
steam-heating systems and steam exhaust pipes. This simplified formula 
is given by Eq. (1237), 



u 2 L, 



(p 2 _p 1 )=2.667Xl0- 5 X-j-B (Eberle) 



(1237) 



A number of these pipe flow formulas have been analyzed and reduced to com- 
mon terms by Gebhardt and the constant used in the following expressions 
are those found by him set together to show the variations : 



(Velocity feet per minute) 



IL 



= B 



(V2~Pi)d 



where A = 



{ 9240 Geipel & Kilgour (a) 

9976 Hawksley (6) 

10350 Martin (c) 

{ 10360 Hurst (d) 



1+3 f' 



15950 
16050 



442 



/(p i 1 - 94 -P2 t 

\ l 2 L 



94 



)d 



Babcock 
Unwin 



Ledoux 



(«) 



(?) 



. (1238) 



1118 ENGINEERING THERMODYNAMICS 

(Flow in pounds per minute) 



~ c 4 



(V2-Pl)< 



where C = 



50.2 Geipel & Kilgour (a) 



=7^i" { 



2M ^zi 



-W-MU5 



)d' 



54.4 Hawksley 

56.5 Martin 
56.5 Hurst 

87 Babcock 
87.5 Unwin 



Ledoux 



(b) 

(«) 
(d) 

(e) 
(/) 



(a) 



(Pressure drop pounds per square inch) 



= (P2-pi)=E^-p, where E= ■ 



.0003960 Geipel & Kilgour (a) 



0003370 Hawksley 
.0003133 Martin 
I .0003126 Hurst 



w 2 L 



= F- 



(•+¥) 



( pi 1.94_ p2 1.94) = >1669 



where F= j ■ 
w 2 L 



.0001321 Babcock 
0010306 Unwin 



d 5 



Ledoux 



(b) 
(c) 
(d) 



(e) 
(/) 

(9) 



(Pipe diameter, inches) 

= d= G\ \- f r^, where G'= ■ 

\(P2-Pl)% 



' .2087 Geipel & Kilgour (a) 

.2010 Hawksley (6) 

.1990 Martin (c) 

.1990 Hurst (d) 



= 69 n / ^ 

•^^(pi 1 - 94 -^ 1 - 94 ) 



Ledoux 



w 



(1239) 



(1240) 



(1241) 



Actual pipe lines have bends, branches and valves, together with enlarge- 
ments and reductions of cross-section, all of which introduce resistances as well 
as the skin friction of the pipe and which are very much more difficult to evalu- 
ate. The general method of procedure is to give the resistance in terms of the 
equivalent length of straight pipe; thus, Hunt gives the following for 90° 
elbows. 

Diameter of pipe 12 3 4 5 6 7 8 9 10 12 14 16 18 20 22 24] 

Elbow resistance in feet 

of straight pipe 1.5 4.9 9.4 14.5 20 25.9 32 38 44.4 50.7 63.7 76.7 90.1 104 117 130 144 



HEAT AND WOEK 1119 

These are of little practical value, as the proportions of elbows differ in every 
make of fittings, and are given here to indicate the appreciable quantities in- 
volved and the necessity for accurate tables of data in commercial work relating 
to some one line of fittings to be used. 

When low-pressure gases like air or furnace products flow in ducts, flues, 
and chimneys, there arises a special case of flow conditions characterized by 
a possibility and often a probability that the whole skin frictional resistance due 
to straight pipe may be entirely overbalanced by the aggregate of the special 
resistances due to bends, branches and changes of cross-sectional areas. There 
are two typical cases of such low-pressure gas flow, first, that in which the flow 
is all horizontal or if vertical the whole system at constant temperature, and 
this is a typical duct-flow condition, second, that in which there is a vertical 
flow not necessarily straight upward and in which the gases are hotter than 
other gases communicating with them, and this is a typical chimney-flow con- 
dition. Each case must be treated by itself in order that workable equations 
and constants may be developed and the former will be taken up first. 

Usually direct flow takes place from, or to, the free atmosphere, or a place 
where at least the pressure is that of the atmosphere whether the temperature 
is the same as it or not, and the problem always takes the form of fixing the 
excess pressure over atmosphere that will deliver a required quantity of air or 
gases through ducts of a given size, or inversely the size for a given capacity, 
when the whole resistance is that due to the system. of passages. The whole 
resistance is, therefore, to be evaluated and this consists of various frictional 
resistances due to straight duct walls, to changes of direction of flow, to changes 
of cross-section of duct, to inlet and to outlet orifice conti actions. To these 
resistances in terms of feet of head of gas, or pounds per square foot, ounces 
per square foot, or inches of water, there is to be added the velocity head to 
get the total equivalent head or pressure that must be applied to dead air, to 
establish flow at the required velocity against the imposed resistances. 

u 2 
Let Hg = tt =head of gas equivalent to velocity of flow; 

u 2 l 
P F = —— = pressure pounds per square foot equivalent to velocity head; 

1 = density of gas in pounds per cubic foot; 
Pf = pounds per square foot static pressure lost in wall friction. 
P R = pounds per square foot static pressure lost in changes of flow 
direction, cross-section, inlets and outlets. 
P = whole static pressure equivalent to velocity and all lesistances; 
F F = friction factor, which, applied to velocity head gives the frictional 

loss of pressure; 
F R — resistance factor, which applied to velocity head gives the loss of 
pressure due to resistances; 
^coefficient of friction for straight walls; 
C = circumference or perimeter of duct in feet; 
A = area of duct in square feet; 
L = length of duct in feet. 



1120 
Then 



ENGINEERING THERMODYNAMICS 



Velocity pressure P v 

p— \ + Frictional pressure P F 

+ Resistance pressure P R 



in pounds per square foot. 



(1242) 



.024 












































































































































































































































































.02* 












































































































































































































































































.020 












































































































































































































































































.018 












































































































































































































































































.016 


































































































































































































































































































































-£.014 
n 

03 






























































1 












































O 










u_ 


-By 


ie'fc 


>v i\ ! 






































M 






















































O pi? 










\ 




















































-\ 












































a 




\ 
















































o 




\ 
















































-4-' 




V 






\ 












































g.010 




v V 




















































\ \ 














































































































\ 






















































\ 
















































.008 






N s 






























































































































































\ 




s 


















































\\ 




\ 














































.006 




V s 


s s 


















































V 


\ 



































-e= 


S 2 - 


















V 


N., 
































o 


HI" 




















*s 


































Q- 


1ft'' 










.C01 














































































































































r 
























































































































.002 



































































































































































































































































































































20 30 

Velocity Feet pec Second 



40 



50 



Fig. 303. — Coefficients of Friction £, for Air in Ducts. 



But 



P=1?(1+*V+F*). 



(1243) 



HEAT AND WORK 



1121 



Hence 



2g\ 



l + &£+F a 



Velocity pressure 



2g 



u 2 li C \ 
+ Maintained resistance -~— f XJj-j + F R J 



(1244) 



This can easily be evaluated numerically, when there are available experimental 
values of the coefficient of friction for straight ducts, and of the flow change 
resistance Fr, but as it seldom if ever happens that a duct is uniform, and as 
these values depend on the dimensions, there must be a summation of the pres- 
sures lost at each point of change and in each straight section. Hence, 

Maintained resistance = ^ | 2 (uKj) + 2u 2 Fr\ . . . (1245) 

Values of the coefficient of friction are given graphically, Fig. 303, for straight 
ducts of brick or iron for velocities up to 50 ft. per second, and for iron ducts 
different values are given for perimeters or circumferences from 8 to 100 ins. 
These values as well as the flow-change factors in the following Table CXLIII 
are those recommended by Ohmes from Rietschel for ventilating ducts, and are 
intended for the usual velocities therein attained, 6 to 24 ft. per second, when 
served by fans, and 3 to 8 ft. per second when the flow is due to natural or con- 
vection draft. 



Table CXLIII 
FLOW CHANGE RESISTANCE FACTORS F R (Reitschel) 



Condition. 



Resistance Factor F R 



Sharp 90° elbow 

" 135° elbow 

Long bend : r = width of duct 

r = 2 to 4 duct widths 

" r = 5 to 6 duct widths 

Long bend 135° 

Long double offset 

Outlet register with valves f free area and 2 Xflue area 

' ' ' ' face at f free area 

• ' wire screens 1.5 Xflue area 

Entrance for square corners 

' ' rounded corners 

" flue extending into header as short pipe . . 

Enlargement of area from A x to A 2 , sharp corners 

Reduction of area from A 2 to A h sharp corners. ...... 

Free discharge into room when velocity becomes zero . 



1.1 
.3 
.25 
.15 

.07 
.15 
.4 to .1 
.6 
.4 
0.0 
1.0 
.5 to .2 

1.5 
(At ^ 2 






Ai\ 2 
A 
1.0 



1122 ENGINEERING THERMODYNAMICS 



om 



These pressures that are lost, due to the resistances, are convertible from 
pounds per square foot to inches of water bj^ the usual factor, but in this branch 
of engineering practice they are usually given in ounces per square foot and 
this value follows directly from the formulas, if the density is used not in terms 
of pounds per cubic foot but in ounces per cubic foot. Of course, there may be 
other resistances in the flow path such as filter screens or pipe coils serving to 
heat or cool the air, and any such resistance is to be separately evaluated and 
added, but as such things are special details they are omitted here and if needed 
are to be looked up in tables of data prepared for engineers practicing this 
specialty. 

When the flow of gases in the system depends on convection, which follows 
the heating of a rising column, which is replaced by a falling cold column, the 
flow characteristics are dependent on the temperatures and vertical heights 
involved and the most important practical case of such flow is that in chimneys, 
though in some ventilating systems a corresponding case is found. 

Let H = vertical distance in feet equal to the difference in level between 

inlet and outlet of the heated column of gases; 
" P = difference in pressure at base of hot column between it and the 

atmosphere, to be called the draft and here given in pounds per 
square foot; 
" hw = draft in inches of water; 

Hg= equivalent head of hot gas; 

T H = temperature of hot column assumed constant throughout; 
T c = temperature of outside air; 
.084 = density of products of combustion at 32° F. and 29.92 ins. Hg; 

492 X. 084 41.43 



$h = density of hot gases, pounds per cubic foot = 



T H 



398 
for products of combustion, or -=- for outside air; 

J- c 

.0807 = density of pure dry air at 32° F. and 29.92 ins. Hg; 

B c =density of pure cold air at 32° F. and 29.92 ins. Hg: 

492 X. 0807 ^39.8 
T c ~ To' 

C = circumference or perimeter of stack in feet; 
A area of stack in square feet; 
w = weight of gases per second. 



Then 



J Weight per cubic foot of 1 v „ = [ Pressure at base of height #1 ^ 24fi v 
[ hot or cold gases J [in pounds per square foot J 



HEAT AND WORK 



1123 



Hence for pure air in the chimney the difference in pressures at the base is 
given by Eq. (1247), 



r, A~A 



(1247) 



For products of combustion of assumed density in the chimney the corre- 
sponding pressure difference is Eq. (1248), 



P = H 



39.8 



41.43\ 
"Th) 



(1248) 



The equivalent head of hot gases is given by Eq. (1249) : 



P P 



39.8 41.43 
T c T H 


= H 


41.43 
TV 



,961 



T c J 



(1249) 



This is in turn equivalent to a draft in inches of water, the usual unit of mea- 
surement, given by Eq. (1250), 



P /7 64 7 95 N 

g^- = H I -jr - -~- ) for chimney gases ; (a) 



r{ ¥ c ~"h ] 



(&) 



(1250) 



Solving for a chimney temperature, ^ = 600° F., or T H =10Q0° F., this gives f 
in. of draft per 100 ft. of height. However, these are the full static pressures or 
their equivalents and after flow is established some pressure will be changed to 
velocity, and some more used up in overcoming the maintained resistances 
in coal beds, boiler tube banks, and economizers in furnace-chimney flow, 
which correspond to tempering coils and filters in heating and ventilating duct 
flow. Calling these special resistances Fs the general relation becomes that of 
Eq. (1251), 



p= £[" 2+ 2 ( uK S + 2(M2Fie)+ 2 



(«%)] 



(1251) 



All these resistances due to flue-wall friction, entrances, outlets, bends or changes 
of cross-section can be evaluated as in ventilating-duct work, but fuel-bed, 



1124 ENGINEERING THERMODYNAMICS 

economizer, and boiler-tube resistances cannot be so evaluated, because in 
them resistance conditions are constantly changing, as is also the velocity and 
temperature of gases due to changes of boiler- load. It is, therefore, necessary 
in dealing with furnace chimney draft to proceed even more empirically than 
for air ducts, and the usual method is to apply a general factor inclusive of all 
interferences and resistances and yielding a numerical result near enough to the 
truth to be of practical value. Thus, Stirling suggests that as an overall cor- 
rection factor is to be applied there is nothing to be gained in assuming the 
gases to be different in composition from air, so that the static maintained 
draft after flow is established, as measured in the base of the stack will be that 
for no flow, less something to account for velocity and friction resistances all 
together, derived somewhat as follows: 

The flue friction loss in head is given by -~- £ T^ or ducts, butwB= (-7 ), 

and for a given weight or volume per second flowing, the velocity is inversely 

w 
proportional to the area, so that u 2 h = (Constant) X-ttj. Therefore, the draft loss 

would be 

(Constant) X^~ -j3 = f~Js> 

where / is given the empiric values of Eq. (1252) and the maintained draft is 
given by Eq. (1253), 

[ .0015 for steel stacks, gas temperature = 600° F. or TV = 1060 1 . 
/= .0011 for steel stacks, gas temperature = 350° F. or T H = . 810 . (1252) 
I -0020 for brick stacks, gas temperature = 500° F. or T H = 960 J 



h w =7MH 






-fw^. ..... (1253) 



As a matter of fact the actual maintained draft, while it is a prime factor in 
the rate of combustion of coal, is not of such great importnace in the problem 
of flow through chimneys as the capacity of chimneys. This capacity is pri- 
marily given in terms of weight of gases per hour, but for practical application it 
is common to express it in terms of fuel-burning capacity, or boiler horse-power 
equivalent to the weight of gases per hour, by assuming values for the con- 
version factors. 

Let G = weight of gases per hour; 
" G m = maximum weight of gases per hour; 
" K = proportionality coefficient. 



HEAT AND WORK 1125 



Then 



G=36QOw = ZmAuZ H = 3600AKy/2gH a SH 
= 3600A2^2^.961^-l) X~^ 



= 149148K^2^(.961^-l) (1254) 



It is clear that as T H becomes greater the draft increases and the velocity cor- 
responding to it, but the density becomes less, so that there would appear to 
be some temperature at which the weight would be a maximum, and this is 
to be found by differentiation. Accordingly, put the above in the form 



G=149148XA^2^(.961^r-yy 



and making -==- = 0, gives 
al h 



•9£>'vn 2T ' T 3 — ^' 



Th 2 T c ' T H 
or 



T H = ^ = 2m\T c (1255) 



This shows that the capacity will be a maximum when the absolute tempera- 
ture of the hot column is, in round numbers, twice that of the cold surrounding 
air. The capacity at this time is found by substituting this internal tempera- 
ture in Eq. (1254), which gives 

G m = U9USK^V2gH (2-1) 

1 H 

= 11962502£ 7 ^V'# (1256.) 

i H 



When the problem is that of stack design, it is this maximum capacity that is 
the important quantity, but it must be remembered that unless the gas tem- 
perature is right, the stack will work at a lesser capacity and the actual or prob- 
able capacity of a stack at a temperature other than that giving maximum 
capacity is a determination quite frequently necessary. In stack designing the 



1126 ENGINEERING THERMODYNAMICS 

gas temperatures are never known, though they may be approximated, but 
instead of making approximations for each quantity separately as involved in 
the stack capacity, this is set down most commonly in terms of coal-burning 
capacity or boiler horse-power, as formulas in the derivation of which general 
assumptions are made. Some of these common assumptions are 

(a) The stack will operate at maximum capacity; 

(6) The air temperature is 50°F., or 7^ = 510, and 7^ = 2.08X510 = 1061, 
or t H =Wl° F. 

(c) Twenty-four pounds of air will be required per pound of coal yielding 

25 lbs. of gases; 

(d) Five pounds of coal per hour will be equivalent to one boiler horse- 

power. 

The justice of these assumptions need not be discussed as they are purely 
arbitrary, and the user of these equations may modify them at will, starting 
from the fundamental relations, but it is on such assumptions that most of the 
existing stacks have been designed and the coefficient K determined to apply 
to them. Introducing the assumptions it follows that 



The value of the coefficient K, which here may be regarded as a general correc- 
tion factor, though it is often called a coefficient of discharge, is most commonly 
taken as, K = .3, but Kent has used another value combined with the 
assumption that next the walls there will be a dead layer of gas so that the 
effective area (KA) will be the actual area less the cross-section of 
this dead gas. This layer is arbitrarily taken as being 2 ins. thick, so that 

2 2 

it will subtract about, ttjcD, and, — X4D sq.ft., from the area A for round and 

square stacks respectively, leaving the effective area as, (A — .59 VA) for round, 
(A — .666V A) for square, or approximately, (A — .6\/A) for both. This sub- 
stituted in Eq. (1258) yields the Kent formula Eq. (1259), 

(Boiler horse-power for the stack) = .333 (A - .WA)VH . . (1259) 

A great many empiric chimney formulas are available, all derivable from the 
same fundamental expressions here developed but with different values for the con- 
stants, and as they are all empiric beyond the first steps they are omitted, and 



HEAT AND WORK 1127 

reference is made to an excellent comparison in Gebhardt's Steam Power Plant 
Engineering. Chimneys intended to carry off the products of combustion of 
gas and oil fires must be proportioned on the same principles except one, as 
control those for coal fires, and that is due to the general absence of anything 
in the former commensurate with the fuel bed resistance in the latter, and 
which is there the main resistance. The general procedure with coal fires is 
to first fix the height that will give the draft needed to burn the coal at the 
desired rate on the grates, and then to select an area sufficient to carry off 
the gases produced. In the latter case the draft is of minor or zero consequence 
so that a shorter, wider stack may be just as serviceable for a gas or oil fire as a 
narrow high one of equal capacity, whereas with coal fires the height is needed 
for, and controlled by the bed resistance or rate of combustion. 

Prob. 1. Simultaneous readings showed the pressure at the ends of a 6-inch 
steam pipe line, 650 ft. long, to be 120 lbs. gage and 105 lbs. gage respectively. What 
was the flow as found by Babcock formula? By the Eberle? 

Prob. 2. It is desired to transmit 1000 cu.ft. of compressed air per minute a dis- 
tance of 2 miles and with pressure drop not to exceed 20 lbs. If the air is initially under 
a pressure of 100 lbs. per square inch gage, what size pipe will be needed? 

Prob. 3. Natural gas is being transmitted at a pressure of 150 lbs. per square inch 
gage through a 10-in. pipe. What will be the amount which may be transmitted with 
a loss of 5 lbs. per sq. inch per mile. What size pipe would be required to carry 
1000 cu.ft. (std.) per minute with this drop? 

Prob. 4. A 4-in. pipe 500 ft. long is carrying dry saturated steam at a pressure of 
100 lbs. per square inch gage initially and with a drop of 5 lbs. Check the weight 
flowing, by the formula of each of the authorities. 

Prob. 5. For the same pipe find the pressure drop should the quantity flowing be 
doubled. 

Prob. 6. A furnace burning 1000 lbs. of coal per hour has a stack 4 ft. in diameter 
and 150 ft. high. If pounds of air per pound of coal is 25, outside temperature 
80° F., and stack temperature 400° F., what will be the draft? 

Prob. 7. What will be the maximum weight of coal which can be burned with this 
stack for the given amount of air per pound of coal? 

Prob. 8. A stack in Glasgow has a height of 435 ft., a diameter of 13 ft. 6 ins. 
and is stated to have a capacity of 10,000 H.P. or 50,000 lbs. of coal per hour. Check 
these figures. 

19. Thermal Efficiency of Compressed Air Engines Alone and in Com- 
bination with Air Compressors. Effect of Preheating and Reheating. Com- 
pressor Suction, Heating and Volumetric Efficiency. Wall Action. For 

certain special service conditions compressed air supplied to piston engines 
is the most desirable method of developing the necessary power, and for this 
reason a brief review of the thermal characteristics not only of the engines 
but of the whole system including the compressors is of value. It is, however, 
important for another reason, and that is the possibility of securing ever so 
much better results than have yet been obtained with air engines, which would 
have the effect of bringing the system into wider use than at present, when it 



1128 ENGINEERING THERMODYNAMICS 

is a sort of last resort. In mines, however, where exhaust steam or gas- 
engine exhaust gases would foul the air that must be breathed by the workers, 
or even set fire to combustible gases in coal workings, compressed-air engines 
are the only available sources of power except electric motors, and are now 
used as haulage locomotives, hoists, pumps, drills and cutters. Practically 
all excavation work must be conducted under conditions that are served by 
compressed-air power and the use of very small units like hammers, riveters, 
drills and clippers is almost universal in shops. Probably nothing has called 
attention to the possibilities of wider large scale use of compressed air, than 
the recent installation of a whole system of compressed-air deep-mine hoists 
in the Butte, Mont., mines. This is especially remarkable in view of the fact 
that the compressors are operated electrically from transmitted water power, 
and it was found better to use electricity to operate compressors discharging 
into storage reservoirs from which the air passed* through heaters on the way 
to hoisting engines, than to operate the hoists electrically. One factor in this 
conclusion is the special construction of the engine which as the cage descends 
operates as a compressor storing air by the energy of the falling hoist, and 
acting as a brake at the same time. Another similar development has recently 
been accomplished in another direction by the perfection of the Porter com- 
pound compressed-air locomotive for use in mines, lumbering operations, or 
other places where fire is much feared and in hauMng cars about the yards of 
large manufacturing plants where cleanliness as well as avoidance of fire risk is 
important, and where the service is so intermittent that a steam locomotive 
might burn more coal to keep up pressure standing, than in hauling loads. 

It seems a very simple operation at first glance to admit to a cylinder of ordi- 
nary steam engine construction, a charge of compressed air instead of steam, 
but two important difficulties or objections appear, peculiar to the substitu- 
tion of the air for steam. In the first place unless there is considerable expan- 
sion the economy will be very poor and the work obtained only a small fraction 
of that expended in compressing the air, and second, as a consequence of the 
expansion the air will become very cold, even a moderate amount of expansion 
lowering its temperature much below the freezing-point and filling cylinder and 
ports with ice from the moisture of the air unless it be dried previously. If, how- 
ever, the air be heated either by internal combustion, as it can well support a 
fire itself, or by external heat application as it passes through chambers or tubes 
heated by steam or the hot products of coal, oil or gas fires, it will increase 
in volume thereby and one cubic foot of cool compressed air will become two or 
three cubic feet of hot compressed air, if its absolute temperature be doubled 
or trebled, and so capable of doing two or three times the work of the cool air. 
By this means, more power can be obtained than is required to run the com- 
pressor, and in fact if compressor and engine be coupled together there will be 
possible a sufficient net output of power to warrant calling the combined unit 
a form of gas engine. Heat so added to compressed air is termed preheat and 
the heat of preheat is converted into work in part, whether the compressor and 
engine work separately at a distance or are coupled together as a single power- 



HEAT AND WORK 1129 

generating unit. It will be shown later that if high pressures are used, the 
efficiency of transformation of the heat of preheat into work may be very high, in 
fact the same as a Brayton gas cycle, which in turn is the same as the modern 
standard Otto gas-engine cycle for equal compressions. This preheat, more- 
over, removes any difficulties of freezing and in some cases only sufficient heat- 
ing is used to accomplish this, and it is right here that a good prospect has been 
neglected. 

Reheating the air between the cylinders of compound compressed-air 
engines in the receiver, is exactly the same in kind and effect as preheating and 
yields efficiencies that correspond to the pressure at which it is taken up; the 
higher the pressure the more return will be available for the heat used, but as 
practiced other factors than thermal efficiency may control the operation. 
One very remarkable case of this sort is that of the Porter locomotive in which 
the air is stored at a pressure of about 800 lbs. per square inch in the tank, 
which gives the tank the large compressed-air weight and low pressure volume 
capacity, needed for a long haul, and also dries the air. This air then passes 
through a reducing valve where it develops kinetic energy of expansion, imme- 
diately losing it by impact and so not suffering any temperature drop, when 
losing pressure from 800 to 250 lbs. per square inch. At this latter pressure and 
at approximately atmospheric temperature, it is admitted to and expanded in 
the high-pressure cylinder, exhausting to the receiver at a temperature as much- 
below that of the atmosphere as practicable, about 30° below zero F. This per 
mits it to be reheated from the atmospheric air itself, which is blasted through the 
receiver tubes, the working air so warmed, subsequently expanding in the low- 
pressure cylinder converts some of the heat of the atmosphere into work. This 
process is therefore, peculiarly interesting, inasmuch as it is a means of utilizing 
for useful work some of the heat of the atmosphere, incidently improving 
economy in the use of the compressed air about 30 per cent, and avoiding at 
the same time cylinder freezing or interference with lubrication. 

It has been shown that so far as economic performance independent of 
cylinder capacity is concerned, compressors give the same result with, as 
without clearance, therefore, in considering the efficiency of combined air 
engines and air compressors, it will be assumed that the compressor has no 
clearance and this assumption materially simplifies the work. With respect 
to the engine, the same would be true if the compression were complete and 
as it can be made so, this assumption will also be made, and the air engine 
treated as if it had zero clearance. If in any case it has clearance without com- 
plete compression it will require more compressed air per unit of work done, 
which is an avoidable waste, most easily eliminated by making compression 
complete. 

Consider Fig. 304, as representing a zero clearance single-stage adiabatic 
compressor card ABCG on which the isothermal is dotted, AD. Then, if the air 
after compression be cooled to its original temperature in the receiver or pipe 
line, which usually happens, the volume of cool compressed air available for 
work is CD. This may be assumed to be admitted to an air engine without 



1130 



ENGINEEKING THERMODYNAMICS 






clearance loss and there acliabatically expanded along DE completely, or used 
non-expansively along DF. These two cases represent limiting economic 
performances, best and worst. The efficiency of the whole process is given by 



P20,000 














































































































c 




D 




B 
























































\ 




















































\ 




\ 

1 












































13,000 








\ 






\ 
















































\ 






\ 










































1 








\ 






\ 










































4 








\ 


\ 




\ 


V 
















































\ 


\\ 






\ 








































$"10,000 










\ 


V 




N 


V 














































\ 








\ 






































2 ' 










\ 


\ 


\ 










































3 

CO 












\ 


\ 


it 
\ 








































u 












\ 




N 


\ 


























































X 


^ 




X 


«N 




















































^ 


o.^ 






























































•-. 


-._ 
























G 








































■ — 


=■= 


















\ 






















b 






















A 




i 

























































5 V 






Volume in cu.f t. 
Fig. 304. — Simple Compressed Air Engine, and Single-stage Compressor, Combined Diagram. 

the ratio of the engine work area, to the compressor work area, whence, for 
single-stage adiabatic compression and simple compressed-air engines, 






_ area CDFG . 

^ 1 = nv> An * or zero en g me expansion 

area Cd^xvjt 



(a) 



Efficiency =\ \. (12CC) 

„ area CDEG , ' . .,. i 

^ 2 = nvt ah lor complete engine expansion (6) 

area Co-ACr 



These areas can be evaluated algebraically. 

Let P b = highest pressure, pounds per square foot; 
" P a = lowest pressure, pounds per square foot; 



IX 



= R F = pressure ratio ; 



" T a = atmospheric temperature absolute; 

" T b = maximum delivery temperature of compressor. 






HEAT AND WORK 
Then according to Eq. (48) making s=y, 

7-1 



1131 



Area CBAG = -~P«7«| 



)■ 



Area CDFG = (P b - P a ) V d = (P b - P a ) V a ^ = V a P a {R P - \)~ 

r b &p 



■■v-m 



Area CDEG = ~^-P e V e ( r 



-'->) 



But 



R P -* 



Therefore, 



Area CDEG = - J -P a V c 
y-1 



7-1 
R P v -1 



7-1 

R P v 



Substituting these algebraic values for the areas in Eq. (1260) the efficiency 
results as in Eq. (1261), 



PaV a 



Ei=- 



Rp-1 
Rp 



•y-l 



jPaVa^R p 7 -1 

7-1 



T-l 

T 



Rp-1 



Ri 



(rJ-* -l) 



(a) 



T 



T-l 



PaV a 



E<X = 



Rp y -1 



T-l 

Rp * 



Y / 1=1 \ 1=1 

-ti PaVa \Rp y -V Rp y 



(b) 



■ . (1261) 



If the compressor were two stage and the engine simple, the diagrams would 
appear as in Fig. 305. The results will be different, as the denominator in the 
efficiency equation, which is the work of the compressor, has a different expres- 
sion given by Eq. (1262). 



2y t-1 

(Work of compressor, two-stage, no clearance) = — ; — P a V a (Rp 2 7 —1). (1262) 



1132 



ENGINEEKING THERMODYNAMICS 



Calling the new efficiencies E3 and E± for two-stage compression compared 
with zero engine expansion and complete engine expansion respectively, 



and 



Ea = 



T-l 
2y 



Rp-1 



T-l' 
#p(#P 2y "I 



(a) 



#4 = i 



1?P v -1 

ff P ? 

x _i 



_ 1 

— 2 



7-1 
# P 2 ^ +1 






(6) 



? 3 

Volume in cu.f t. 



. (1263) 




Fig. 305. — Simple Compressed Air Engine, and Two-stage Compressor, Combined Diagram. 



It is clear that practically any combination of compressor, single, two-stage 
and three-stage, with engine simple, compound, triple, having an infinite, a 
finite or no receiver, and any phase relations, may be made, so a complete 
investigation of all cases would be very lengthy. To avoid spreading out the 
work of analysis, no more combinations will be examined, but the methods 
used can be applied by any one to other combinations, by the aid of the equa- 



HEAT AND WORK 1133 

tions of Chapter II for compressors, and Chapter III for engines with expo- 
nential laws for the expansion. 

The effect of preheating the air between compressor and engine may be 
shown in two different ways. According to the first method the fraction of heat, 
so supplied that is ultimately converted into work, can be determined by divid- 
ing the net work area between the compression and expansion lines by the foot- 
pound equivalent of the heat. This will give the efficiency of the combination 
as a transformer of heat of preheat into work, and is the thermal efficiency of 
a gas-engine cycle consisting of phases made up by the lines in question. The 
other way compares the whole work of the air engine taken as useful output, 
to the sum of the compressor work and the foot-pound equivalent of the heat 
supplied, and this gives the efficiency of the whole combination as it is. Both 
results are of value, each, however, for what it shows. 

In Fig. 306, the single compressor with simple air engine is reproduced with 
various amounts of preheat, and the T$ diagram drawn to correspond to the 
thermal cycle thus created. The no-expansion case of the air engine is omitted 
because the primary purpose is to show how good an effect of reheat may be 
obtained and obviously, failure to expand after preheating involves pure avoid- 
able waste. The preheat is added from D to D', D to B, or D to D" , the first 
being small in amount, insufficient to bring the temperature and volume of the 
cooled air to the values corresponding to compressor delivery, the second just 
reproduces these values, the third makes the volumes and temperatures 
larger. 

Preheat to compressor delivery temperature will, of course, result in an 
engine exhaust temperature equal to that of the atmosphere or compressor 
suction. Less preheat than this will cause the exhaust to be cooler than the 
atmosphere and the minimum allowable amount should be that which will pre- 
vent freezing or interference with lubrication. If the air be very dry such as 
would be the case if compressed to 1000 lbs. per square inch or even less, the 
exhaust may be 20 to 30° F. below zero without making much trouble, but with 
more moisture, such as would be carried by air at 100 to 200 lbs. per square inch 
pressure, the final temperature may be as low as 10 to 15° F., since the high 
initial temperatures will prevent ice sticking to the walls. While in many cases 
the preheat used is just sufficient to prevent freezing, that is, the initial tem- 
perature absolute, at the beginning of expansion, high enough for the ratio 
of expansion to give the above final temperatures, the real economic 
value of preheat is obtained only with larger amounts. There is no reason why 
temperatures as high as the hottest superheated steam should not be used, 
500° to 600° F. in engines of exactly the same structure, or in excess of this to 
any degree if gas-engine structures are used as a model. These large preheats 
give no higher per cent thermal return than the lower ones, but make it possible 
to operate larger engines with smaller compressors, materially reducing the 
first cost of the equipment. 

The efficiency of the heat of preheat alone is that of a cycle bounded by the 
maximum and minimum pressure lines and two adiabatics, the first adiabatic 



1134 



ENGINEERING THERMODYNAMICS 



corresponding to the work that would be obtained without preheat, and the 
second that corresponding to expansion from the temperature and volume 
produced by the preheat. On the diagram, Fig. 306, this cycle is represented 
for the three ^different amounts of preheat by, EDD'E'E, EDBAE, and 
EDD"E'E to both PV, and T$> coordinates. These are, of course, all Brayton 
gas cycles, having established efficiencies given by Eq. (1264) below. 



1 T 

(Efficiency of heat of preheat) = 1 — [ = 1 — 



Ri 



Ta 



(1264) 



This is true no matter how much heat is added and no matter what the temper- 
atures. It is exactly the same as an Otto gas cycle efficiency for equal com- 



p 
































T 










































18000 


— 


DD'B 


D"| 




















4) 






































1 




r \ 




























































D 




<o 




1 




A 


















































B 












8 




\ 




























<! 
1 

a 










































£ 12000 

i 

u 

ft 9000 

to 




\ 


\ 




































































\ 


A 








































n 


s 




























V> 




























D 




































, 






\ 






























_. 




























: 




g cooo 

g 

I 






\ 


























g • 

a 

'/ 






























/ 


\ 


















\ 


s V 


































"f 






































■>» 


■ ^«. 


















E 












































At 


m, 










..^ 


~»~ 






















































r 




r 


/ 






A 




C 












































































2 3 4 

Volume in Cubic Feet 



.1 .15 

Entropy 



Fig. 306. — Effect of Preheat on Compressed Air Engine Performance. 

pressions, and is very high, yielding not less than twice as high a return for the 
heat, as is obtained from the heat of steam in a steam-driven compressor cylinder, 
and may be three or four times as high, depending entirely on the conditions of 
steam and air pressures. It proves the desirability of using very high air pres- 
sures for high thermal per cents of return from the heat of preheat, and with 
compressed air at 13 atmospheres or 191 lbs. per square inch absolute for a 
14.7 lbs. standard atmosphere, the thermal efficiency would be, ideally, 



E=l- 



(13)- 



= 52 per cent. 



If the diagram factors for gas engines applied to these cases the realizable 
amount would be about half f this, or 26 per cent, but without water jackets 



HEAT AND WORK 



1135 



it is likely that fully f would be realized, so the probable thermal efficiency of 
the heat of preheat would be 39 per cent, a truly high value that will bear exper- 
imental investigation. 
The other standard of efficiency, that for the whole system, is given by Eq. 

(1265): 



Efficiency of compressor and 1 
engine with preheat J 



Work of the engine 



(Work of compressor) 

+ (Ft. Lb. equiv. to heat of preheat) 



(1265) 



The foot pound equivalent of the heat of preheat is, of course, equal to, 778 X (the 
heat of preheat) = 778 X wC v (Temperature after preheat — Temperature before). 
If the temperature before preheat be taken as equal to that of the atmosphere, 
which corresponds to complete cooling of the compressed air in receiver and 
pipe line, this becomes, calling the final temperature T H and air temperature 
T A and w the weight of air 



(Ft. Lb. equivalent to heat of preheat) = 77SwC p (T H -T A )^ 



(1266) 



The work of the engine with preheat is equal to that without, plus that gained. 
The work without preheat was given by 



T-l 



PaVa 



7-1 

R P -v -1 
I R P y 



while the work done by the preheat is 

(Work done by preheat) = 778 (Efficiency of preheat) X (Heat of preheat) 

X778wC P (T H -T A ). 



1 



T-l 

R P v 



Hence 



J Work of engine 
with preheat 



T-l 



^PaV 



T-l 

R P y -1 



T-l 

L R P * . 



+ 



r y=1 
R P y -1 



R P y 



X778wC P (T H -T A ). (1267) 



1136 
But 

and 



ENGINEERING THERMODYNAMICS 



P a V a = wRT a = 77$w(C p -C v )T A , 



Op 



C. 



y 1 Up .. Op 0» 



so that — '— P a Va = 77SwC p T A , whence 
Y-l 



(Work of engine with preheat) = 77SwC t 



7-1 

R P v -1 



Ri 



(T a +Th-T a ) 



= 77SwC p T h 



7-1 

# P * -1 



Ri 



(1268) 



f Efficiency of compressor and! 
[ engine with preheat 



778wC P T H 



7-1 
R P v -1 



Ri 



-^PaV a [Rp y y -i\+77SwC p (Th-T a ) 



7-1 

#p > -1 



[•g[*?u] + (,-g)T 



(1269) 



The same methods apply, but of course, different equations result, for the more 
complex cases of any number of compressor stages, combined with any number 
of expansions with both preheat and receiver reheat. To illustrate, one case 
will be worked out, that of two-stage compression with compound engine, first 
with reheat without preheat, and later with preheat also, as in Fig. 307. The 
first case is that for two-stage compression with perfect intercooling ABCD, 
followed by engine expansion E to F after the delivered air has reacquired the 
atmospheric temperature by cooling from D to E. After this high-pressure 
cylinder expansion the air is reheated to the atmospheric or original tempera- 
ture F to C in the engine receiver, which may be done by the atmosphere itself 
warming the working air as is done in the Porter air locomotive, before sub- 
sequent expansion C to G in the low-pressure cylinder. The other case has the 
same compression line ABCD, without final cooling after delivery, followed by 
preheating D to E, high-pressure cylinder expansion E to F, receiver reheating 






HEAT AND WORK 



1137 



F to G and low-pressure cylinder expansion G to H. In every case the efficiency 
of the use of the heats, whether of preheat or reheat, is given by Eq. (1270) : 

/Efficiency of preheat or\ = /Work done per unit\ = 1 1 ^ ^ QSIG) 

\ reheat / \ of heat added / myT 



or 



/Work added by preheatX _77g[x 1 1 
\ or reheat / L D ~-J 

lip t 



X 



[Heat added at con 
stant pressure 



•]■ 



(1271) 



Of course, this is true only for complete expansion, but as complete expansion 
is attainable and as this is the best possible result it is the only one worth spend- 
ing time to analyze. 




12 3 4 

Volume in Cu. Ft. 



4 5 

Volume in Cu. Ft. 



Fig. 



307 —Two-etage Compressor, Compound Compressed Air Engine with and without 
Preheat, and with Receiver Reheat. 



Evaluation of the work of the engine is best done by the pressure-volume 
methods, assuming no preheat or reheat, and to it may be added the work 
derived from preheat and reheat as thermally determined. The net work of 
engine and compressor together represented by the enclosed areas is best evalu- 
ated thermally, because any such area is made up of a number of Brayton 
cycles for which the pressure ranges and temperatures and quantities of heat 
are known. Thus taking the right-hand diagram, Fig. 307, its work by areas is 
given by Eq. (1272) : 



r Work of Brayton cycle, area &uu& n, i 
/Total work area\ = +Work of Brayton cycle ABB'EFF'A, II 
V ABCDEFGHA ) , WnrV _ f T^vton cvcle F'FGHF', III 



(1272) 



1138 ENGINEERING THERMODYNAMICS 

For both compressor and engine the receiver pressure should be 



so that 



P c = Pb = P/=Pg = v (max. pressure) X (min. pressure) . 



^« VKFa" P Pr 



As the efficiency is a function of these pressure ratios only 



Work I = (Heat added D to B') X / 1 - 






= C v w{Tv-T d ) I 



y-J. 



=w^-w(i~) 



—4-1 



= C P w(Te-T v ) 1- 



7V 



#P 2 ^ 



Work III = 0(!F„ - Z 7 ,) /l - ~ 
\ Rp 1 

= CMT 6 -T f )(l-f^ 



(a) 

(&) 

(a) 

(b) 
(a) 

(b) 



. (1273) 



(1274) 



The heat of preheat = C p w(T e -T a ). 
The heat of reheat = C p w(T g - T f ). 



(1275) 

(1276) 
(1277) 



These results are not carried further, to save space, and they are introduced 
only to illustrate a method of procedure that is useful and easily applied to any 
special case that may arise. It should be noted, however, that if the air after 
compressor delivery be cooled down to its original temperature before pre- 
heating, it will have the volume V&j and the heat so lost must be made up by 
preheating before the net work evaluation as above is made. If it is just made 
up and no more, the line EF will fall on DC and no net work results in the high- 
pressure range, if it- is not all made up, EF will fall to the left of DC and a nega- 
tive work area will result. 



HEAT AND WORK 1139 

Just as in steam and gas engines there is a heat exchange between walls and 
gases due to the difference in temperature, sometimes in one and sometimes 
in the other direction, so also is there a heat exchange between air and cylinder 
walls in both an engine and an air compressor, which must be taken into account 
in any discussion of their performance, when passing from the cyclic possibili- 
ties to the probabilities for real machines. Two general methods of attack are 
available, but have been used to only a small extent however, in evaluating the 
effect or fixing its nature. The first is that developed for gas engines and is 
concerned with heat exchanges during expansion and compression. It was 
shown that the rate of heat gain or loss by the gas per unit of volume change 



given by, I — J, could be expressed as a function of pressures and volumes meas- 
ured directly from an indicator card if the clearance is accurately known and if 
the cylinder and valves are tight. The only net effect of this exchange that is of 
any practical value is to change the value of the exponent " s " giving it some 
value other than the adiabatic. While for compressors and air engines this sort 
of analysis shows' heat exchanges that are alternately positive and negative, the 
net effect is not serious enough to warrant the use of any exponent other than the 
adiabatic value, s — y = 1.4 for air. Other values have been frequently reported, 
but with the exception of cases confined to small cylinders at low piston speeds, 
all are open to suspicion as they are in the direction that the effects of leakage 
would be. It was shown in Fig. 5 that for expansion lines, the smaller the 
value of " s " the higher the curve will lie, the adiabatic being the lowest, and 
gain of heat such as would be expected as a net result in cold compressed air 
engines, if anything, would be shown by a lesser value of " s " than 1.4. Con- 
versely, for compression lines, Fig. 6, the smaller " s," the lower the curve, 
and heat loss as would be expected as the net result in compressors if anything 
would also lower the curve. Therefore, in both cold-air engines and com- 
pressors such net heat changes as might be expected would lower the value of 
" s, " but it is also found that leakage may affect the diagrams as much as a 
considerable heat exchange. It requires only a moment's thought in view of 
the very low conductivity of air and the quiescent condition of the layers of 
air next the walls to realize that, in the short time available for heat transfer, 
the amount that could be exchanged during expansion or compression must 
be very small and negligibly so in normal cases. The discussion of heat transfer 
in Chapter IV also bears this out, for there it was shown how very low and 
consistently low, was the rate of heat transfer whenever a gas was involved as 
giving or receiving the heat. Furthermore, it was shown that increased rates 
were only obtainable with very vigorous surface scrubbing such as corresponds 
to high velocity of flow and with gases divided up into thin layers or streams. 
In cylinders the air is in about the best condition that could by found for pre- 
vention of heat transfer, so it is difficult to conceive»of " s " having any materially 
different value than the adiabatic s=y = 1.4. 

There is, however, another period in compressors and air engines where a 
material heat exchange occurs, and that is during entrance, corresponding in 



ENGINEERING THERMODYNAMICS 



engines to admission at the high pressure up to cut-off, and in compressors to 
suction at the low pressure. The effect in air engines is to cool the admitted 
air, so that for each cubic foot measured at conditions external to the cylinder 
and admitted, there will be in the cylinder less than a cubic foot before expan- 
sion begins, and at a tempera tuie lower than that of the air supply. This is 
similar in kind to the initial condensation of steam engines and has the effect 
of increasing the consumption per hour per I.H.P. over the indicated value, 
but in air engines it has the additional effect of bringing the exhaust air 
to a lower temperature because expansion starts with air colder than the supply. 
In compressors the effect is to prevent the cylinder taking in as much as cor- 
responds to the product of its apparent volumetric efficiency ^and displacement, 
measuring volumes at pressures and temperatures external to the cylinder. 
Determination of the effect of both compressor suction heating and engine 
initial cooling have been made by measuring the volume supplied or delivered 
at pressures and temperatures external to the cylinder and comparing these 
volumes with those indicated. The results can be expressed in terms of volumes 
or in terms of the temperature changes that correspond to the volume differ- 
ences, and as cylinders work at such widely different temperatures and the effect 
is one of temperature, the latter is to be preferred. The problem may then be 
stated, how much will the volume change during admission to air compressor 
or engine cylinders, when the difference between initial and final temperatures 
of the air in passing through has a known value? Some general conclusions 
are available and some specific examples may be cited to support them, 
that will make it possible to pretty closely approximate this effect. The 
effect is generally to change the volume from 10 to 30 per cent, reducing it 
for engine admission and increasing it for compressor suction, or better still, 
the temperature rise is from 25 to 50 per cent of the difference between the tem- 
peratures of supply and delivery air, entering and leaving the cylinders. One 
series of long-run tests of an ammonia compressor using weighed anhydrous 
gave the following results in round numbers, which show the rise in temperature 
during admission to average about 33 per cent of the difference between maxi- 
mumland minimum fluid temperature, excluding an abnormal case: 



Temperature before suction, degrees F . . 
Temperature after delivery, degrees F . . . 
Temp. diff. between delivery and suction. 

Temperature rise during suction 

/ Temperature rise during admission ^ 
Temperature difference in fluid / " ' 



{ 



15 


14 


15 


14 


18 


216 


218 


246 


253 


243 


201 


204 


231 


239 


225 


68 


67 


76 


59 


71 



34 



.33 



.33 



25 



.32 



A small air engine working on an atmospheric temperature supply, of compressed 
air, showed a volumetric reduction of 10 per cent on entrance, which is equivalent 
to a reduction of temperature of about 30 per cent of the difference between 
supply and exhaust-air temperatures. One valuable contribution, that of 
Trinks, that the extent of the surface exposed to the suction heating has no 
material influence on the heating effect, was established by measuring air 



HEAT AND WORK 1141 

for an 8X8 in. compressor fitted with different clearance surfaces but with 
constant clearance volume. The whole range of results show an increase of 
20 to 25 per cent in the volume of air while entering. Many examples of separate 
data of the same sort all lead to the conclusion that the volumes will be affected 
as much as corresponds to a temperature rise during entrance of 25 to 50 per 
cent of the difference between initial and final temperature, but it is impossible 
to relate the figures to speed, cylinder size, construction, kind of cylinder cooling, 
temperature range or any other factor, that might seem to affect it but as 
experimentally proved frequently fails to produce an anticipated effect. 

This being the case the volume relations may be set down as follows for air 
engines : 

/ Cubic feet air indicated at cut-off \ _ / Absolute temp, air at cut-off \ /i97^ 

\ Cubic feet air supplied / \ Absolute temp, air supplied / 

But if the temperature after entrance, is equal to that at supply, less 25 to 50 
per cent of the difference between discharge and supply air, then 

(Temp, at cut-off) = (Temp, supply) - (.25 to .50) X (_ ( ^Zp .exhaust)) (1279) 



/ Cu.ft. air in'dicated \ _ , 9 , *nw fi Z ^bs. ^ em P- exhaust \ 
\Cu. ft. air supplied/ [_ \Abs. temp, supply / 



(1280) 



Similarly, for compressors, 

/Cu.ft. air indicated on suction line\ _ /Abs. temp, air after suction\ ^ 9R . 

\ Cu.ft. air supplied / \ Abs. temp, air supplied ■/'" ' 



But 



(Temp, after suction) = (Temp, supply) + (.25 to .50) X (i.T^ m * ^ )) (1282) 



Hence 



/ Cu.ft. air indi C ated \r 25to / Abs. temp, air del. NJ 
\Cu.ft. air supplied / [_ \Abs. temp, air sup./ J 

It is customary in compressors to use the term volumetric efficiency defined 
by Eqs. (1284) and (1285) : 

. ,,.«.. ti Cu.ft. air indicated on suction line ,- ftOJN 
Apparent volumetric efficiency = E v = Displacement — ' ^ ' 



1142 ENGINEEKING THERMODYNAMICS 

-o , t — • m Cubic feet air supplied 

Ileal volumetric emciency = E 1 V = =pn — -, — — 

-Displacement 



(1285) 



l + (.25to.50)[( f! >8 - f temp -? rdel - )-l1 
L \Abs.temp. air sup./ J 



Prob. 1. Air at 135 lbs. pressure per square inch gage and 70° F. is admitted to 
a cylinder in which the cut-off is §. What will be the temperature after expansion? 
What will it be for -J cut-off, with, and without considering initial cooling? 

Prob. 2. At what temperature must air be initially, to prevent freezing of exhaust, 
if the expansion is from 80 lbs. per inch gage to atmosphere? 

Prob. 3. Air is compressed adiabatically in one stage from atmosphere to 10 atmos- 
pheres absolute, and then used in a single-cylinder steam without preheat or 
expansion. What will be the efficiency of the system, and to what value would it be 
raised by using complete expansion in the engine? The compressor may be assumed to 
have zero clearance and the engine to have complete compression. 

Prob. 4. Should the high pressure be only 5 atmospheres instead of 10, how would 
the results differ? 

Prob. 5. Should the compressor of Problem 3 be two-stage would the results be 
changed, and if so, to what extent? 

Prob. 6. What would be the emciency of the heat of preheat in a case where air was 
used at a pressure of 120 lbs. per square inch gage, the low pressure being zero gage, 
the air being preheated 100° F. above atmosphere. Would the result be different if 
the preheating were doubled and if so, how much? 

Prob. 7. What would be the efficiency of compressor and engine when the high 
pressure is 140 lbs. per square inch absolute, the low pressure 14.1 lbs. per square inch 
absolute, and the air preheated 300° F. above that of the original temperature? Would 
the result be different if the amount of preheat were doubled? If halved? If so, by 
how much? What is the efficiency of the heat of preheat alone? 

Prob. 8. In an air engine it was found that the temperature of the exhaust was 
10° F. and that of the supply air 350° F. What would be the likely limit of the ratio - 
of cubic feet of air indicated to cubic feet supplied, and the horse-power expected of the 
air engine per horse-power of compressor. 

Prob. 9. Air is being adiabatically compressed from atmosphere to 100 lbs. per square 
inch gage. The atmospheric temperature is 50° F. and apparent volumetric efficiency 
is 95 per cent; what will be a fair value for the real volumetric efficiency? 

20. Mechanical Refrigeration, General Description of Processes and 
Structures. Thermal Cycles and Refrigerating Fluids. Limiting Tempera- 
tures and Pressures. It is just as important that some substances be 
cooled to temperatures lower than the surroundings and kept cool for long 
periods of time, as is the heating and keeping hot of other substances, and until 
the thermal processes here discussed were understood this could be done only 
to the limited degree corresponding to the use of natural ice with or without 
salts to lower the melting-point. The preservation of foods including meat, 
fish, fruits, poultry, eggs and butter to prevent destruction between the periods 



HEAT AND WORK 1143 

cf production and consumption, the cooling of beverages, waters, beers and 
wines to render them more palatable, the manufacture of ice in places where 
natural ice is too costly, impure or impossible, are all illustrations of important 
cooling processes, involving the abstraction of heat from a great variety of 
substances and such as would suggest themselves to anyone. The perfection 
of apparatus, however, to accomplish these obviously desirable ends has 
resulted in a very much wider application of that same equipment with suit- 
able modifications to other equally important, though more purely industrial 
purposes, the refrigerating processes becoming one step in the general scheme 
of production of some result not itself a cooled body. For example, the dry- 
ing of air is best accomplished by cooling it to as low a temperature as will 
reduce the moisture in saturated air to the desired low value, and dry air is 
very necessary in some places; just as moist air is necessary in cotton 
spinning, for example, so is dry air in the manufacture of photographic films. 
Dried air is useful also in the drying and dry storing of wood, fabrics, 
crystals and powders, and in the blowing of blast furnaces, where each 
pound of moisture by dissociation into hydrogen and oxygen consumes heat 
of coke combustion that by its removal becomes unavailable for reducing iron 
from the ore. Some other examples of the use of refrigeration as a process 
are found in the manufacture of chocolates, the storage of furs against moths, 
the working of tobacco, the recovery of paraffine from oil, shaft sinking through 
soft earth by freezing the whole mass to permit of the removal of the frozen 
material by the ordinary methods of excavation, and finally the control of the 
rate of fermentation in various liquors and the rate of germination of seeds or 
growth of bacteria. 

To accomplish the cooling of any given substance and to maintain it at 
a low temperature continuously, requires primarily, that some other still colder 
substance to be termed the refrigerating fluid be brought close to it, so heat 
may flow by reason of the temperature difference to the colder refrigerating 
fluid through plates or pipes separating them. It is evident, therefore, that 
any sort of contact conditions that may be suitable or convenient will serve 
to cool anything, if only a still colder refrigerating fluid be available, so that 
thermally considered, the problem of refrigeration really consists in the dis- 
covering of means for making some fluid as cold as may be necessary at the 
lowest cost, and without danger to operators or injury to the apparatus. As 
the absorption of heat by any cold refrigerating fluid would ultimately raise 
its temperature as high as the body giving up heat to it, continuity of the 
refrigerating process involves as another essential element, the circulation 
of the refrigerating fluid, removing continuously that which has taken up some 
heat and replacing it with fresh quantities. As a result, there would be accu- 
mulating somewhere beyond the refrigerator large amounts of refrigerating fluid 
colder than the cooled substances and, of course, colder than the surrounding 
atmosphere and earth, so the disposition of this fluid must be provided for. 
Such refrigerants as have been found suitable are expensive so they cannot 
be thrown away after once performing a refrigerating function, but must 



1144 ENGINEERING THERMODYNAMICS 

be restored to their original condition to be used again and again, with- 
out end. 

These thermal and work processes on the refrigerating fluid itself which 
make possible the absorption by it of heat in the low-temperature ranges, and 
its ultimate restoration to a condition which will permit a repetition, consti- 
tute the so-called refrigerating cycles, and as the fluid may be either a gas, or 
a liquid-vapor, there may be a very great variety of refrigerating cycles, at 
least as many as there are of cycles for transforming heat into work by both 
vapor and gas systems. It is not, however, necessary or desirable here to 
analyze all possible refrigerating cycles, as the conclusions of the past fifty 
years may be accepted as proving the superiority of a few survivors from the 
numerous proposals. 

In general, whether the fluid be a gas or a vapor, it may be imagined as 
passing through a cycle consisting of heat absorption in low ranges of tempera- 
ture as one phase, and if the original condition is to be restored, heat absorbed 
must be given up to surroundings which will constitute another phase; between 
absorption of heat at low temperatures and its discharge to air or water later 
at temperatures higher than their own, the fluid itself must be raised in tem- 
perature and this can be done by doing work on it, so a third phase will be'one 
of compression. At least one more phase is needed to close the cycle, but there 
may be more than one. If the fluid be a gas, the absorption of heat by it in 
the refrigerator as it circulates is most feasibly accomplished at constant pres- 
sure, so heating at constant pressure, A to B, would be the first phase, Fig. 308, 
the next, adiabatic compression, B to C, followed by heat abstraction or cooling, 
C to D, again at constant pressure, the fuel temperature always above that 
of the available water which is to carry it away, and closing the cycle by 
adiabatic expansion D to A to lower the temperature to the original value. 

It will be noted that this is a Brayton gas-engine cycle executed in the 
reverse direction, and therefore all the thermal relations established for that 
cycle in Section 11 of this Chapter may be used with appropriate modifica- 
tions for this reverse refrigerating cycle when its analysis is taken up later. 

Should a liquid-vapor be used, absorption of heat by it at a temperature 
lower than in the refrigerator will evaporate liquid, and this is the first phase; 
compression adiabatically will raise temperature and pressure to such values 
as will permit condensation as heat is given up to the circulating water, and 
these steps constitute the second and third phases, while liquid cooling accom- 
panied by reduction of pressure closes the cycle, though there are modifications 
to be noted depending on quality of the vapor at various points and the 
manner in which the drop in pressure after condensation takes place. If 
after condensation is complete at D, Fig. 309, the liquid adiabatically 
expands D to A, it will have the quality at A at the low temperature. If, 
however, it merely escapes through a valve it will follow the line of constant 
total heat D to i' and be more wet, and this is more nearly the process as 
practiced. From the point A or A f , evaporation proceeds at a constant tempera- 
ture lower than that in the refrigerator until no more liquid remains, B", after 



HEAT AND WOEK 



1145 



which it may be superheated, B"B ,n '. At any time, that is when the quality 
has any value, such for example as corresponds to points B, B r , B" or B"\ 
the heat absorption may be stopped by removal of the vapor-liquid from the 
refrigerator to the compressor, for adiabatic compression to condenser pres- 
sure. Here, four different compression lines are shown, the first, BC, illustrating 
adiabatic compression from wet vapor to less wet vapor; B'C, from wet to dry 
saturated vapor; B"C", from dry saturated to superheated; B"C'" } from 
















Tern 


p. of Oir. Wa 


ter 




C 


u 












— 


























^B 


A 


Tem 


p. of B 


efrige 


rator 





































































































Cubic Feet 



.01 .02 .03 

Entropy 



Fig. 308. — Refrigeration Cycle for a Gaseous Fluid. 



superheated to more superheated states. When the vapor after adiabatic 
compression is superheated the first action in the condenser is not condensation, 
but a sort of precooling for the removal of superheat, C'"C"C , after which 
true condensation proceeds C to D, at which it is complete. 

As the constant total heat line DA' for the liquid passage from condenser 
to refrigerating evaporator is fairly close to the liquid line DA" for the common 
refrigerating fluids, it is customary to consider the heats involved as equal in nu- 
merical value for the small temperature ranges involved, and this procedure is 



1146 



ENGINEEKING THEEMODYNAMICS 



















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HEAT AND WORK 1147 

justified by the fact that the error is very small, smaller than the lines are 
correctly located or the properties of these fluids accurately known. There- 
fore, the refrigerating cycle involving the use of a vapor-liquid fluid may be 
taken as the Rankine cycle reversed and all the methods and conclusions 
developed for this cycle as a heat to work transforming process, are available 
with suitable modifications when analyzing it as a refrigerating cycle. 

It is clear from the preceding discussion that any gas or any liquid may be 
used as a refrigerating fluid when subjected to these cycles of operation with only 
one limitation, and that is they must not pass through the critical state. This 
does not mean, however, that all are equally available, for they are not, and the 
first consideration of importance that reduces the number to a very few is 
the pressure necessary to cause evaporation at such low temperatures as refrig- 
erants must assume, a high limit for which may be taken at 0° F. and a low 
limit at —50° F. If the pressures corresponding to the saturated vapor at such 
temperatures are too low, there would be needed very large pipes and com- 
pressors to handle the necessary quantity. This consideration with others con- 
sidered later, limits the available fluids to the following, having the boiling- 
points at atmospheric pressure corresponding and in order: sulphuric ether, 
C 2 H 5 0, 34° C. = 93° F.; sulphur dioxide, S0 2 , -10° C. = 14° F.; methyl 
chloride, CH 3 C1, -22° C. = -8°F.; methyl ether, C 2 H 6 0, -24°C, = -11° F.; 
ammonia, NH 3 , -33° C,= -27° F.; carbon dioxide, C0 2 , -79° C.= -110° F.; 
nitrous acid, N 2 0, -88° C.= -126° F.; ethylene, C 2 H 4 , -103° C.= -152° F.; 
oxygen, 2 , -181° C.= -294° F.; air, -192° C.= -314° F.; nitrogen, N 2 , 
-198° C.= -324° F.; hydrogen, H 2 , -243° C.= -416° F. 

The other items in the elimination of the less desirable fluids are; (a) the 
condenser pressure corresponding to ordinary circulating water temperatures 
which, if too high, require thick and expensive pipes and unduly expensive 
compressors; (&) the latent heat of evaporation at the refrigerator or evaporat- 
ing pressures which, if too small, require the circulation of excessive quantities 
of fluid to produce a given amount of refrigeration; (c) the volume of vapor 
per pound at the low pressure which, if too large, requires too large a com- 
pressor to circulate the required quantity, the compressor size being a function 
of the cubic feet of vapor per B.T.U. of latent heat taken by it from the 
refrigerator; (d) the cost per lb. of the fluids which must be replaced to 
supply leakage ; (e) the corrosiveness of the fluid on the metal parts. All 
these things carefully studied have resulted in the placing of anhydrous 
ammonia in the first rank of refrigerating fluids, it being used in the pro- 
duction of over 90 per cent of all the mechanical refrigeration now in 
service, with carbonic acid and sulphur dioxide next in order, and practically 
all others eliminated except air, which is used entirely as a gas and which is 
the only gas so used. 

These fluids as the media for the execution of the refrigerating cycles must be 
treated in mechanical apparatus of suitable form, and this apparatus in type 
at least is now pretty well standardized and falls in one of three so-called sys- 
tems of mechanical refrigeration which must be understood in principle before 



1148 ENGINEERING THERMODYNAMICS 

attempting any analysis or calculations of performance. These three systems 
are: 

A. The dense-air system, so called because the air which is the medium 

is never allowed to fall to atmospheric pressure, so as to reduce 
the size of the cylinders and pipes through which a given weight 
is circulating. 

B. The compression system, using ammonia, carbon dioxide and sulphur 

dioxide, and so called to distinguish it from the third system, 
because a compressor is used to raise the pressure of the vapor and 
deliver it to the condenser after removing it from the evaporator. 

C. The absorption system, using ammonia, and so called because a 

weak water solution removes vapor from the evaporator by absorp- 
tion, the richer aqua ammonia so formed being pumped into a 
high-pressure chamber called a generator in communication with 
the condenser, where the ammonia is discharged from the liquid 
solution to the condenser by heating the generator, to which the 
solution is delivered by the pump. 

No matter what the system may be, it will require circulating water to receive 
its heat, and the temperature of the water determines the highest temperature 
allowable in the system and indirectly the highest pressure, so one structural 
element is a cooler or condenser or both or several of each, to which circulating 
water is supplied. In some cases it will be found that the same water may be 
used in series over two different parts when the temperature in one is 
higher than in the other. Another structural element common to all is the refrig- 
erator proper, in which the air of the air system is warmed and the ammonia 
or carbon dioxide evaporated, and its most common form is a coil of pipe. 
This pipe may be placed directly in a room to be kept cool, in which case the 
arrangement is called direct expansion, but more usually the coil or an equivalent 
structure forms part of a brine-cooling chamber, the cold brine from which is 
circulated by pumps through room coils or through ice-making tanks, holding 
cans that in turn contain the water to be frozen, though a great variety of 
combinations of direct cooling and brine circulation are in use. It will be 
assumed in what follows that in all cases a brine cooler will be used with the 
understanding that the cooling coils may instead, cool anything other than 
brine directly. 

The air system is illustrated diagrammatically in Fig. 310, for a closed 
system, in which air previously dried of moisture is continuously circulated, 
showing a compressing cylinder delivering air through a water cooler to an 
expansion cylinder, which in turn sends the air through refrigerating coils back 
to the compressing cylinder. As the work done by the expansion of the cooled 
compressed air is less than that needed to compress the air from the refrig- 
erating coils through the same pressure range, there will be necessary a source 
of power to supply the difference, so a steam or gas-engine cylinder is added. 



HEAT AND WORK 



1149 



While any range of temperatures desired may easily be obtained with reason- 
able pressures, some standard is desirable for purposes of comparison, so it 
will be assumed that the air must leave the refrigerating coils at 0° F. and 
enter at —80° F., also that the cooling water available will permit of an initial 
temperature in the expansion cylinder of 70° F. If expansion in this cylinder 
is complete, as it may be if the cut-off is suitably adjusted, the pressure ratio is 
given by Eq. (1286): 



Rp= k = {k) y 1= (IS) 346=(L4)3 ' 46=3 - 2 ( a pp rox -)- • ( 1286 ) 



1 '^Circulating Waters 



Cool Compressed Air 



Air Cooler 



Expanding 
Cylinder 



Cold Low Pressure Air - 



-Hot Compressed Air 



Compressing 
Cylinder 



Brine Cooler 



"< Brine 



Power 
Cylinder 



Warmed Low Pressure Air 



Fig. 310. — Diagrammatic Arrangement of the Air System of Refrigeration. 



If the pressure in the refrigerating coils is one atmosphere that in the water- 
cooling coils must be 3.2 atmospheres absolute, but if high pressures are carried 
in the refrigerating coils to reduce the size of equipment, four atmospheres, for 
example, the high pressure will be thirteen atmospheres absolute. The indicator 
diagram of the compressor without losses is shown for this latter case by ABCD 
of Fig. 311, on which BC represents the volume of hot air actually delivered 
by the compressor due to the clearance volume V c . If this be cooled and 
admitted to an expansion cylinder having a clearance volume Vn, and working 
with complete compression GH and expansion EF, the admission volume HE, 
is to the compressor delivery volume BC, as the absolute temperatures after 



1150 



ENGINEERING THERMODYNAMICS 



and before cooling. These two indicator cards when combined become 
equivalent to the PV cycle of Fig. 312, on which CB represents the delivered 
volume of hot compressed air, and CE, the admitted volume of resultant 
cooled air, or EB the reduction in volume due to cooling, EF the expansion, 
FA the refrigerator heating of the air and A B the compression, substantially, 
as described previously. Of course, compressor suction and delivery valve 
losses, and suction heating, as well as incomplete or overexpansion and com- 





c 


H ^ 








E B 




• 














p 












\\ 


















12 












\ 


\ 




























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> 


v\ 
















10 




i 










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*> 










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A 


































2 


























At 


ms. 

































.5 1 1.5 2 2.5 3 V 3-5 

Volumes in Cubic Feet 
Fig. 311. — Dense Air Refrigeration, Compressor and Engine Combined Indicator Cards. 

pression in the expansion cylinder, with admission cooling of the air therein, 
will distort the diagram of actual machines, but the effects of such distortion 
are easily understood and need not be subjected to analysis, the cycle above 
representing the perfect conditions will give best possible results that are to 
be determined analytically. 

The compression system for ammonia, carbon dioxide or other similar con- 
densible vapors is illustrated in Fig. 313, diagrammatically, the arrangement 
of equivalent parts being maintained as in the air system, Fig. 310, to bring out 



HEAT AND WORK 



1151 



differences more clearly. Here the condenser replaces the air cooler, and the 
vapor delivered to it by the compressor is changed therein to liquid at the same 
pressure, which liquid escapes through a throttle, generally termed an expansion 
valve, to the refrigerating coils where the pressure is low and evaporation 
thereby induced, the discharge vapor from the coils being regularly drawn off by 
the compressor suction. Of course, other elements are added in practice, for 
example, a bypass piping, so the condenser may be pumped out into the refrig- 
erator for cleaning or repairs, a charging connection for taking in fresh ammonia, 
a forecooler at the condenser to remove superheat and which may use warmer 
water than the condenser itself, a liquid receiver beyond the condenser to store 







































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fcP 


















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12 






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8 
G 


































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2 










F 


* A 


Atms. 





000 
.500 
400 
3C0 
200 
100 

















































B 


E. ■ 

-70° F 




















•^^A 


^60°F 


F 

















































Volumes in Cubic Feet 

Fig. 312. — Dense Air Refrigerating Thermal Cycle. 



.01 .02 .03 .04 .05 <p 
Entropy 



a quantity of fluid to meet sudden fluctuations of demand while vapor returns 
slowly, a purge valve at the highest point of the condenser to blow off non- 
condensible gases from the system, a liquid separator on the compressor suction 
to prevent the liquid of wet vapors entering the cylinder, oil or glycerine traps 
between compressor and condenser to prevent entrance into the latter of any 
such matter used in the cylinder and discharged from it, and other similar 
elements often in large numbers, none of which, however, affect the funda- 
mental thermal process. 

For equal temperatures maintained at delivery from condenser and refrig- 
erator coils as were assumed reasonable for the air system, that is 70° F. and 
0° F., the other temperatures and pressures will be different for this system 
and likewise different for each fluid used in it. Only two fluids, ammonia and 



1152 



ENGINEEEING THERMODYNAMICS 



carbon dioxide, will be here considered, and it will be assumed for convenience 
that the vapor leaving the coils in which pressure is assumed constant through- 
out, is dry saturated, and that the condenser is so constructed that the pres- 
sure therein corresponds to liquid leaving at 70° F.; actually it will be higher. 
Under these circumstances the compressor suction pressure will be 29.7 lbs. 
per square inch absolute for ammonia, and 313.7 lbs. per square inch for carbon 



High Pressure y ' 
Liquid 



Liquid Throttie ory^r; 
Expansion Valve 

Low Pressure Liquid 




/ 



3 



Brine Cooler 



Brine 



High Pressure 
Vapor 



Power 
Cylinder 



Low Pressure Vapor 



A 



Fig. 313. — Diagrammatic Arrangement of Ammonia or Carbon Dioxide Compression System 

of Refrigeration. 



dioxide, while the corresponding condenser pressures will be 128.4 lbs. per 
square inch for ammonia and 851.5 lbs. per square inch for carbon dioxide, all 

absolute and taken directly from Tables XLIX, and L, of the properties of 

p 
saturated vapors, Chapter IV. Accordingly, the pressure ratios are, -- = 4.32 for 

* L 
p 

ammonia, and — = 2.72 for carbon dioxide, and neglecting wall action in the 

" L 



HEAT AND WOEK 



1153 



compressor as well as pipe heating or cooling, the temperature of the gas after 
delivery from compressor to condenser will be as given by Eq. (1287), taking 
Y = 1.3 for both vapors superheated. 



Th=T l 






J 460(4.32)- 231 
| 460(2.72)- 231 



645° F. abs. for ammonia (a) 

580° F. abs. for carbon dioxide (b) 



L . (1287) 



For these pressures the two compressor indicator cards are drawn, Fig. 314, 
for the assumption of no losses and for equal per cents of clearance, ABCD 

















































P 
800 

CD 

i 

i 

Sioo 

u 
200 


























































c 




B' 


















\ 


p 




















\ 












c 


B 


















\co 


2 










I 


120 

a 






























u 
o 

ft on 


1 


\ 


\NH 3 






































\ 












A' 


o 
u 

w 

% 


\ 












D' 














A lu 


\ 




























D 










A 
Atms. 















































4 8 

Volumes in Cu. Ft. 



200 



.1 .2 .3 

Volumes in Cu. Ft. 



Fig. 314. — Ammonia and Carbon Dioxide Compressor Diagrams between Equal Temperature 

Limits. 



representing ammonia and A'B'CD' carbon dioxide. On each, the suction 
volume is DA or D'A', and the delivery volume BC or B'C, so they are equiv- 
alent to no clearance cards with these bottom and top lengths. As the func- 
tion of the condenser is that of volume reduction at constant pressure, its action 
may be equally well represented by the delivery line BC and B'C, Fig. 315, 
and similarly, the coil evaporation at constant pressure is also represented by 
the suction line DA and D'A f , so that the whole process is represented by the 
PV cycles, ABCD and A'B'CD', for which the corresponding T$ cycles are 
shown at the right, each for one pound of substance. 



1154 



ENGINEERING THERMODYNAMICS 



P 

800 



;600 



•J 400 
& 



200 



CO< 



500 



-D L 



-400 



300 



-200- 



L00 



CJZ 



ro 



0°F. 



^ 



F. 



X 



CO* 










1 
Vol. 


3u. Ft 


2 




V 

T 

600 
/ 




-05 







1 
Entropy 


2 


4> 


p 


c 


B 
















NH 3 






B 




\ 


/ 




120 




\ 








> 


/ 


70°F.- 






\ 


/ 




3 










\ 




\ 




NH 3 
















N 




d 1 
O3 80 

ft 


D 
400 

a 

< 

04 

a 

200 






o j f/ 








i 


\ 


























CO 
CO 

£40 


























D 


































A 












Aim. 






























lT" 


.4 
Vo 


1. Cu. I 




j 


V 








.4 


Entrop 


.8 

y 




.12 


</> 



Fig. 315. — Ammonia and Carbon Dioxide Refrigerating Cycles. 



HEAT AND WOEK 1155 

The absorption system of apparatus for ammonia is represented in Fig. 316 
diagrammatically, which illustrates only one of several possible arrangements, 
but a common one that serves as a basis for analysis. Here the condenser, 
anhydrous ammonia liquid line, expansion valve, evaporating coils and brine 
cooler are the same as before, all differences being concerned with the drawing 
off of the vapor produced in the refrigerating coils and its delivery at the higher 
pressure to the condenser. This is accomplished by treating an aqueous solu- 
tion of ammonia, alternately allowing it to absorb ammonia from the evaporat- 
ing coils in an absorber at a rate determined by a regulating valve and the tem- 
perature and concentration of the solution, and later heating this richer solution 
in a generator by means of steam coils. Thus, referring to the diagram, the 
low-pressure vapor passes from the evaporating coils through the suction- 
regulating valve to the absorber chamber where it meets a weaker aqueous 
solution, being absorbed and so developing heat, which is carried off by the 
circulating water coil and which latter controls the temperature of absorption. 
A more concentrated and cool solution of strong liquor is then pumped from this 
low-pressure absorption chamber to the high-pressure heating chamber or 
generator, passing on the way through a heat exchanger where it gets warmer 
by cooling weak liquor flowing in the other direction. The heating in the 
generator will drive off both ammonia and water vapor but mainly the former, 
and this vapor mixture rises through a double cooling chamber or analyzer 
and rectifier. It is first cooled a little by coming into contact with the incoming 
strong liquor in the analyzer, and later more cooled in the rectifier by circulating 
water that has been used in the condenser. At each cooling step some water 
vapor is thrown down and returns with such ammonia as it absorbs on the way 
to the generator, while the ammonia not so reabsorbed passes to the con- 
denser, as the cooling has not been carried far enough in analyzer and rectifier 
to condense it. Of course, a little water vapor will pass over to the condenser, 
making its delivery consist not of pure anhydrous ammonia, but rather of an 
aqueous solution, mainly NH3, which behaves nearly enough like anhydrous 
for practical purposes, unless by mismanagement it finally attains too much 
water, which must be removed before continuing the refrigeration effectively. 

It will be assumed that the condenser discharge is pure anhydrous and in 
this case the coil and condenser pressures will be the same as for the compressing 
system. Obviously, a considerable range of liquor strengths may be used and 
a still wider range of temperatures maintained in both absorber and generator, 
so for the purpose of calculation some common values must be assigned to these 
quantities for preliminary calculation. Accordingly, it will be assumed that the 
generator has the uniform temperature of 250° F., which corresponds to 
saturated steam at 29".82 lbs. per square in absolute or 15 lbs. gage pressure 
in round numbers, receiving rich liquor 35 per cent ammonia by weight, at 
215° F. It will also be assumed that the weak liquor, 25 per cent, enters the 
absorber at 125° F. so that the exchange has reduced it 125°, while the 35 per 
cent liquor leaves the absorber at 100° F. and is warmed in the exchanger 
10° less than the weak is cooled because of superior specific heat, so that 



1156 



ENGINEERING THERMODYNAMICS 



Circulating Water 



Condenser 



Rectifier 



High Pressure 
- Liquid N H 3 






Circulating Water 

* L_ 



Circulating Water 



Analyzer 



Ammonia & 
1 ' Water Vapor 



Strong Liquor 



Weak Liquor 
Regulating Valve 



Suction 

Regulating 

Valve 

:/ 



Low Pressure / 
Liquid N H 3 







Fig. 316. — Diagrammatic Arrangement of the Ammonia Absorption System of Refrigeration. 



HEAT AND WORK 1157 

its rise of temperature is 125° — 10° = 115°, entering the generator at 100° F. 
+ 115° = 215° F. Finally, it will be assumed that the pressure in the absorber 
is 5 lbs. per square inch gage lower than in the evaporating coils by reason 
of the action of the regulating valve, which makes the absorber pressure 
29.7 — 5 = 24.7 lbs. per square inch absolute. Referring to the table of proper- 
ties of ammonia water solutions, Chapter IV, and the corresponding charts, it 
appears that a solution at the pressure 24.7 lbs. per square inch absolute in the 
absorber and at a temperature of 100° F. would if saturated contain 00 per cent by 
weight, so that the 35 per cent solution which leaves is not saturated. Similarly, 
in the generator the condenser pressure of 128.4 lbs. per square inch is main- 
tained and at the temperature of the 25 per cent solution, leaving it 250° F., 
would if saturated contain 00 per cent by weight. 

These quantities are all subject to almost infinite variation, yet something 
must be assumed approximately corresponding to practice as a basis of com- 
parative computation, and in this case it is especially important, as no thermal 
diagrams or cycles can be conveniently established to represent in general 
terms the whole process as is so easy for the other systems. 

This general description of the methods and the functions of the structural 
elements of those mechanical refrigerating systems as are to-day reduced to 
standard engineering practice, will serve as a basis of both absolute and com- 
parative calculations concerning them, in which the important quantities to 
be determined are, the quantity of refrigerating fluid that must be circulated 
per minute to produce a unit of refrigerating effect, the size and displacement 
of cylinders necessary, the work that must be done per unit of refrigeration 
and the general relations between heats gained and lost by the fluid and the 
work that must be done on it. 

21. Performance of Mechanical Refrigerating Cycles and Systems. Quan- 
tity of Fluid Circulated per Minute per Ton Refrigeration, Horse-power, and 
Heat Supplied per Ton. Refrigeration per Unit of Work Done and its Relation 
to Thermal Efficiency of the System. As in the cases of the steam boiler and gas 
producer where capacity is to be measured in terms of a quantity of heat per min- 
ute delivered, a capacity unit has been established and given the arbitrary term 
" horse-power," with a different meaning in each case, so in dealing with refrig- 
erating systems engineering practice has developed and adopted a somewhat 
similar term but with a more rational basis as the unit of capacity, and this is 
the ton refrigeration. If the latent heat of fusion of ice be taken as 144 B.T.U., 
which is nearly correct, and the ton as 2000 lbs., then one ton of ice in melting 
would absorb from the surroundings 2000X144 = 288,000 B.T.U., and if this 
were accomplished in a day of 24 hours, heat would be flowing at the rate of 
288,000 B.T.U. per 24 hours, or 12,000 B.T.U. per hour, or 200 B.T.U. per 
minute. Whenever the refrigerating fluid absorbs heat at these rates it is said 
to be developing a ton refrigeration, or when brine absorbs heat at the same 
rate it likewise is said to be developing a ton refrigeration, so that capacity 
may be measured, as in the refrigerating fluid, as in the brine, and also as from 
the body giving up heat such as beef, or water ice. When at one point of the 



1158 



ENGINEEEING THEEMODYNAMICS 



system, heat is being transferred at the ton refrigeration rate it will not be 
necessarily, at othe^points in the system, so it is necessary to establish a standard 
that shall be invariable and that is in the fluid itself. Therefore, a system has 
one ton refrigerating capacity when the refrigerating fluid is absorbing heat in the 
primary refrigerating coils at the rate of 288,000 B.T.U. per 24 hours, 12,000 
B.T.U. per hour, or 200 B.T.U. per minute though the useful effect in cooling or 
freezing will be less if there are any gains or losses between. 

Each pound of fluid is capable of taking up in the refrigerating coils a definite 
amount of heat to be called refrigeration per pound of fluid, so that to produce 
the ton refrigeration a definite weight of fluid in pounds per minute must be 
circulated according to the general formula, Eq, (1288) ; 



(Lbs. of fluid per minute per ton) = 



200 



(Refrigeration per lb. of fluid)' 



. (1288) 



This is a general expression for which there is a volumetric equivalent given 
by Eq. (1289), where the volume is that of delivery from the refrigerating coils. 



[ Cu.ft. fluid deliv- 
ered from refrig- 
I erating coils per 
{ min. per ton 



— 9nn /^ U * ^' P er ^°' fl u id a ^ CQ il delivery \ 
\ Refrigeration per lb. of fluid / 






\ (1289) 



200 



^Refrigeration per cu.ft. fluid at coil del 



:) (6) 



Assuming no heating or cooling of the fluid between the points of coil delivery 
and compressor entrance, the same volume will approach the compressor in com- 
pression systems, as left the coils, but on entering the compressor it will be 
heated and expanded; it will also suffer a loss of pressure, and reexpansion in 
the compressor will delay entrance. Defining the true volumetric efficiency. 
Ey, as the ratio of the volume drawn from the suction measured at conditions 
of pressure and temperature external to the cylinder, to the displacement 
volume, the compressor displacement per ton capacity may be set down as in 
Eq. (1290), for the air and compression systems, 



Compressor dis- 
placement cu.ft. 
per min. per ton 



/ Cu.ft. per lb. fluid at coil delive ry\ "1 ( 
\ Refrigeration per lb. of fluid / J 



-[■ 



200 



EV^Refrig- per cu.ft. fluid at coil del.) 



(&) 



(1290) 



That which corresponds in the absorption system to compressor displacement 
in the other systems, is the displacement of the pump which is equal to the 



HEAT AND WORK 1159 

cu.ft. of strong liquor per minute that is capable o£ carrying off the quantity of 
ammonia circulating, if its volumetric efficiency is 100 per cent, which it may 
be substantially if of proper type and properly run. 

Let w R = pounds rich liquor to be circulated per pound of NH 3 taken up in 

absorber; 
11 Cr = per cent NH3 in rich liquor; 
11 C w = per cent NH3 in weak liquor. 

Then (tqTj)> an ^ (iTjTj) = pounds NH 3 per pound solution weak and rich; 

" ( — Ton - /' anc * (' — mn / = P ouno ^ s water per pound solution weak 

and rich; 

( ifto — n ) ' anc * ( 100 — C ~ ) = P oun< ^ s NH 3 per pound water in weak 

and rich solution. 

Therefore, the weight of ammonia gained per pound of water when the 
solution passes from the weak to the rich state is, ( inn _^ ) — ( in o-Tr ) * 

But 1 lb. of water is associated with f _ ^ - J pounds of rich solution, 

therefore, the pounds of NH3 taken up per pound of rich liquor formed, is 
equal to the pounds NH3 gained per pound of water, divided by pounds of 
rich liquor per pound of water; and the pounds rich liquor formed per pound 
of NH 3 taken up is the reciprocal of this and given by Eq. (1291) : 

( 10 ° ) 

yioo -c R J 100(100 -c w ) ^ ioo-(V ( n 

Wr ( Qr \ ( C w \ Cr(100-C w )-C w (100-Cr) Cr-Cw' { } 
\100 -C R ) \100 -Cw) 

The displacement of the pump per ton will, therefore, be the product of the 
weight of rich liquor per pound of NH3 absorbed, into the cu.ft. of liquor per 
pound, and the pounds of NH3 per minute per ton. 



Rich liquor pump 



". H T /100-CV\ /Cu.ft. rich\ 

displacement per = ^^^^j X ^ u lb J 



X 



minute per ton 



200 



Refrigeration 
per lb. NH 3 



. (1292) 



1160 



ENGINEERING THERMODYNAMICS 



Solution of all these equations for weights and volumes of fluids and the corre- 
sponding displacements per minute per ton, depends on the evaluation of the 
term, (Refrigeration per pound of fluid). In the case of the air system this is, 
of course, nothing more than the product of specific heat at constant pressure 
into the temperature rise in the coils. For the liquid vapor system it is made 
up of parts depending on the final quality of vapor discharge which may be set 
down as in Eq. (1293) : 



Refrig- 
eration 
per lb. 
fluid 



C, 



(Temp, air leaving coils) — 
(Temp, air entering coils) 

" (Latent heat) X (Quality" 
of vapor, leaving coils) 
— (Heat of liquid bet. 

_ supply and coil temp.)_ 

(Latent heat) 
— (Heat of liquid bet. 
supply and coil temp.) 

" (Latent heat) 
— (Heat of liquid bet. 
supply and coil temp.) 
-\-C v [(Final vapor 
temp.) — (Sat. vapor 

_ temp, in coils)] 



For air system (a) 



Final wet vapor 
for vapor sys- \ (b) 
terns 



Final dry sat. ] 
vapor for va- j 
por systems J 

Final super- 
heated vapor 
for vapor sys- 
tems 



(c) 



► (d) 



(1293) 



These are all tabular quantities except the heat of air and of vapor superheat, 
which are easily evaluated, but to facilitate determinations the charts of Figs. 
317 for ammonia and 318 for carbon dioxide have been calculated and are 
somewhat similar to those given for boiler capacity, where B.T.U. per pound of 
steam for boilers and pounds evaporation per hour per boiler horse-power, 
correspond to B.T.U. refrigeration per pound of fluid and pounds of fluid per 
minute per ton here. 

From the data of these charts applied to the equations above the displace- 
ments of compressors and pumps can be computed by the slide rule, intro- 
ducing the chart quantities in the equations. When superheated vapor 
densities are to be evaluated either vapor may be assumed to behave as a 
perfect gas, volumes being directly, and density inversely proportional to 
absolute temperatures. The volume per pound of ammonia solutions is to be 
evaluated from that of water in the steam tables, dividing the water value 
by the specific gravity of the solution table, at the end of Chapter IV, or 
read off directly on the additional chart, Fig. 319. 

With a given weight and volume of fluid per minute per ton passing through 
the cylinders, the maximum and minimum pressures for which are determined by 
the temperature desired in the coils and that available by the cooling water in 
coolers or condensers, it follows that a definite horse-power per ton refrigeration 
will result for each system and each operative condition of any one. The cycle 



HEAT AND WORK 



1161 



of operations to PV or T$ coordinates offers a means of evaluating the work 
in foot-pounds per pound of fluid which, when multiplied by the pounds per 
minute per ton, gives work in foot-pounds per minute per ton and this divided 
by 33,000 is the horse-power indicated, per ton refrigeration. These same 
cycles, but more particularly those drawn to T$ coordinates, also offer means 
of directly evaluating the work per B.T.U. absorbed by the fluid in the refrig- 



I 60 



o 

-e 40 



20 



25 



20 



C.3 
u 15 



4J 10 

H 



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serf J 


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W3 
























































































































































.S75 


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540 
















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40 



30 



20 



10 -10 10 

Refrigerating Coil Press. Lbs. per Sq. Lb. Ga. 



20 



30 



40 



Fig. 317. — Chart to Determine Available Refrigerating Effect per Pound of Ammonia for 
Any Refrigerator Pressure and Any Refrigerator or Liquid Temperature. 



erator, the procedure being similar to that used with engine cycles in obtaining 
the ratio of work to heat supplied which was the efficiency of the work develop- 
ing cycle. Both of these determinations are worth while. 

As compression is always actually or substantially adiabatic and as vapors 
may be wet, dry saturated or superheated and as the value of " s " may and 



1162 



ENGINEERING THERMODYNAMICS 



usually is a variable as quality changes, making the PV method uncertain, the 
T$ method of evaluating the work of compression for wet vapors is the only 
one that can be relied upon for accurate results. The Mollier diagram for NH3 
and C.O2 gives these results directly, so in Figs. 320 and 322 are given the T$ 
diagrams for NH 3 and CO2, and in Figs. 321 and 323 the corresponding Mollier 
diagrams. It must be remembered that the results from these Mollier diagrams 













































































































































































































































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-1. 





500 



400 



300 200 300 

Refrigerator Coil Press. Ubs. Sq. In. Ga. 



400 



500 



Fig. 318. — Chart to Determine Available Refrigerating Effect per Pound of Carbon Dioxide 
for any Refrigerator Pressure and any Refrigerator or Liquid Temperature. 



give the work in B.T.U. for the whole cycle, which is similar in form to the 
Rankine for steam. 

• The diagrams give also the B.T.U. absorbed per pound at the lower tem- 
perature, so the ratio of the work to refrigeration can be taken directly from 
the diagrams. This is the measure of the goodness of the process for refrigera- 
tion, sometimes though improperly termed efficiency, improper in that it is a 
different meaning for the word than that heretofore used. To facilitate 



HEAT AND WORK 



1163 



calculation still further but also to show most clearly the influence of the 
several factors that control it, this ratio, 

B.T.U. work of refr igerating cycle 
B.T.U. absorbed at low temp. 

is plotted, Figs. 324 and 325 for ammonia and Fig. 326 and 327 for carbon 
dioxide, to coordinates similar to those used for the Rankine cycle efficiency, 
which is a similar ratio, the work there, however, being divided by the heat 
taken in at the high temperature. 



.0188 .0186 .0184 . 



Cu.Ft. per Lb. 
.018 .0178 .0176 .0174 .0172 



.01(58 .0166 b' 




53 



50 



56 57 58 

Lbs. per Cu. Ft. 



Fig. 319. — Density and Specific Volume of Ammonia-water Solutions. 

The evaluation of this performance ratio for the gas cycle is best made alge- 
braically by methods similar to those used in the analysis of the Brayton work 
cycle. This cycle is shown in Fig. 328, ABCD, to both PV and T$ coordinates. 
Here heat is taken up from D to A in the refrigerator and abstracted from BtoC 
in the cooler, so the work done between the two adiabatics is equal to the dif- 
ference on the T$ diagram between the areas under CB and DA : 



Work of air refrigerat- 
ing cycle in B.T.U. 
per pound of air 



/Heat given up to\ /Heat taken in at\ . . 
\ the water cooler/ \the refrigerator / 



C p (T b -T c )-C p (T a -T d ) 
C p (T. b -T e -T a +T d ) 



(6) 
(c) 



(1294) 



1164 



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HEAT AND WORK 



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5 10 15 20 25 30 

Refrigerating Coil Pressure, Lbs.per sq.in. Gauge 



35 



40 



Fig. 324. — Work in B.T.U., per B.T.U. Absorbed in Refrigeration, by Ammonia kSupplied as 
Liquid at any Temperature and Vaporizing at any Coil Pressure to Dry Saturated 
Vapor. 



Work of air refrigerating cycle 
Heat given to water cooler 



= 1 



T a -T d 
T b -T c 



= 1 



H) 



Tall- 

T b 



1 —TjL — 1 —TJl 

T b ~~ T c 



-y-l 
P d \~ 



P C/ 

VcV- 1 



= 1 



-(9, 



7-1 

PAT" 



® 



r 



= Eb — 



Efficiency 
of Bray- 
ton work 
cycle 



(a) 

(b) 
(c) 



(1295) 



HEAT AND WORK 



1169 



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Degrees F. Superheat 



Initial Quality 



Fig. 325. — Work in B.T.U.,per B.T.U. Absorbed in Refrigeration, by Ammonia Supplied as 
Liquid at any Temperature and Vaporizing to any Quality or Superheat at 15 Pounds 
per Square Inch Gage. 



T ( 1 — — 
/W ork of air refrigerating cycle X _ T b —T c _ b \ T b 

\Heat absorbed in refrigeration/ T a —T d T L _Td 






7-1 

PA" 



Pc 

VaV 



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(b) 


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(c) 



(1296) 



1170 



ENGINEERING THERMODYNAMICS 



The importance of expressing the ratio of work done to refrigeration effect for 
the air cycle in terms of the Brayton work-cycle efficiency, is due purely to 
convenience of calculation as this efficiency is the same as for the Otto cycle for 
the same compression, curves for which have been given in Section 1 1 of this 
Chapter and which can be used to solve these refrigerating problems. 

Pressure- volume determinations are, of course, equally available for the 
determination of the work of the compressors of the air and the compression- 



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"^S^, ""^s^, ^->. "— «««^ 


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"*"*■— «~.^ ~" B '"---.^. ==>I ""-""•-- — B, 


■—--l — 1 __ "-" 


r-^™.™ 



200 250 300 350 400 450 500 

Pressure of Saturated CO 2 Vapor leaving Coils 
lbs. per Sq. In. Gauge 



550 



Fig. 326. — Work in B.T.U., per B.T.U. Absorbed in Refrigeration, by Carbon Dioxide Supplied 
as Liquid at any Temperature and Vaporizing at any Coil Pressure to Dry Saturated Vapor. 



vapor systems, of the engine of the air system, and of the pump of the absorp- 
tion system, but these methods nee^l not be repeated here as they are completely 
presented in Chapters II and III, except that for pump work, which is to be 
evaluated as for expansive fluids with no cut-off or compression, but without any 
reexpansion of clearance fluid. It will be found for the vapors even if they are 
superheated, that the work as determined by PV methods does not check the 



HEAT AND WORK 



1171 



work as determined by T$ methods and the difference is a measure of the uncer- 
tainty of the physical properties, specific heat of liquid and of superheated vapor, 
latent heat and densities of liquid and vapor. Of course, these data might 
be manipulated to give checking results for work, but this is an improper pro- 





























































































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Quality of C0 2 leaving 1 Coils (for a Constant Coil Pressure of 300 Lbs. per Sq. In. Gage) 

Fig. 327.— Work in B.T.U., per B.T.U. Absorbed in Refrigeration, by Carbon Dioxide Supplied 
as Liquid at any Temperature and Vaporizing to any Quality or Superheat at 300 
Pounds per Square Inch Gage. 



cedure; the best practice is to use both methods and assume the correct result 
to be somewhere between but nevertheless indeterminate with precision, until 
a redetermination of properties experimentally has been made, such as is about 
to be undertaken by the U. S. Bureau of Standards. , 



1172 



ENGINEERING THERMODYNAMICS 



From the values of the work in heat units, per unit of heat absorbed in 
refrigeration the horse-power per ton refrigeration follows by the simple 
transformations below, which show a direct proportionality, for the com- 
pression systems. 







































































































































































p 


























T 






















B 












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1 2 

Vol. in Cu. El. 



3w0 .01 .02 .03 .04 .05 

Entropy 



Fig. 328. — Air System. Diagram for Evaluating Work in B.T.U., per Unit of Heat 
Absorbed in Refrigeration. 






778 
(I.H.P. per ton refrigeration) = r^ X (Work in B.T.U. per min. per ton) 



778X200 
33000 



= 4.71X 



X 



Work in B.T.U. per minut e \ 
B.T.U. absorbed per minute/ 



Work in B.T.U. 



B.T.U. absorbed at low temp 



■) (1297) 



The heat supplied to the system per ton refrigeration is as important a 
determination as the work or horse-power per ton, and in the case of the absorp- 
tion system more so, as the rich liquor pump work is very small while the heat 
supplied to the generator is large, both together representing the heat equiva- 
lent of the energy necessary to get the low-pressure vapor into the high-pressure 
condition. Such heat as is supplied to develop power in steam or gas engines 
is given by the following Eq. (1298) : 



( 



B.T.U. per hour per\ = 2545 

I.H.P, of engine / Engine thermal efficiency. 



HEAT AND WORK 



1173 



But 



En ffid e nc hermal ) = ( Efficienc y of en S ine c y° le ) x (Efficiency factor), 



Hence 

/B.T.U. per hour per\ = 2545^ 

\ I.H.P. of engine / (Efficiency of engine cycle) X (Efficiency factor)' ' 

Introducing the ratio of, I.H.P. of compressor to I.H.P. of engine, which is less 
than unity, the heat consumption per I.H.P. of compressor follows by Eq. 
(1299): 



B.T.U. sup- 
plied per hr. 
to engine per 
I.H.P. o f 
compressor 



2545 



/Efficiency of \ v /Efficiency] 
\ engine cycle/ \ factor / 



r i.H.P. of engine ] 
[I.H.P. of comp. J 



The actual I.H.P. of the compressor per ton refrigeration will be the cycle H.P. 
per ton divided by the diagram factor of the refrigerating cycle, which includes 
all losses, but which is nearly 100 per cent, certainly in any good compression 
not less than 95 per cent. 

Therefore, 

(B.T.U. supplied per hour to engine, per ton refrig.) 



2545 



Efficiency of \ /Efficiency 
_ \ engine cycle/ \ factor 



r m p P.en g . 1 x r C y cl.H.P.pert on1 
LI.H.P. comp. J [Refrig- diag. fact.J 



These quantities have all been evaluated except the ratio of compressor to 
engine horse-power, which is the mechanical efficiency of the drive and which 
depends on mechanical construction, speed, size, and must be estimated by 
good judgment. 

For the absorption system the heat supplied to the pump is to be evaluated 
in a manner equivalent to the above, but the heat of steam supplied to the 
generator is a very much larger amount indeed, though not necessarily at so 
high a temperature as is necessary for economical power generation, and must 
be determined separately from the properties of aqueous solutions of ammonia. 
It is a little surprising to find that this cannot be done with precision 
because of lack of data on the thermal properties of the solutions, but estimates 
may be made on various assumptions and often these estimates are taken by 
engineers as representing the truth when they really do not. The most com- 



1174 ENGINEERING THERMODYNAMICS 

mon assumption made in evaluating the heat supplied to the generator per ton 
refrigeration or per pound of anhydrous ammonia circulating, to which it is 
directly related, is that nothing but NH3 vapor will be discharged and that 
the heat of its liberation is equal to the heat of absorption of the same amount 
under inverse circumstances. For example, the curves and equations of Section 
12, Chapter IV, give the heat of absorption per pound of ammonia absorbed, 
when a solution of one strength becomes of higher ammonia content. Inversely, 
if a rich liquor becomes weaker through the loss of the same amount 
as was absorbed as above, there will be required an equal amount of heat. 
It is, however, quite impossible to discharge ammonia vapor from a solution 
without also discharging water vapor, and the difficulty involved here lies 
in the fixing of the amount of water vapor so carried off and the evaluation of 
heat of its evaporation, which must be added to that of the ammonia discharged. 
With an analyzer above the generator, condensing and returning some water 
vapor with its heat by incoming rich liquor as in the diagram, Fig. 316, it is 
necessary only to consider the water content of the NH3 vapor beyond this 
point and this is less than at generator discharge, so the error of neglecting the 
heat of vaporization of the water vapor is less when an analyzer is present 
than .when it is not, and appreciably less. If data were available to give the 
amount of water vapor in a saturated mixture beyond the analyzer and the 
temperature or vapor evolution at a given pressure in the generator itself, the 
heat equivalent of the water- vapor discharge could be found, but neither of 
these things are known. The law of molecular rise of boiling-point does not 
ajpply to ammonia water solutions, so the boiling-point or temperature of 
vapor evolution for a given liquor cannot be calculated for a given pressure. 
The best that can be done at present is to neglect the water vapor entirely and 
to remember that the heat required by the generator will be something 
(#) more than thus found. On this basis 

/B.T.U. supplied to generatorX _ /Heat of absorption per pound NHa\ 
\ per lb. NH3 circulating / \ between C w and C R / 



B.T.U. supplied to gen- 
erator per hour per 
ton refrigeration 



(Heat of absorption\ 
per lb. NH 3 be-| + £ 
tween Cw and Cr j 



X 



Lbs.N H 3 
per hr. 
per ton 



(1301) 



The heat balance for refrigerating systems might be set down as for power- 
generating systems equating energy taken in to that given out but as such heat 
balances are long and complicated without the introduction of vitiating assump- 
tions they will be omitted. For any particular case in which they may be 
required the methods discussed in previous sections provide the necessary 
material as far as it is at present possible to do. 

Example 1. Construction and use of Diagrams, Figs. 317 and 31&, These diagrams 
are for the purpose of finding the refrigerating effect per pound of fluid, which is made up 
of the latent heat, or as much of it as is available, less the heat necessary to cool the 



HEAT AND WORK 1175 

liquid from its original temperature to that due to the pressure in the coils, plus the 
heat absorbed in superheating the vapor. 

A horizontal scale of pressures is laid off in both directions for a vertical axis carrying 
a B.T.U. scale. In the section to the right of the center axis curves are drawn repre- 
senting various temperatures of the liquid before entering the refrigerator coils. These 
are so drawn that the vertical scale opposite the intersection of a vertical from 
any pressure with any curve gives the latent heat for that pressure less the heat 
required to cool the liquid. This is the available heat for refrigerating if the vapor 
leaves the coils dry and saturated. In the section to the left of the center axis are 
two sets of curves, the lower, representing temperatures of the vapor leaving the 
coils, is so drawn that the value of the left-hand vertical scale opposite a point of 
intersection of a vertical from any pressure with any curve, gives the heat absorbed in 
superheating the vapor. The sum of this and the value found in the first section gives 
the total refrigerating effect for the case when the vapor leaves the coils in a super- 
heated state. The upper curves in this section represent quality of the vapor if the 
liquid has not been entirely evaporated and are so drawn that the value on the vertical 
scale opposite the point of intersection of a vertical from any pressure with any curve, 
shows the heat unavailable for refrigerating, due to incomplete evaporation of the liquid, 
and the difference between this value and that found in the first section gives the total 
refrigerating effect for the case of wet vapor leaving the coils. 

As an example let it be required to find the refrigerating effect per pound of ammonia 
when the pressure in the coils is 20 lbs. gage, the temperature of the liquid before enter- 
ing the coil is 70° F. and 

(a) Vapor leaves dry and saturated; 

(b) Vapor leaves 90 per cent dry; 

(c) Vapor leaves at a temperature of 30° F. 

From 20 in the right-hand section project up to curve 70°. The value on the vertical 
scale at this point is 502 B.T.U., which is the value for case (a). From 20 in the left- 
hand section project to curve 90 per cent; the value on the left-hand vertical scale is 57, 
therefore, for case (b) the result is 502 —57 =445 B.T.U. For case (c), project from 20 to 
curve 30°, the value on the vertical scale corresponding to which is 12.5, hence the result 
for this case is 502+12.5=514.5. 

Prob. 1. In a cold-storage room it is estimated that 350 lbs. of ice melt every minute. 
What would be the capacity of a machine to do an equal amount of refrigeration? What 
will be the capacity of a machine required to make 100 tons of ice per day from water at 
60° F., the ice being at a temperature of 25° F.? 

Prob. 2. Three types of machines are under consideration, (a) dense air; (b) ammonia 
compression; (c) carbon dioxide compression. In every case the temperature of the 
gas or vapor leaving the coils is 20° F. In the air system the air enters the coils at a 
temperature of —50° F. In the case of the ammonia system the pressure in the coils 
is 15 lbs. per square inch absolute, the liquid temperature before entering is 60° F., and 
in the carbon dioxide system the pressure is 300 lbs. per square inch gage, and the tem- 
perature before entering the coils is 80° F. What must be the compressor displacement 
for each case for a common true volumetric efficiency of 80 per cent? I 

Prob. 3. If the ammonia should leave the coils in Prob. 2 with 5 per cent unevapor- 
ated, what effect would there be on the temperature of the vapor and on the compressor 
size? If the carbon dioxide should leave dry and saturated what would be the change 
in these two quantities. 



1176 ENGINEERING THERMODYNAMICS 

Prob. 4. Should an ammonia absorption system work so that the temperature of the 
anhydrous liquid entering the coils was 70° F. and that of the vapor leaving the coils was 
30° F. with a pressure of 1 atmosphere, what would the displacement of the rich liquor 
pump, the rich liquor being 30 per cent NH 3 , and the weak liquor 20 per cent by weight? 

Prob. 5. In a refrigerating system it is desired to maintain a temperature of 10° F. 
in the coils, and there is available for condensing purposes water having a temperature of 
70° F. What will be the required cylinder displacement per minute of the compressor, 
and the horse-power per ton refrigeration for an ammonia-compression system, and a 
similar system using C0 2 ? What would be the relative worth of the two systems? 

Prob. 6. Air is used for refrigeration in a system where the highest pressure which 
is desirable to carry is 150 lbs. gage, the cooling water temperature is 60° F., and the 
cold air is desired to have a temperature of —50°. What will be the temperature of the 
air leaving the refrigerator if each pound absorbs 15 B.T.U.? What will be the low 
pressure, work done, ratio of work to heat absorbed in refrigerator and given to cooler? 
Draw the PV and T§ diagrams. 

Prob. 7. In an ice-making plant making 100 tons of ice per day from water at 40° F. 
the ice is at 20° F. when removed from the cans; the cooling water available has a 
temperature of 60° F. and the pressure in the coils is 10 lbs. gage. If the engine driving 
the compressor has a thermal efficiency of 12 per cent, the work required to overcome 
friction in the engine and compressor is 15 per cent of the engine horse-power, and the 
compressor-diagram factor is 90 per cent, how many B.T.U. must be supplied to the 
engine per hour, for, (a) an ammonia plant; (b) a carbon dioxide plant? 

Prob* 8. An absorption system is in operation so as to produce 50 tons refrigera- 
tion. The cooling water for the condenser is at 50° F. and the liquid NH 3 is cooled 
to within 10° of this value. The temperature of the vapor leaving the coils is 20° F. and 
the pressure in them is 10 lbs. gage. The strength of the ammonia liquor varies between 
15 and 30 per cent. Assuming that 25 per cent more heat must be supplied to the gen- 
erator than is needed to liberate the ammonia, what will be the B.T.U. supplied to gen- 
erator per hour per ton refrigeration? 



/***^ys&vpL 



